How do P-V diagrams work in thermodynamics
https://www.khanacademy.org/science/physics/thermodynamics/laws-of-thermodynamics/a/what-are-pv-diagrams
What are PV diagrams? (article) | Khan Academy
2015-11-25
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<div><p>Learn what PV diagrams are and how to use them to find the change in internal energy, work done, and heat.</p><div><div><p></p><h2>What are PV diagrams?</h2><p></p><div><p>Consider a gas sealed in a container with a tightly fitting yet movable piston as seen below. We can do work on the gas by pressing the piston downward, and we can heat up the gas by placing the container over a flame or submerging it in a bath of boiling water. When we subject the gas to these thermodynamics processes, the pressure and volume of the gas can change.</p></div><div><p>A convenient way to visualize these changes in the pressure and volume is by using a Pressure Volume diagram or <strong>PV diagram</strong> for short. Each point on a PV diagram corresponds to a different state of the gas. The pressure is given on the vertical axis and the volume is given on the horizontal axis, as seen below.</p></div><div><p>Every point on a PV diagram represents a different state for the gas (one for every possible volume and pressure). As a gas goes through a thermodynamics process, the state of the gas will shift around in the PV diagram, tracing out a path as it moves (as shown in the diagram below).</p></div><div><p>Being able to decode the information shown in a PV diagram allows us to make statements about the change in internal energy delta, U, heat transferred Q, and work done W on a gas. In the sections below, we'll explain how to decipher the hidden information contained in a PV diagram. </p></div><div><p><em>Note: Unless otherwise specified, we will assume that the work W refers to the work done on the gas.</em></p></div></div><div><p></p><h2>How do we determine the sign of the work done from a PV diagram?</h2><p></p><div><p>Let's say our gas starts out in the state shown in the PV diagram below. </p></div><div><p>If we <strong>press the piston downward</strong>, the volume of the gas will decrease, so the <strong>state must shift to the left</strong> toward smaller volumes (as seen in the diagram below). Since the gas is being compressed we can also say for sure that <strong>positive work W is being done on the gas</strong>. </p></div><div><p>Similarly, if we let the gas expand, <strong>pushing the piston upward</strong>, the volume of the gas will increase, so the <strong>state must shift to the right</strong> toward larger volumes (as seen in the diagram below). Since the gas is expanding we can also say for sure that <strong>negative work W is being done on the gas.</strong></p></div><div><p>So if we ever see a state shifting to the left on a PV diagram we can say for sure that the work done on the gas was positive. Similarly, if we ever see a state shifting to the right on a PV diagram we can say for sure that the work done on the gas was negative.</p></div></div><div><p></p><h2>How do we determine the magnitude of the work done from a PV diagram?</h2><p></p><div><p>The <strong>work done</strong> during a thermodynamic process is equal to the <strong>area under the curve</strong> as seen in the diagram below. </p></div><div><p>The reason why work is equal to the area under the curve is that,</p></div><div><p>W, equals, F, delta, x, equals, left parenthesis, P, A, right parenthesis, delta, x, equals, P, left parenthesis, A, delta, x, right parenthesis, equals, P, delta, V</p></div><div><p>And since P, delta, V is just the start text, h, e, i, g, h, t, end text, times, start text, space, w, i, d, t, h, end text of the rectangle shown above, the work is equal to the area. If we use pressure units of start text, p, a, s, c, a, l, s, end text and volume units of start text, m, end text, cubed then the energy we find will be in units of start text, j, o, u, l, e, s, end text. </p></div><div><p>We have to be really careful with signs though. If the path on a PV diagram is directed to the left, the volume is decreasing, and positive work is being done on the gas. If the path on a PV diagram is directed to the right (as in the diagram above), the volume is increasing, and negative work is being done on the gas since W, start subscript, start text, b, y, space, g, a, s, end text, end subscript, equals, minus, W, start subscript, start text, o, n, space, g, a, s, end text, end subscript.</p></div><div><p>It doesn't matter what shape the path takes, the area under the curve will still represent the work done. For any curved path we can imagine breaking the area into an infinite amount of infinitesimally thin rectangles. </p></div><div><p>The area of each rectangle would represent the work done during each infinitesimal step, and the sum of the areas would represent the total work done for the entire process.</p></div><div><p>It should be said that we are always going to assume these processes are taking place slowly enough that the entire gas can be at thermodynamic equilibrium at every moment (i.e. the same temperature throughout the gas). If this seems dubious to you, you're right to question it. However, even though basically no real world processes will exactly satisfy this requirement, our ability to model many thermodynamic processes are not fatally jeopardized by this lack of adherence to ideal circumstances.</p></div></div><div><p></p><h2>How do we determine the sign of delta, U from a PV diagram?</h2><p></p><div><p>Remember that internal energy and temperature are proportional U, \propto, T. So if the temperature increases, the internal energy must also increase. </p></div><div><p>Now, if the gas we're considering is an ideal gas we also know that, </p></div><div><p>P, V, equals, N, k, start subscript, B, end subscript, T</p></div><div><p>And if no gas is allowed to escape (so the number of molecules N is constant) we can say that P, V, \propto, T. All of this means that,</p></div><div><p>U, \propto, T, \propto, P, V</p></div><div><p>So if the quantity of pressure times volume left parenthesis, P, times, V, right parenthesis increases, the temperature T and internal energy U must also increase (which makes delta, U positive). This idea is represented in the diagram shown below.</p></div><div><p>This means that anytime the state in a PV diagram ends up further up and right than where it started, delta, U is a positive number. Similarly, anytime the state in a PV diagram ends up further down and left than where it started, delta, U is a negative number.</p></div><div><p>Now if the state in the PV diagram moves up and left (pressure increases and volume decreases), or down and right (pressure decreases and volume increases), it is a little ambiguous whether the quantity left parenthesis, P, times, V, right parenthesis actually increased or decreased (since one variable increased and the other variable decreased). To be sure, one would have to check the exact values of the initial and final P and V on the axes of the graph to tell if the quantity left parenthesis, P, times, V, right parenthesis actually increased or decreased.</p></div><div><p>It is also good to note that if the quantity left parenthesis, P, times, V, right parenthesis does not change, then the temperature T and internal energy U do not change either. For instance, if the pressure doubles, and the volume is cut in half, left parenthesis, P, times, V, right parenthesis remains the same value (since 2, P, times, start fraction, V, divided by, 2, end fraction, equals, P, V). the temperature T and internal energy U will end the process with the same values they started with.</p></div></div><div><p></p><h2>How do we determine the sign of Q from a PV diagram?</h2><p></p><div><p>Given a PV diagram, we typically have to rely on the first law of thermodynamics delta, U, equals, Q, plus, W to determine the sign of the net heat that enters or exits a gas. If we solve this equation for the heat Q we get,</p></div><div><p>Q, equals, delta, U, minus, W</p></div><div><p>Now that we know this, we can use what we know about finding the sign of delta, U and W to find the sign of Q in many cases. So for instance, if the change in internal energy is positive and the work done is negative, </p></div><div><p>Q, equals, left parenthesis, plus, right parenthesis, minus, left parenthesis, minus, right parenthesis, equals, plus ...the net heat must be positive.</p></div><div><p>Which makes sense, since if the internal energy increased even though work was done by the gas, that implies that more heat must have entered the gas than energy lost due to the work done by the gas.</p></div><div><p>Or for example, if the internal energy decreases and the work is positive,</p></div><div><p>Q, equals, left parenthesis, minus, right parenthesis, minus, left parenthesis, plus, right parenthesis, equals, minus ...the net heat must be negative.</p></div><div><p>Which makes sense, since if the internal energy decreased even though work was done on the gas, that implies that more heat must have left the gas than energy gained by the gas from work being done on it.</p></div></div><div><p></p><h2>What do solved examples involving PV diagrams look like?</h2><p></p><p></p><h3>Example 1: Finding signs</h3><p></p><div><p>An ideal gas in a sealed container is taken through the process shown in the PV diagram below. </p></div><p></p><h3>Example 2: Finding area</h3><p></p><div><p>An ideal gas in a sealed container is taken through the process shown in the PV diagram below. The initial volume of the gas is V, start subscript, i, end subscript, equals, 0, point, 25, start text, m, end text, cubed and the final volume of the gas is V, start subscript, f, end subscript, equals, 0, point, 75, start text, m, end text, cubed. The initial pressure of the gas is P, start subscript, i, end subscript, equals, 70, comma, 000, start text, space, P, a, end text and the final pressure of the gas is P, start subscript, f, end subscript, equals, 160, comma, 000, start text, space, P, a, end text.</p></div><div><p><strong>What is the work done on the gas during the process shown?</strong></p></div><div><p>We can find the work done by determining the total area under the curve on a PV diagram. We have to make sure we use the <strong>total</strong> area, all the way down to the volume axis. For instance, we can imagine viewing the area under the curve in the example shown above as a triangle and a rectangle (as seen below). </p></div><div><p>Now we just find the sum of the areas of the triangle and rectangle. The height of the rectangle is the pressure P, start subscript, i, end subscript and the width of the rectangle is the change in volume delta, V, equals, V, start subscript, f, end subscript, minus, V, start subscript, i, end subscript. So,</p></div><div><p>start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, start text, h, e, i, g, h, t, end text, times, start text, w, i, d, t, h, end text (area of the rectangle)</p></div><div><p>start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, P, start subscript, i, end subscript, times, delta, V (height is P, start subscript, i, end subscript and width is delta, V)</p></div><div><p>start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, left parenthesis, 70, comma, 000, start text, space, P, a, end text, right parenthesis, times, left parenthesis, 0, point, 75, start text, m, end text, cubed, minus, 0, point, 25, start text, m, end text, cubed, right parenthesis (plug in values)</p></div><div><p>start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, 35, comma, 000, start text, space, J, end text (calculate)</p></div><div><p>We can find the area of a triangle by using A, equals, start fraction, 1, divided by, 2, end fraction, b, h.</p></div><div><p>start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, b, h (area of the triangle)</p></div><div><p>start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, b, left parenthesis, 160, comma, 000, start text, P, a, end text, minus, 70, comma, 000, start text, space, P, a, end text, right parenthesis (the height of the triangle is the difference in pressures P, start subscript, f, end subscript, minus, P, start subscript, i, end subscript)</p></div><div><p>start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, 0, point, 75, start text, m, end text, cubed, minus, 0, point, 25, start text, m, end text, cubed, right parenthesis, left parenthesis, 160, comma, 000, start text, P, a, end text, minus, 70, comma, 000, start text, space, P, a, end text, right parenthesis (the base of the triangle is the difference in volumes V, start subscript, f, end subscript, minus, V, start subscript, i, end subscript)</p></div><div><p>start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, 22, comma, 500, start text, space, J, end text (calculate)</p></div><div><p>So the total area under the curve is 35, comma, 000, start text, space, J, end text, plus, 22, comma, 500, start text, space, J, end text, equals, 57, comma, 500, start text, space, J, end text</p></div><div><p>This area represents the absolute value of the total work done during the process. To determine the sign of the work done on the gas we notice that the process moves the state to the right, causing the gas to expand. When gas expands the <strong>work done on the gas is negative</strong>. So,</p></div><div><p>W, start subscript, start text, o, n, space, g, a, s, end text, end subscript, equals, minus, 57, comma, 500, start text, space, J, end text (celebrate)</p></div></div></div>||||I|||| If you're seeing this message, it means we're having trouble loading external resources on our website.





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 Physics library

 Unit 10: Lesson 3

 Laws of thermodynamics
 * Macrostates and microstates
 * Quasistatic and reversible processes
 * First law of thermodynamics / internal energy
 * More on internal energy
 * What is the first law of thermodynamics?
 * Work from expansion
 * PV-diagrams and expansion work
 * What are PV diagrams?
 * Proof: U = (3/2)PV or U = (3/2)nRT
 * Work done by isothermic process
 * Carnot cycle and Carnot engine
 * Proof: Volume ratios in a Carnot cycle
 * Proof: S (or entropy) is a valid state variable
 * Thermodynamic entropy definition clarification
 * Reconciling thermodynamic and state definitions of entropy
 * Entropy intuition
 * Maxwell's demon
 * More on entropy
 * Efficiency of a Carnot engine
 * Carnot efficiency 2: Reversing the cycle
 * Carnot efficiency 3: Proving that it is the most efficient
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 What are PV diagrams?

 Google Classroom
 Learn what PV diagrams are and how to use them to find the change in internal energy, work done, and heat.

 What are PV diagrams?

 Consider a gas sealed in a container with a tightly fitting yet movable piston as seen below. We can do work on the gas by pressing the piston downward, and we can heat up the gas by placing the container over a flame or submerging it in a bath of boiling water. When we subject the gas to these thermodynamics processes, the pressure and volume of the gas can change.
 A convenient way to visualize these changes in the pressure and volume is by using a Pressure Volume diagram or PV diagram for short. Each point on a PV diagram corresponds to a different state of the gas. The pressure is given on the vertical axis and the volume is given on the horizontal axis, as seen below.
 Every point on a PV diagram represents a different state for the gas (one for every possible volume and pressure). As a gas goes through a thermodynamics process, the state of the gas will shift around in the PV diagram, tracing out a path as it moves (as shown in the diagram below).
 Being able to decode the information shown in a PV diagram allows us to make statements about the change in internal energy ΔU\Delta U Δ U delta, U , heat transferred QQ Q Q , and work done WW W W on a gas. In the sections below, we'll explain how to decipher the hidden information contained in a PV diagram.
 Note: Unless otherwise specified, we will assume that the work WW W W refers to the work done on the gas.

 How do we determine the sign of the work done from a PV diagram?

 Let's say our gas starts out in the state shown in the PV diagram below.
 If we press the piston downward, the volume of the gas will decrease, so the state must shift to the left toward smaller volumes (as seen in the diagram below). Since the gas is being compressed we can also say for sure that positive work WW W W is being done on the gas.
 Similarly, if we let the gas expand, pushing the piston upward, the volume of the gas will increase, so the state must shift to the right toward larger volumes (as seen in the diagram below). Since the gas is expanding we can also say for sure that negative work WW W W is being done on the gas.
 So if we ever see a state shifting to the left on a PV diagram we can say for sure that the work done on the gas was positive. Similarly, if we ever see a state shifting to the right on a PV diagram we can say for sure that the work done on the gas was negative.

 How do we determine the magnitude of the work done from a PV diagram?

 The work done during a thermodynamic process is equal to the area under the curve as seen in the diagram below.
 The reason why work is equal to the area under the curve is that,
 W=FΔx=(PA)Δx=P(AΔx)=PΔVW=F\Delta x =(PA)\Delta x=P(A\Delta x)=P\Delta V W = F Δ x = ( P A ) Δ x = P ( A Δ x ) = P Δ V W, equals, F, delta, x, equals, left parenthesis, P, A, right parenthesis, delta, x, equals, P, left parenthesis, A, delta, x, right parenthesis, equals, P, delta, V
 And since PΔVP\Delta V P Δ V P, delta, V is just the height× width\text{height} \times \text{ width} height × width start text, h, e, i, g, h, t, end text, times, start text, space, w, i, d, t, h, end text of the rectangle shown above, the work is equal to the area. If we use pressure units of pascals\text{pascals} pascals start text, p, a, s, c, a, l, s, end text and volume units of m3\text{m}^3 m 3 start text, m, end text, cubed then the energy we find will be in units of joules\text{joules} joules start text, j, o, u, l, e, s, end text .
 [Why are the units J?]
 We have to be really careful with signs though. If the path on a PV diagram is directed to the left, the volume is decreasing, and positive work is being done on the gas. If the path on a PV diagram is directed to the right (as in the diagram above), the volume is increasing, and negative work is being done on the gas since Wby gas=−Won gasW_\text{by gas}=-W_\text{on gas} W by gas ​ = − W on gas ​ W, start subscript, start text, b, y, space, g, a, s, end text, end subscript, equals, minus, W, start subscript, start text, o, n, space, g, a, s, end text, end subscript .
 It doesn't matter what shape the path takes, the area under the curve will still represent the work done. For any curved path we can imagine breaking the area into an infinite amount of infinitesimally thin rectangles.
 The area of each rectangle would represent the work done during each infinitesimal step, and the sum of the areas would represent the total work done for the entire process.
 [What if there is a closed path for a cyclic process?]
 It should be said that we are always going to assume these processes are taking place slowly enough that the entire gas can be at thermodynamic equilibrium at every moment (i.e. the same temperature throughout the gas). If this seems dubious to you, you're right to question it. However, even though basically no real world processes will exactly satisfy this requirement, our ability to model many thermodynamic processes are not fatally jeopardized by this lack of adherence to ideal circumstances.

 How do we determine the sign of ΔU\Delta U Δ U delta, U from a PV diagram?

 Remember that internal energy and temperature are proportional U∝TU \propto T U ∝ T U, \propto, T . So if the temperature increases, the internal energy must also increase.
 Now, if the gas we're considering is an ideal gas we also know that,
 PV=NkBTPV=Nk_BT P V = N k B ​ T P, V, equals, N, k, start subscript, B, end subscript, T
 And if no gas is allowed to escape (so the number of molecules NN N N is constant) we can say that PV∝TPV \propto T P V ∝ T P, V, \propto, T . All of this means that,
 U∝T∝PV\Large U \propto T \propto PV U ∝ T ∝ P V U, \propto, T, \propto, P, V
 So if the quantity of pressure times volume (P×V)(P \times V) ( P × V ) left parenthesis, P, times, V, right parenthesis increases, the temperature TT T T and internal energy UU U U must also increase (which makes ΔU\Delta U Δ U delta, U positive). This idea is represented in the diagram shown below.
 This means that anytime the state in a PV diagram ends up further up and right than where it started, ΔU\Delta U Δ U delta, U is a positive number. Similarly, anytime the state in a PV diagram ends up further down and left than where it started, ΔU\Delta U Δ U delta, U is a negative number.
 Now if the state in the PV diagram moves up and left (pressure increases and volume decreases), or down and right (pressure decreases and volume increases), it is a little ambiguous whether the quantity (P×V)(P \times V) ( P × V ) left parenthesis, P, times, V, right parenthesis actually increased or decreased (since one variable increased and the other variable decreased). To be sure, one would have to check the exact values of the initial and final PP P P and VV V V on the axes of the graph to tell if the quantity (P×V)(P \times V) ( P × V ) left parenthesis, P, times, V, right parenthesis actually increased or decreased.
 It is also good to note that if the quantity (P×V)(P \times V) ( P × V ) left parenthesis, P, times, V, right parenthesis does not change, then the temperature TT T T and internal energy UU U U do not change either. For instance, if the pressure doubles, and the volume is cut in half, (P×V)(P \times V) ( P × V ) left parenthesis, P, times, V, right parenthesis remains the same value (since 2P×V2=PV2P \times \dfrac{V}{2}=PV 2 P × 2 V ​ = P V 2, P, times, start fraction, V, divided by, 2, end fraction, equals, P, V ). the temperature TT T T and internal energy UU U U will end the process with the same values they started with.

 How do we determine the sign of QQ Q Q from a PV diagram?

 Given a PV diagram, we typically have to rely on the first law of thermodynamics ΔU=Q+W\Delta U=Q+W Δ U = Q + W delta, U, equals, Q, plus, W to determine the sign of the net heat that enters or exits a gas. If we solve this equation for the heat QQ Q Q we get,
 Q=ΔU−WQ=\Delta U-W Q = Δ U − W Q, equals, delta, U, minus, W
 Now that we know this, we can use what we know about finding the sign of ΔU\Delta U Δ U delta, U and WW W W to find the sign of QQ Q Q in many cases. So for instance, if the change in internal energy is positive and the work done is negative,
 Q=(+)−(−)=+Q=(+)-(-)=+\qquad Q = ( + ) − ( − ) = + Q, equals, left parenthesis, plus, right parenthesis, minus, left parenthesis, minus, right parenthesis, equals, plus ...the net heat must be positive.
 Which makes sense, since if the internal energy increased even though work was done by the gas, that implies that more heat must have entered the gas than energy lost due to the work done by the gas.
 Or for example, if the internal energy decreases and the work is positive,
 Q=(−)−(+)=−Q=(-)-(+)=-\qquad Q = ( − ) − ( + ) = − Q, equals, left parenthesis, minus, right parenthesis, minus, left parenthesis, plus, right parenthesis, equals, minus ...the net heat must be negative.
 Which makes sense, since if the internal energy decreased even though work was done on the gas, that implies that more heat must have left the gas than energy gained by the gas from work being done on it.

 What do solved examples involving PV diagrams look like?

 Example 1: Finding signs

 An ideal gas in a sealed container is taken through the process shown in the PV diagram below.
 Select the correct statement about the signs of the following quantities: change in internal energy of the gas ΔU\Delta U Δ U delta, U , net work done on the gas W,W, W , W, comma and net heat that enters the gas QQ Q Q
 Choose 1 answer:
 Choose 1 answer:
 * (Choice A)
  ΔU\Delta U Δ U delta, U is positive, WW W W is negative, QQ Q Q is positive
 A
  ΔU\Delta U Δ U delta, U is positive, WW W W is negative, QQ Q Q is positive
 * (Choice B)
  ΔU\Delta U Δ U delta, U is negative, WW W W is positive, QQ Q Q is positive
 B
  ΔU\Delta U Δ U delta, U is negative, WW W W is positive, QQ Q Q is positive
 * (Choice C)
  ΔU\Delta U Δ U delta, U is negative, WW W W is positive, QQ Q Q is negative
 C
  ΔU\Delta U Δ U delta, U is negative, WW W W is positive, QQ Q Q is negative
 * (Choice D)
  ΔU\Delta U Δ U delta, U is negative, WW W W is negative, QQ Q Q is positive
 D
  ΔU\Delta U Δ U delta, U is negative, WW W W is negative, QQ Q Q is positive

 Check
 [What is the solution?]

 Example 2: Finding area

 An ideal gas in a sealed container is taken through the process shown in the PV diagram below. The initial volume of the gas is Vi=0.25m3V_i=0.25 \text{m}^3 V i ​ = 0 . 2 5 m 3 V, start subscript, i, end subscript, equals, 0, point, 25, start text, m, end text, cubed and the final volume of the gas is Vf=0.75m3V_f=0.75 \text{m}^3 V f ​ = 0 . 7 5 m 3 V, start subscript, f, end subscript, equals, 0, point, 75, start text, m, end text, cubed . The initial pressure of the gas is Pi=70,000 PaP_i=70,000 \text{ Pa} P i ​ = 7 0 , 0 0 0 Pa P, start subscript, i, end subscript, equals, 70, comma, 000, start text, space, P, a, end text and the final pressure of the gas is Pf=160,000 PaP_f=160,000 \text{ Pa} P f ​ = 1 6 0 , 0 0 0 Pa P, start subscript, f, end subscript, equals, 160, comma, 000, start text, space, P, a, end text .
 What is the work done on the gas during the process shown?
 Solution:
 We can find the work done by determining the total area under the curve on a PV diagram. We have to make sure we use the total area, all the way down to the volume axis. For instance, we can imagine viewing the area under the curve in the example shown above as a triangle and a rectangle (as seen below).
 Now we just find the sum of the areas of the triangle and rectangle. The height of the rectangle is the pressure PiP_i P i ​ P, start subscript, i, end subscript and the width of the rectangle is the change in volume ΔV=Vf−Vi\Delta V=V_f-V_i Δ V = V f ​ − V i ​ delta, V, equals, V, start subscript, f, end subscript, minus, V, start subscript, i, end subscript . So,
 area 1=height×width\blueD{\text{area 1}}=\text{height} \times \text{width} \quad area 1 = height × width start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, start text, h, e, i, g, h, t, end text, times, start text, w, i, d, t, h, end text (area of the rectangle)
 area 1=Pi×ΔV\blueD{\text{area 1}}=P_i \times \Delta V \quad area 1 = P i ​ × Δ V start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, P, start subscript, i, end subscript, times, delta, V (height is PiP_i P i ​ P, start subscript, i, end subscript and width is ΔV\Delta V Δ V delta, V )
 area 1=(70,000 Pa)×(0.75m3−0.25m3)\blueD{\text{area 1}}=(70,000\text{ Pa}) \times (0.75\text{m}^3-0.25\text{m}^3)\quad area 1 = ( 7 0 , 0 0 0 Pa ) × ( 0 . 7 5 m 3 − 0 . 2 5 m 3 ) start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, left parenthesis, 70, comma, 000, start text, space, P, a, end text, right parenthesis, times, left parenthesis, 0, point, 75, start text, m, end text, cubed, minus, 0, point, 25, start text, m, end text, cubed, right parenthesis (plug in values)
 area 1=35,000 J\blueD{\text{area 1}}=35,000 \text{ J} \quad area 1 = 3 5 , 0 0 0 J start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, 35, comma, 000, start text, space, J, end text (calculate)
 We can find the area of a triangle by using A=12bhA=\dfrac{1}{2}bh A = 2 1 ​ b h A, equals, start fraction, 1, divided by, 2, end fraction, b, h .
 area 2=12bh\greenD{\text{area 2}}=\dfrac{1}{2}bh \quad area 2 = 2 1 ​ b h start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, b, h (area of the triangle)
 area 2=12b(160,000Pa−70,000 Pa)\greenD{\text{area 2}}=\dfrac{1}{2}b(160,000\text{Pa}-70,000\text{ Pa}) \quad area 2 = 2 1 ​ b ( 1 6 0 , 0 0 0 Pa − 7 0 , 0 0 0 Pa ) start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, b, left parenthesis, 160, comma, 000, start text, P, a, end text, minus, 70, comma, 000, start text, space, P, a, end text, right parenthesis (the height of the triangle is the difference in pressures Pf−PiP_f-P_i P f ​ − P i ​ P, start subscript, f, end subscript, minus, P, start subscript, i, end subscript )
 area 2=12(0.75m3−0.25m3)(160,000Pa−70,000 Pa)\greenD{\text{area 2}}=\dfrac{1}{2} (0.75\text{m}^3-0.25\text{m}^3)(160,000\text{Pa}-70,000\text{ Pa}) \quad area 2 = 2 1 ​ ( 0 . 7 5 m 3 − 0 . 2 5 m 3 ) ( 1 6 0 , 0 0 0 Pa − 7 0 , 0 0 0 Pa ) start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, 0, point, 75, start text, m, end text, cubed, minus, 0, point, 25, start text, m, end text, cubed, right parenthesis, left parenthesis, 160, comma, 000, start text, P, a, end text, minus, 70, comma, 000, start text, space, P, a, end text, right parenthesis (the base of the triangle is the difference in volumes Vf−ViV_f-V_i V f ​ − V i ​ V, start subscript, f, end subscript, minus, V, start subscript, i, end subscript )
 area 2=22,500 J\greenD{\text{area 2}}=22,500 \text{ J} \quad area 2 = 2 2 , 5 0 0 J start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, 22, comma, 500, start text, space, J, end text (calculate)
 So the total area under the curve is 35,000 J+22,500 J=57,500 J35,000 \text{ J} + 22,500 \text{ J}=57,500 \text{ J} 3 5 , 0 0 0 J + 2 2 , 5 0 0 J = 5 7 , 5 0 0 J 35, comma, 000, start text, space, J, end text, plus, 22, comma, 500, start text, space, J, end text, equals, 57, comma, 500, start text, space, J, end text
 This area represents the absolute value of the total work done during the process. To determine the sign of the work done on the gas we notice that the process moves the state to the right, causing the gas to expand. When gas expands the work done on the gas is negative. So,
 Won gas=−57,500 JW_\text{on gas}=-57,500 \text{ J}\quad W on gas ​ = − 5 7 , 5 0 0 J W, start subscript, start text, o, n, space, g, a, s, end text, end subscript, equals, minus, 57, comma, 500, start text, space, J, end text (celebrate)
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 * joe blogs
 7 years ago
 Posted 7 years ago. Direct link to joe blogs's post “If work done by a system ...”
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 If work done by a system is equal to change in pressure-volume. Why does an upward and rightward shift in the pressure-volume diagram cause an increase in internal energy? Work done by the system should reduce the systems internal energy shouldn't it?
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  + davidsantopietro
  7 years ago
  Posted 7 years ago. Direct link to davidsantopietro's post “The work done in expandin...”
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 The work done in expanding does in fact take energy out of the system, so the only way the state can shift to a higher PV value while the gas does work is to have even more heat Q added to the gas during the process so that overall the gas gains energy.
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 * PREDATOR
 7 years ago
 Posted 7 years ago. Direct link to PREDATOR's post “maybe this is not a good ...”
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 maybe this is not a good doubt .. but if the gas has been compressed why is the work done on the gas positive? is nt work done = PdeltaV
 so the work done by the gas should be negative if im not wrong
 ...
 correct me if im wrong
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  + Rodrigo Campos
  7 years ago
  Posted 7 years ago. Direct link to Rodrigo Campos's post “I would say is more a mat...”
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 I would say is more a matter of semantics
 - The work done on the gas is positive
 - The work done by the gas is negative
 The conservation of energy (first law of thermodynamics) states:
 Internal Energy Change = Heat added into the System + Work done on the System
 or
 Internal Energy Change = Heat added into the System - Work done by the System
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 * humem005
 3 years ago
 Posted 3 years ago. Direct link to humem005's post “If we were to forcefully ...”
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 If we were to forcefully pull the piston up instead of push down - would this be considered work in or out (+ or -). Just a little confused because we are applying a force (doing work) to move the piston but the volume is increasing and pressure decreasing so according to the convention the gas is doing work?? :)
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 * Miteqcomp
 7 years ago
 Posted 7 years ago. Direct link to Miteqcomp's post “What if there is a closed...”
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 What if there is a closed path for a cyclic process?- This is confusing me.
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  + C Hart
  7 years ago
  Posted 7 years ago. Direct link to C Hart's post “This is now explained in ...”
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 This is now explained in a little pop-up box in the middle of the article, but you may have missed it or it may have been added after your question.

 While the cycle is moving L to R (increasing volume) the work done BY the gas (area under the curve) is considered positive. While the cycle is moving R to L (decreasing volume) the work done BY the gas is considered negative. The total work of the cycle is the sum of these two works (areas under the curves), but since one is negative, it's essentially the difference. The difference between the two areas is, by definition, the area within the cycle and that is the work done BY the gas during the cycle.

 As for the sign, if the cycle is clockwise, that means the higher line (higher work area) was L to R (positive work), so the total W is positive. if the cycle is counterclockwise, that means the higher line (higher W area) was R to L (negative work), so the total W is negative (work was done ON the gas). But in the either case, the magnitude of the work is still the area inside the PV cycle.
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 * Hafsa Kaja Moinudeen
 6 years ago
 Posted 6 years ago. Direct link to Hafsa Kaja Moinudeen's post “The slope of the PV diagr...”
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 The slope of the PV diagram has the units N/m^5, does that mean anything?
 Thanks
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  + Jay
  6 years ago
  Posted 6 years ago. Direct link to Jay's post “The slope tells you how f...”
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 The slope tells you how fast the pressure is changing with respect to volume. Slope m, is equal to Pressure/Volume, or dimensionally: m=(F/A)/V in SI units, it will be kg/(m^5) or kg*(m^-4) s^(-2). This tells you how fast the pressure is changing with volume for every kg of gas/atoms in the system per volume. If you think about classical mechanics, for example, a ball heading towards the earth, (neglecting air resistance and approximating the distance near the surface of the earth), the force is F=N=kg (m/s^2). Intuitively, you can think of it as I need to apply this force to move this object with this mass. And to see how long it takes, you divide by the time interval (1/s). The extra s is to account for the extra time needed to move the object that is already traveling with some velocity.

 Sorry if I ended up confusing you more... The slope essentially tells you how fast the dependent variable (y-axis) is changing when you let the independent (x-axis) vary.

 - (senior) physics/mechanical engineering
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 * Tiago Dias
 6 years ago
 Posted 6 years ago. Direct link to Tiago Dias's post “I solved example 2 using ...”
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 I solved example 2 using integral Calculus. I found the equation of the line and then I calculated its definite integral from 0.25 to 0.75. Alas, I gost -45778 joules as final answer. Does this have to do with thwe fact that I did not account for the constant of the indefinite integral of the equation of the line? I guess so, but could refer me to some videos where Sal takes this into account? (Please do not bother to xplain the matematical technique here. Refer me some videos)
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  + jbradbury994
  6 years ago
  Posted 6 years ago. Direct link to jbradbury994's post “When using calculus to ca...”
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 When using calculus to calculate these changes you really have to take into account that the pressure isn't constant.
 From ideal gas law, P = nRT/V
 If you sub this in for P when integrating this takes the pressure changes into account.
 Leading to
 W = -PdV
 = -nRT X integral of 1/v
 Giving final answer of, W = -nRT ln(Vf/Vi) (Vf = v final, Vi = v initial)
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 * Fardeen Ashraf
 3 years ago
 Posted 3 years ago. Direct link to Fardeen Ashraf's post “When a gas is doing work ...”
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 When a gas is doing work yet increasing its pressure, can we say that the pressure increase is definitely due to heat being added?
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  + taurepton
  3 years ago
  Posted 3 years ago. Direct link to taurepton's post “Unless gas is added, it i...”
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 Unless gas is added, it is. If gas is doing work, its volume is increasing. As pressure and volume increases, T=PV/nR is increasing too. Thus, Delta U = Q - W is positive. As W is positive, Q should be positive.
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 * Dean Pavlovic
 6 years ago
 Posted 6 years ago. Direct link to Dean Pavlovic's post “why does the volume chang...”
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 why does the volume change determines work sign in the first place? to me, pressure change would be much more intuitive since thats what causing the imbalance - increasing alerady higher pressure in containter, while decreasing already low pressure in surrounding
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  + Andrew M
  6 years ago
  Posted 6 years ago. Direct link to Andrew M's post “Work is force*distance. ...”
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 Work is force*distance. The volume change occurs because a wall of the container is being displaced. If you just increase the pressure and there is no volume change (think about how you would do that), no work is being done.
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 * Juan Eduardo Ynsil Alfaro
 5 years ago
 Posted 5 years ago. Direct link to Juan Eduardo Ynsil Alfaro's post “What if we use energy to ...”
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 What if we use energy to push the piston up? Wouldn't it mean that we are losing energy?
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  + Mark Geary
  5 years ago
  Posted 5 years ago. Direct link to Mark Geary's post “If you add external energ...”
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 If you add external energy to the system by performing work on the system, then the system is gaining energy.
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 * miARNr
 4 years ago
 Posted 4 years ago. Direct link to miARNr's post “why In the PV diagram we ...”
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 why In the PV diagram we are using the internal pressure to calculate the work ? shouldn't it be the external pressure as we have done in the previous videos ?
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  + Koof
  2 years ago
  Posted 2 years ago. Direct link to Koof's post “since it's a quasistatic ...”
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 since it's a quasistatic pressure, every intermediate state is in equilibrium i.e. internal pressure is almost equal to external pressure so they can be considered interchangable
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Learn what PV diagrams are and how to use them to find the change in internal energy, work done, and heat.     What are PV diagrams?    Consider a gas sealed in a container with a tightly fitting yet movable piston as seen below. We can do work on the gas by pressing the piston downward, and we can heat up the gas by placing the container over a flame or submerging it in a bath of boiling water. When we subject the gas to these thermodynamics processes, the pressure and volume of the gas can change.   A convenient way to visualize these changes in the pressure and volume is by using a Pressure Volume diagram or PV diagram for short. Each point on a PV diagram corresponds to a different state of the gas. The pressure is given on the vertical axis and the volume is given on the horizontal axis, as seen below.   Every point on a PV diagram represents a different state for the gas (one for every possible volume and pressure). As a gas goes through a thermodynamics process, the state of the gas will shift around in the PV diagram, tracing out a path as it moves (as shown in the diagram below).   Being able to decode the information shown in a PV diagram allows us to make statements about the change in internal energy delta, U, heat transferred Q, and work done W on a gas. In the sections below, we'll explain how to decipher the hidden information contained in a PV diagram.     Note: Unless otherwise specified, we will assume that the work W refers to the work done on the gas.       How do we determine the sign of the work done from a PV diagram?    Let's say our gas starts out in the state shown in the PV diagram below.    If we press the piston downward, the volume of the gas will decrease, so the state must shift to the left toward smaller volumes (as seen in the diagram below). Since the gas is being compressed we can also say for sure that positive work W is being done on the gas.    Similarly, if we let the gas expand, pushing the piston upward, the volume of the gas will increase, so the state must shift to the right toward larger volumes (as seen in the diagram below). Since the gas is expanding we can also say for sure that negative work W is being done on the gas.    So if we ever see a state shifting to the left on a PV diagram we can say for sure that the work done on the gas was positive. Similarly, if we ever see a state shifting to the right on a PV diagram we can say for sure that the work done on the gas was negative.      How do we determine the magnitude of the work done from a PV diagram?    The work done during a thermodynamic process is equal to the area under the curve as seen in the diagram below.    The reason why work is equal to the area under the curve is that,   W, equals, F, delta, x, equals, left parenthesis, P, A, right parenthesis, delta, x, equals, P, left parenthesis, A, delta, x, right parenthesis, equals, P, delta, V   And since P, delta, V is just the start text, h, e, i, g, h, t, end text, times, start text, space, w, i, d, t, h, end text of the rectangle shown above, the work is equal to the area. If we use pressure units of start text, p, a, s, c, a, l, s, end text and volume units of start text, m, end text, cubed then the energy we find will be in units of start text, j, o, u, l, e, s, end text.    We have to be really careful with signs though. If the path on a PV diagram is directed to the left, the volume is decreasing, and positive work is being done on the gas. If the path on a PV diagram is directed to the right (as in the diagram above), the volume is increasing, and negative work is being done on the gas since W, start subscript, start text, b, y, space, g, a, s, end text, end subscript, equals, minus, W, start subscript, start text, o, n, space, g, a, s, end text, end subscript.   It doesn't matter what shape the path takes, the area under the curve will still represent the work done. For any curved path we can imagine breaking the area into an infinite amount of infinitesimally thin rectangles.    The area of each rectangle would represent the work done during each infinitesimal step, and the sum of the areas would represent the total work done for the entire process.   It should be said that we are always going to assume these processes are taking place slowly enough that the entire gas can be at thermodynamic equilibrium at every moment (i.e. the same temperature throughout the gas). If this seems dubious to you, you're right to question it. However, even though basically no real world processes will exactly satisfy this requirement, our ability to model many thermodynamic processes are not fatally jeopardized by this lack of adherence to ideal circumstances.      How do we determine the sign of delta, U from a PV diagram?    Remember that internal energy and temperature are proportional U, \propto, T. So if the temperature increases, the internal energy must also increase.    Now, if the gas we're considering is an ideal gas we also know that,    P, V, equals, N, k, start subscript, B, end subscript, T   And if no gas is allowed to escape (so the number of molecules N is constant) we can say that P, V, \propto, T. All of this means that,   U, \propto, T, \propto, P, V   So if the quantity of pressure times volume left parenthesis, P, times, V, right parenthesis increases, the temperature T and internal energy U must also increase (which makes delta, U positive). This idea is represented in the diagram shown below.   This means that anytime the state in a PV diagram ends up further up and right than where it started, delta, U is a positive number. Similarly, anytime the state in a PV diagram ends up further down and left than where it started, delta, U is a negative number.   Now if the state in the PV diagram moves up and left (pressure increases and volume decreases), or down and right (pressure decreases and volume increases), it is a little ambiguous whether the quantity left parenthesis, P, times, V, right parenthesis actually increased or decreased (since one variable increased and the other variable decreased). To be sure, one would have to check the exact values of the initial and final P and V on the axes of the graph to tell if the quantity left parenthesis, P, times, V, right parenthesis actually increased or decreased.   It is also good to note that if the quantity left parenthesis, P, times, V, right parenthesis does not change, then the temperature T and internal energy U do not change either. For instance, if the pressure doubles, and the volume is cut in half, left parenthesis, P, times, V, right parenthesis remains the same value (since 2, P, times, start fraction, V, divided by, 2, end fraction, equals, P, V). the temperature T and internal energy U will end the process with the same values they started with.      How do we determine the sign of Q from a PV diagram?    Given a PV diagram, we typically have to rely on the first law of thermodynamics delta, U, equals, Q, plus, W to determine the sign of the net heat that enters or exits a gas. If we solve this equation for the heat Q we get,   Q, equals, delta, U, minus, W   Now that we know this, we can use what we know about finding the sign of delta, U and W to find the sign of Q in many cases. So for instance, if the change in internal energy is positive and the work done is negative,    Q, equals, left parenthesis, plus, right parenthesis, minus, left parenthesis, minus, right parenthesis, equals, plus ...the net heat must be positive.   Which makes sense, since if the internal energy increased even though work was done by the gas, that implies that more heat must have entered the gas than energy lost due to the work done by the gas.   Or for example, if the internal energy decreases and the work is positive,   Q, equals, left parenthesis, minus, right parenthesis, minus, left parenthesis, plus, right parenthesis, equals, minus ...the net heat must be negative.   Which makes sense, since if the internal energy decreased even though work was done on the gas, that implies that more heat must have left the gas than energy gained by the gas from work being done on it.      What do solved examples involving PV diagrams look like?     Example 1: Finding signs    An ideal gas in a sealed container is taken through the process shown in the PV diagram below.     Example 2: Finding area    An ideal gas in a sealed container is taken through the process shown in the PV diagram below. The initial volume of the gas is V, start subscript, i, end subscript, equals, 0, point, 25, start text, m, end text, cubed and the final volume of the gas is V, start subscript, f, end subscript, equals, 0, point, 75, start text, m, end text, cubed. The initial pressure of the gas is P, start subscript, i, end subscript, equals, 70, comma, 000, start text, space, P, a, end text and the final pressure of the gas is P, start subscript, f, end subscript, equals, 160, comma, 000, start text, space, P, a, end text.    What is the work done on the gas during the process shown?    We can find the work done by determining the total area under the curve on a PV diagram. We have to make sure we use the total area, all the way down to the volume axis. For instance, we can imagine viewing the area under the curve in the example shown above as a triangle and a rectangle (as seen below).    Now we just find the sum of the areas of the triangle and rectangle. The height of the rectangle is the pressure P, start subscript, i, end subscript and the width of the rectangle is the change in volume delta, V, equals, V, start subscript, f, end subscript, minus, V, start subscript, i, end subscript. So,   start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, start text, h, e, i, g, h, t, end text, times, start text, w, i, d, t, h, end text (area of the rectangle)   start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, P, start subscript, i, end subscript, times, delta, V (height is P, start subscript, i, end subscript and width is delta, V)   start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, left parenthesis, 70, comma, 000, start text, space, P, a, end text, right parenthesis, times, left parenthesis, 0, point, 75, start text, m, end text, cubed, minus, 0, point, 25, start text, m, end text, cubed, right parenthesis (plug in values)   start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, 35, comma, 000, start text, space, J, end text (calculate)   We can find the area of a triangle by using A, equals, start fraction, 1, divided by, 2, end fraction, b, h.   start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, b, h (area of the triangle)   start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, b, left parenthesis, 160, comma, 000, start text, P, a, end text, minus, 70, comma, 000, start text, space, P, a, end text, right parenthesis (the height of the triangle is the difference in pressures P, start subscript, f, end subscript, minus, P, start subscript, i, end subscript)   start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, 0, point, 75, start text, m, end text, cubed, minus, 0, point, 25, start text, m, end text, cubed, right parenthesis, left parenthesis, 160, comma, 000, start text, P, a, end text, minus, 70, comma, 000, start text, space, P, a, end text, right parenthesis (the base of the triangle is the difference in volumes V, start subscript, f, end subscript, minus, V, start subscript, i, end subscript)   start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, 22, comma, 500, start text, space, J, end text (calculate)   So the total area under the curve is 35, comma, 000, start text, space, J, end text, plus, 22, comma, 500, start text, space, J, end text, equals, 57, comma, 500, start text, space, J, end text   This area represents the absolute value of the total work done during the process. To determine the sign of the work done on the gas we notice that the process moves the state to the right, causing the gas to expand. When gas expands the work done on the gas is negative. So,   W, start subscript, start text, o, n, space, g, a, s, end text, end subscript, equals, minus, 57, comma, 500, start text, space, J, end text (celebrate)
What are PV diagrams? (article) | Khan Academy
https://www.khanacademy.org/science/physics/thermodynamics/laws-of-thermodynamics/a/what-are-pv-diagrams
https://energyeducation.ca/encyclopedia/Pressure_volume_diagram
Pressure volume diagram - Energy Education
2018-12-10
None
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<div><p>The <b>PV diagram</b> models the relationship between <a href="https://energyeducation.ca/encyclopedia/Pressure">pressure</a> (P) and <a href="https://energyeducation.ca/encyclopedia/Volume">volume</a> (V) for an <a href="https://energyeducation.ca/encyclopedia/Ideal_gas_law">ideal gas</a>. An ideal gas is one that never condenses regardless of the various changes its <a href="https://energyeducation.ca/wiki/index.php?title=State_variable&amp;action=edit&amp;redlink=1">state variables</a> (pressure, volume, <a href="https://energyeducation.ca/encyclopedia/Temperature">temperature</a>) undergo. <sup><a href="#cite_note-knight-1">[1]</a></sup> In addition, the processes plotted on PV diagrams only work for a closed <a href="https://energyeducation.ca/encyclopedia/System">system</a> (in this case the ideal gas), so there is no exchange of matter, but still an exchange of energy. 
</p><p>Pressure and volume exhibit a causal relationship, meaning that the change of one variable will cause the change in the other. To understand how pressure directly affects volume (and vice versa)—imagine a sealed container, containing an ideal gas (the system), that has a moving <a href="https://energyeducation.ca/encyclopedia/Piston">piston</a>. If a <a href="https://energyeducation.ca/encyclopedia/Force">force</a> is applied, the piston moves down, and the gas would compress—decreasing the volume in the system and causing an increase in pressure. Moreover, if the piston moves up, the volume of the system would increase, decreasing the pressure of the system. Therefore, an increase in one variable will cause the decrease in the other, and vice versa. However, if an increase (or decrease) in pressure and/or volume is desired, an external heat source (or a cooling source) from its surroundings must be added. 
</p><p>In addition, these diagrams not only model the relationship between pressure and volume for an ideal gas, but can also be used to calculate <a href="https://energyeducation.ca/encyclopedia/Work">work</a> done (on or by the <a href="https://energyeducation.ca/encyclopedia/System">system</a>). This is done by calculating the <i>negative of the area under the curve</i> which can be done geometrically or by integration.<sup><a href="#cite_note-knight-1">[1]</a></sup> The general formula to calculate work done by an ideal gas is the integral:<sup><a href="#cite_note-knight-1">[1]</a></sup>
</p>
<p>where:
</p><p>: means that work is being done <i>on the system</i> (compression)
</p><p>: means that work is being done <i>by the system</i> (expansion)
</p><p>: no work is done (isochoric process—see figure below)
</p><p><br/>
The fundamental thermodynamic processes modelled on PV diagrams (isochoric, isobaric, and isothermal processes) all follow the <a href="https://energyeducation.ca/encyclopedia/Ideal_gas_law">ideal gas law</a> except for <a href="https://energyeducation.ca/encyclopedia/Adiabatic">adiabatic</a> processes—which will be discussed in detail on its main page. The following are the examples of each process modelled on the PV diagram. Each of their pages will describe their unique characteristics in detail: 
</p>
<ul>
<li>Four main ideal gas law processes modelled on PV diagrams</li>
<li><div>
<div><p><a href="https://energyeducation.ca/encyclopedia/File:Isobar.png"></a></p></div>
<p><b><a href="https://energyeducation.ca/encyclopedia/Isobar_(pressure)">Isobaric</a> Process:</b> <br/> Are occurrences when the volume of the system is changing (induced by external heating or cooling sources) but pressure is kept constant<sup><a href="#cite_note-2">[2]</a></sup>
</p>
</div></li>
<li><div>
<div><p><a href="https://energyeducation.ca/encyclopedia/File:Isochore.png"></a></p></div>
<p><b><a href="https://energyeducation.ca/encyclopedia/Isochore">Isochoric</a> Process:</b> <br/> This models a system where the volume stays constant (ex. locking the piston in place) but pressure is changing. This will require an external heating/cooling source to increase/decrease pressure. <sup><a href="#cite_note-3">[3]</a></sup>
</p>
</div></li>
<li><div>
<div><p><a href="https://energyeducation.ca/encyclopedia/File:Isotherm.png"></a></p></div>
<p><b><a href="https://energyeducation.ca/encyclopedia/Isothermal">Isothermal</a> Process:</b> <br/> This diagram is showing an ideal gas exhibiting constant temperature. This is possible because <i><a href="https://energyeducation.ca/encyclopedia/Heat">heat</a> is being exchanged</i> with its surroundings.<sup><a href="#cite_note-4">[4]</a></sup>
</p>
</div></li>
<li><div>
<div><p><a href="https://energyeducation.ca/encyclopedia/File:Adiabat.png"></a></p></div>
<p><b><a href="https://energyeducation.ca/encyclopedia/Adiabatic">Adiabatic</a> Process:</b> <br/> Similar to an isothermal process at first glance, adiabats represent an ideal gas exhibiting a change in temperature, because there is <i>no heat exchange</i> between the system and surroundings. The purple line represents an adiabat, while the dashed line (--) are isotherms. <sup><a href="#cite_note-5">[5]</a></sup>
</p>
</div></li>
</ul>
<h2></h2>
<p>The various processes seen above can be combined to create cycles found in most <a href="https://energyeducation.ca/encyclopedia/Internal_combustion_engine">internal</a> and <a href="https://energyeducation.ca/encyclopedia/External_combustion_engine">external</a> combustion engines. These diagrams are showing how pistons in engines (powered by fuel) or the various processes in a <a href="https://energyeducation.ca/encyclopedia/Power_plant">power plant</a>, change the volume and pressure of a working fluid (ex. steam water for turbines, fuel-air mixture for engines) to create <a href="https://energyeducation.ca/encyclopedia/Work">work</a>. Then this work can be used to create electricity or move a vehicle. The work done would be calculated by finding the area inside the closed cycle (ex. the yellow shaded area of the Rankine cycle below). 
</p><p>The following are the main cycles used in various power plants and transportation mechanisms: 
</p>
<ul>
<li>Four common PV cycles</li>
<li><div>
<div><p><a href="https://energyeducation.ca/encyclopedia/File:Otto_PV_cycle.png"></a></p></div>
<p><b><a href="https://energyeducation.ca/encyclopedia/Otto_cycle">Otto cycle</a></b>: <br/> This is used in some automobiles and machines that use gasoline as fuel. <sup><a href="#cite_note-6">[6]</a></sup>
</p>
</div></li>
<li><div>
<div><p><a href="https://energyeducation.ca/encyclopedia/File:Brayton-cycle-crop.JPG"></a></p></div>
<p><b><a href="https://energyeducation.ca/encyclopedia/Brayton_cycle">Brayton cycle</a></b>: <br/>This cycle is used for gas turbine engines and some jet engines. <sup><a href="#cite_note-7">[7]</a></sup>
</p>
</div></li>
<li><div>
<div><p><a href="https://energyeducation.ca/encyclopedia/File:Diesel-cycle-crop.JPG"></a></p></div>
<p><b><a href="https://energyeducation.ca/wiki/index.php?title=Diesel_cycle&amp;action=edit&amp;redlink=1">Diesel cycle</a></b>: <br/> This process is for vehicles that use diesel as fuel. <sup><a href="#cite_note-8">[8]</a></sup>
</p>
</div></li>
<li></li>
</ul>
<h2></h2>
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 Pressure volume diagram

 The PV diagram models the relationship between pressure (P) and volume (V) for an ideal gas. An ideal gas is one that never condenses regardless of the various changes its state variables (pressure, volume, temperature) undergo. [1] In addition, the processes plotted on PV diagrams only work for a closed system (in this case the ideal gas), so there is no exchange of matter, but still an exchange of energy.

 Pressure and volume exhibit a causal relationship, meaning that the change of one variable will cause the change in the other. To understand how pressure directly affects volume (and vice versa)—imagine a sealed container, containing an ideal gas (the system), that has a moving piston. If a force is applied, the piston moves down, and the gas would compress—decreasing the volume in the system and causing an increase in pressure. Moreover, if the piston moves up, the volume of the system would increase, decreasing the pressure of the system. Therefore, an increase in one variable will cause the decrease in the other, and vice versa. However, if an increase (or decrease) in pressure and/or volume is desired, an external heat source (or a cooling source) from its surroundings must be added.

 In addition, these diagrams not only model the relationship between pressure and volume for an ideal gas, but can also be used to calculate work done (on or by the system). This is done by calculating the negative of the area under the curve which can be done geometrically or by integration.[1] The general formula to calculate work done by an ideal gas is the integral:[1]

 where:

 : means that work is being done on the system (compression)

 : means that work is being done by the system (expansion)

 : no work is done (isochoric process—see figure below)


 The fundamental thermodynamic processes modelled on PV diagrams (isochoric, isobaric, and isothermal processes) all follow the ideal gas law except for adiabatic processes—which will be discussed in detail on its main page. The following are the examples of each process modelled on the PV diagram. Each of their pages will describe their unique characteristics in detail:


 * Four main ideal gas law processes modelled on PV diagrams

 * Isobaric Process:
 Are occurrences when the volume of the system is changing (induced by external heating or cooling sources) but pressure is kept constant[2]

 * Isochoric Process:
 This models a system where the volume stays constant (ex. locking the piston in place) but pressure is changing. This will require an external heating/cooling source to increase/decrease pressure. [3]

 * Isothermal Process:
 This diagram is showing an ideal gas exhibiting constant temperature. This is possible because heat is being exchanged with its surroundings.[4]

 * Adiabatic Process:
 Similar to an isothermal process at first glance, adiabats represent an ideal gas exhibiting a change in temperature, because there is no heat exchange between the system and surroundings. The purple line represents an adiabat, while the dashed line (--) are isotherms. [5]


 Applications of the PV diagram

 The various processes seen above can be combined to create cycles found in most internal and external combustion engines. These diagrams are showing how pistons in engines (powered by fuel) or the various processes in a power plant, change the volume and pressure of a working fluid (ex. steam water for turbines, fuel-air mixture for engines) to create work. Then this work can be used to create electricity or move a vehicle. The work done would be calculated by finding the area inside the closed cycle (ex. the yellow shaded area of the Rankine cycle below).

 The following are the main cycles used in various power plants and transportation mechanisms:


 * Four common PV cycles

 * Otto cycle:
 This is used in some automobiles and machines that use gasoline as fuel. [6]

 * Brayton cycle:
 This cycle is used for gas turbine engines and some jet engines. [7]

 * Diesel cycle:
 This process is for vehicles that use diesel as fuel. [8]

 * Rankine cycle
 This cycle is widely used by power plants, such as coal-fired power plants or nuclear reactors. [9]

 References

 1. ↑ 1.0 1.1 1.2 R. Knight, Physics for scientists and engineers, 2nd ed. San Francisco: Pearson Addison Wesley, 2008, p. 491.
 2. ↑ Wikimedia Commons [Online], Available: http://wiki.mitsted.dk/?page=Billede:Isobar-proces.png
 3. ↑ Wikimedia Commons [Online], Available: http://wiki.mitsted.dk/?page=Billede:Isokor-proces.png
 4. ↑ Wikimedia Commons [Online], Available: https://commons.wikimedia.org/wiki/File:Isothermal_process.svg
 5. ↑ Wikimedia Commons [Online], Available: https://commons.wikimedia.org/wiki/File:Adiabatic_process.png
 6. ↑ Made internally by a member of the Energy Education team
 7. ↑ Made internally by a member of the Energy Education team
 8. ↑ Made internally by a member of the Energy Education team
 9. ↑ Made internally by a member of the Energy Education team
 Retrieved from "https://energyeducation.ca/wiki/index.php?title=Pressure_volume_diagram&oldid=9801"
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Navigation menu
 The PV diagram models the relationship between pressure (P) and volume (V) for an ideal gas. An ideal gas is one that never condenses regardless of the various changes its state variables (pressure, volume, temperature) undergo.  [1]  In addition, the processes plotted on PV diagrams only work for a closed system (in this case the ideal gas), so there is no exchange of matter, but still an exchange of energy. 
 Pressure and volume exhibit a causal relationship, meaning that the change of one variable will cause the change in the other. To understand how pressure directly affects volume (and vice versa)—imagine a sealed container, containing an ideal gas (the system), that has a moving piston. If a force is applied, the piston moves down, and the gas would compress—decreasing the volume in the system and causing an increase in pressure. Moreover, if the piston moves up, the volume of the system would increase, decreasing the pressure of the system. Therefore, an increase in one variable will cause the decrease in the other, and vice versa. However, if an increase (or decrease) in pressure and/or volume is desired, an external heat source (or a cooling source) from its surroundings must be added. 
 In addition, these diagrams not only model the relationship between pressure and volume for an ideal gas, but can also be used to calculate work done (on or by the system). This is done by calculating the negative of the area under the curve which can be done geometrically or by integration. [1]  The general formula to calculate work done by an ideal gas is the integral: [1] 
where:
 : means that work is being done on the system (compression)
 : means that work is being done by the system (expansion)
 : no work is done (isochoric process—see figure below)
  
The fundamental thermodynamic processes modelled on PV diagrams (isochoric, isobaric, and isothermal processes) all follow the ideal gas law except for adiabatic processes—which will be discussed in detail on its main page. The following are the examples of each process modelled on the PV diagram. Each of their pages will describe their unique characteristics in detail: 
Four main ideal gas law processes modelled on PV diagrams
 
     
  Isobaric Process:  Are occurrences when the volume of the system is changing (induced by external heating or cooling sources) but pressure is kept constant [2] 
 
 
     
  Isochoric Process:  This models a system where the volume stays constant (ex. locking the piston in place) but pressure is changing. This will require an external heating/cooling source to increase/decrease pressure.  [3] 
 
 
     
  Isothermal Process:  This diagram is showing an ideal gas exhibiting constant temperature. This is possible because  heat is being exchanged with its surroundings. [4] 
 
 
     
  Adiabatic Process:  Similar to an isothermal process at first glance, adiabats represent an ideal gas exhibiting a change in temperature, because there is no heat exchange between the system and surroundings. The purple line represents an adiabat, while the dashed line (--) are isotherms.  [5] 
 
 
The various processes seen above can be combined to create cycles found in most internal and external combustion engines. These diagrams are showing how pistons in engines (powered by fuel) or the various processes in a power plant, change the volume and pressure of a working fluid (ex. steam water for turbines, fuel-air mixture for engines) to create work. Then this work can be used to create electricity or move a vehicle. The work done would be calculated by finding the area inside the closed cycle (ex. the yellow shaded area of the Rankine cycle below). 
 The following are the main cycles used in various power plants and transportation mechanisms: 
Four common PV cycles
 
     
  Otto cycle :  This is used in some automobiles and machines that use gasoline as fuel.  [6] 
 
 
     
  Brayton cycle : This cycle is used for gas turbine engines and some jet engines.  [7] 
 
 
     
  Diesel cycle :  This process is for vehicles that use diesel as fuel.  [8]
Pressure volume diagram - Energy Education
https://energyeducation.ca/encyclopedia/Pressure_volume_diagram
https://www.e-education.psu.edu/png520/m3_p3.html
PV Diagram for Pure Systems
2020-01-01
None
<div><div><p>In the previous discussion, we used the P-T diagram and were not concerned about changes in the <em>volume</em> of the system. If we want to follow changes in volume, we may construct P-v or T-v diagrams, in which we hold temperature (T) or pressure (P) constant. Let us consider the case of a P-v Diagram (Figure 3.4).</p>
<div><p></p><p>Figure 3.4: P-v Diagram For A Pure Component</p>
</div>
<p>In this case temperature is being held constant; our substance is undergoing an <em>isothermal compression process.</em> Starting at E (all-vapor condition), an increase in pressure will result in a rather significant reduction in volume since the gas phase is compressible. If we keep compressing isothermally, we will end up at point F, where the gas will be saturated and the first droplet of liquid will appear. We have come to the two-phase condition, where liquid (L) and vapor (V) co-exist in equilibrium, for the first time, albeit an infinitesimal amount of liquid.</p>
<p>Once we reach the two-phase condition, what happens is not intuitive. While we keep on compressing by decreasing the volume (path F-G,) the pressure of the system remains constant; this condition continues until <em>all</em> the vapor has become liquid. Point G represents the last condition of liquid and vapor (L+V) coexistence, saturated liquid condition (liquid in equilibrium with an infinitesimal amount of vapor.) Once we have only liquid, if we keep on compressing (i.e., attempting to reduce liquid volume) we will observe a rapid increase in pressure, as indicated by the steep slope in the P-v diagram. This is because liquid is virtually incompressible, hence, a great deal of pressure is needed to cause a small reduction in volume.</p>
<p>It is important to recognize some points of this process. If we recall our previous definitions of basic concepts, we will recognize point F, where only a tiny quantity of liquid exists in an otherwise completely gaseous system (the <em>dew point </em>of the system at the given temperature). Similarly, Point G is the <em>bubble point</em>; only an infinitesimally small bubble of vapor exists in an otherwise liquid system.</p>
<p>But wait a second. Let us try to compare Figure 3.4 with Figure 3.2. Can we relate them to each other? Where is path F-G in Figure 3.4 represented in Figure 3.2 (repeated below)?</p>
<div><p></p><p>Figure 3.2 <em>(repeated)</em>: Vapor pressure curve and ACB Path</p>
</div>
<p>The answer is, path F-G is represented by <em>one</em> point in Figure 3.2; that is, point C. Recall, for a single-component system, dew points and bubble points are identical. During a phase transition, both pressure and temperature must remain constant for pure components.</p>
<p>Now, if we want to generate all the possible points that make up the vapor pressure curve in Figure 3.2, we would need to repeat the experiment for different temperatures. We would end up with a family of isotherms (each similar to the one presented in Figure 3.4). This is represented in Figure 3.5.</p>
<div><p></p><p>Figure 3.5: Family of P-v isotherms for a pure component</p>
</div>
<p>The zone where the isotherms become flat delineates the two-phase region. It is clearly seen that by plotting all the pairs in that zone (P<sub>1</sub>,T<sub>1</sub>), (P<sub>2</sub>,T<sub>2</sub>)… (P<sub>c</sub>, T<sub>c</sub>) we will be able to reproduce Figure 3.2.</p>
<p>If we now draw a line through all the <em>Bubble Points </em>in Figure 3.5, and then draw a line connecting all the <em>Dew Points</em>, we will end up with the <em>Bubble Point Curve</em> and the <em>Dew Point Curve, </em>respectively. It is clear that the two curves meet at the critical point (P<sub>c</sub>, T<sub>c</sub>). Furthermore, the two curves delineate the phase envelope, which contains the 2-phase region inside. If we “clean” Figure 3.5 a little, we end up with the <em>phase envelope</em> that is shown in Fig. 3.6.</p>
<div><p></p><p>Figure 3.6: P-v Diagram And Phase Envelope Of A Pure Substance</p>
</div>
<p>If you carefully follow the trend of the critical isotherm (@ T = T<sub>c</sub> in Fig. 3.5), you will realize that it has a point of inflexion (change of curvature) at the critical point. Furthermore, the critical point also represents the maximum point (apex) of the P-v envelope. Mathematically, this information is conveyed by the expressions:</p>
<p>$$\left(\frac{\partial P}{\partial \vec{V}}\right)_{P C, T C}=\left(\frac{\partial^{2} P}{\partial \vec{V}^{2}}\right)_{P C T C}=0$$</p>
<p>which are usually known as the <em>criticality conditions</em>. These conditions are always satisfied at the critical point. We will comment more on this after we begin the discussion on Equations of State (EOS) — semi-empirical relationships that mathematically model the P-v-T behavior of fluids.</p>
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 PV Diagram for Pure Systems

 Print

 In the previous discussion, we used the P-T diagram and were not concerned about changes in the volume of the system. If we want to follow changes in volume, we may construct P-v or T-v diagrams, in which we hold temperature (T) or pressure (P) constant. Let us consider the case of a P-v Diagram (Figure 3.4).

 Figure 3.4: P-v Diagram For A Pure Component

 In this case temperature is being held constant; our substance is undergoing an isothermal compression process. Starting at E (all-vapor condition), an increase in pressure will result in a rather significant reduction in volume since the gas phase is compressible. If we keep compressing isothermally, we will end up at point F, where the gas will be saturated and the first droplet of liquid will appear. We have come to the two-phase condition, where liquid (L) and vapor (V) co-exist in equilibrium, for the first time, albeit an infinitesimal amount of liquid.

 Once we reach the two-phase condition, what happens is not intuitive. While we keep on compressing by decreasing the volume (path F-G,) the pressure of the system remains constant; this condition continues until all the vapor has become liquid. Point G represents the last condition of liquid and vapor (L+V) coexistence, saturated liquid condition (liquid in equilibrium with an infinitesimal amount of vapor.) Once we have only liquid, if we keep on compressing (i.e., attempting to reduce liquid volume) we will observe a rapid increase in pressure, as indicated by the steep slope in the P-v diagram. This is because liquid is virtually incompressible, hence, a great deal of pressure is needed to cause a small reduction in volume.

 It is important to recognize some points of this process. If we recall our previous definitions of basic concepts, we will recognize point F, where only a tiny quantity of liquid exists in an otherwise completely gaseous system (the dew point of the system at the given temperature). Similarly, Point G is the bubble point; only an infinitesimally small bubble of vapor exists in an otherwise liquid system.

 But wait a second. Let us try to compare Figure 3.4 with Figure 3.2. Can we relate them to each other? Where is path F-G in Figure 3.4 represented in Figure 3.2 (repeated below)?

 Figure 3.2 (repeated): Vapor pressure curve and ACB Path

 The answer is, path F-G is represented by one point in Figure 3.2; that is, point C. Recall, for a single-component system, dew points and bubble points are identical. During a phase transition, both pressure and temperature must remain constant for pure components.

 Now, if we want to generate all the possible points that make up the vapor pressure curve in Figure 3.2, we would need to repeat the experiment for different temperatures. We would end up with a family of isotherms (each similar to the one presented in Figure 3.4). This is represented in Figure 3.5.

 Figure 3.5: Family of P-v isotherms for a pure component

 The zone where the isotherms become flat delineates the two-phase region. It is clearly seen that by plotting all the pairs in that zone (P1,T1), (P2,T2)… (Pc, Tc) we will be able to reproduce Figure 3.2.

 If we now draw a line through all the Bubble Points in Figure 3.5, and then draw a line connecting all the Dew Points, we will end up with the Bubble Point Curve and the Dew Point Curve, respectively. It is clear that the two curves meet at the critical point (Pc, Tc). Furthermore, the two curves delineate the phase envelope, which contains the 2-phase region inside. If we “clean” Figure 3.5 a little, we end up with the phase envelope that is shown in Fig. 3.6.

 Figure 3.6: P-v Diagram And Phase Envelope Of A Pure Substance

 If you carefully follow the trend of the critical isotherm (@ T = Tc in Fig. 3.5), you will realize that it has a point of inflexion (change of curvature) at the critical point. Furthermore, the critical point also represents the maximum point (apex) of the P-v envelope. Mathematically, this information is conveyed by the expressions:

 $$\left(\frac{\partial P}{\partial \vec{V}}\right)_{P C, T C}=\left(\frac{\partial^{2} P}{\partial \vec{V}^{2}}\right)_{P C T C}=0$$

 which are usually known as the criticality conditions. These conditions are always satisfied at the critical point. We will comment more on this after we begin the discussion on Equations of State (EOS) — semi-empirical relationships that mathematically model the P-v-T behavior of fluids.

 ‹ Vapor Pressure Curve up Action Item ›
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In the previous discussion, we used the P-T diagram and were not concerned about changes in the volume of the system. If we want to follow changes in volume, we may construct P-v or T-v diagrams, in which we hold temperature (T) or pressure (P) constant. Let us consider the case of a P-v Diagram (Figure 3.4).
   Figure 3.4: P-v Diagram For A Pure Component
In this case temperature is being held constant; our substance is undergoing an isothermal compression process. Starting at E (all-vapor condition), an increase in pressure will result in a rather significant reduction in volume since the gas phase is compressible. If we keep compressing isothermally, we will end up at point F, where the gas will be saturated and the first droplet of liquid will appear. We have come to the two-phase condition, where liquid (L) and vapor (V) co-exist in equilibrium, for the first time, albeit an infinitesimal amount of liquid.
Once we reach the two-phase condition, what happens is not intuitive. While we keep on compressing by decreasing the volume (path F-G,) the pressure of the system remains constant; this condition continues until all the vapor has become liquid. Point G represents the last condition of liquid and vapor (L+V) coexistence, saturated liquid condition (liquid in equilibrium with an infinitesimal amount of vapor.) Once we have only liquid, if we keep on compressing (i.e., attempting to reduce liquid volume) we will observe a rapid increase in pressure, as indicated by the steep slope in the P-v diagram. This is because liquid is virtually incompressible, hence, a great deal of pressure is needed to cause a small reduction in volume.
It is important to recognize some points of this process. If we recall our previous definitions of basic concepts, we will recognize point F, where only a tiny quantity of liquid exists in an otherwise completely gaseous system (the dew point of the system at the given temperature). Similarly, Point G is the bubble point; only an infinitesimally small bubble of vapor exists in an otherwise liquid system.
But wait a second. Let us try to compare Figure 3.4 with Figure 3.2. Can we relate them to each other? Where is path F-G in Figure 3.4 represented in Figure 3.2 (repeated below)?
   Figure 3.2 (repeated): Vapor pressure curve and ACB Path
The answer is, path F-G is represented by one point in Figure 3.2; that is, point C. Recall, for a single-component system, dew points and bubble points are identical. During a phase transition, both pressure and temperature must remain constant for pure components.
Now, if we want to generate all the possible points that make up the vapor pressure curve in Figure 3.2, we would need to repeat the experiment for different temperatures. We would end up with a family of isotherms (each similar to the one presented in Figure 3.4). This is represented in Figure 3.5.
   Figure 3.5: Family of P-v isotherms for a pure component
The zone where the isotherms become flat delineates the two-phase region. It is clearly seen that by plotting all the pairs in that zone (P1,T1), (P2,T2)… (Pc, Tc) we will be able to reproduce Figure 3.2.
If we now draw a line through all the Bubble Points in Figure 3.5, and then draw a line connecting all the Dew Points, we will end up with the Bubble Point Curve and the Dew Point Curve, respectively. It is clear that the two curves meet at the critical point (Pc, Tc). Furthermore, the two curves delineate the phase envelope, which contains the 2-phase region inside. If we “clean” Figure 3.5 a little, we end up with the phase envelope that is shown in Fig. 3.6.
   Figure 3.6: P-v Diagram And Phase Envelope Of A Pure Substance
If you carefully follow the trend of the critical isotherm (@ T = Tc in Fig. 3.5), you will realize that it has a point of inflexion (change of curvature) at the critical point. Furthermore, the critical point also represents the maximum point (apex) of the P-v envelope. Mathematically, this information is conveyed by the expressions:
$$\left(\frac{\partial P}{\partial \vec{V}}\right)_{P C, T C}=\left(\frac{\partial^{2} P}{\partial \vec{V}^{2}}\right)_{P C T C}=0$$
which are usually known as the criticality conditions. These conditions are always satisfied at the critical point. We will comment more on this after we begin the discussion on Equations of State (EOS) — semi-empirical relationships that mathematically model the P-v-T behavior of fluids.
PV Diagram for Pure Systems
https://www.e-education.psu.edu/png520/m3_p3.html
https://www.fourmilab.ch/scanalyzer/archives/2021/06/why-do-crowned-pulleys-work-the-second-law-of-thermodynamics.html
SCANALYZER
2021-06-04
None
<div><div>
<p>
</p><h2>CONTEXT â¢ THE HAPPENING WORLD â¢ CONTINUITY â¢ TRACKING WITH CLOSEUPS</h2>
<p></p>
<div>
<p>
<a href="https://www.fourmilab.ch/scanalyzer/archives/2021/06/nasa-were-going-back-to-venus-the-davinci-and-veritas-missions.html">« NASA: âWe're Going Back to Venusâ, the DAVINCI+ and VERITAS Missions</a> |
<a href="https://www.fourmilab.ch/scanalyzer/">Main</a>
| <a href="https://www.fourmilab.ch/scanalyzer/archives/2021/06/evidence-for-active-volcanoes-on-mars.html">Evidence for Active Volcanoes on Mars? »</a>
</p>
<h2>Friday, June 4, 2021</h2>
<h3>CONTINUITY: Why Do Crowned Pulleys Work? The Second Law of Thermodynamics!</h3>
<p>Posted at June 4, 2021 14:22</p>
</div>
</div>||||I|||| SCANALYZER

 CONTEXT â¢ THE HAPPENING WORLD â¢ CONTINUITY â¢ TRACKING WITH CLOSEUPS

 « NASA: âWe're Going Back to Venusâ, the DAVINCI+ and VERITAS Missions | Main | Evidence for Active Volcanoes on Mars? »

 Friday, June 4, 2021

 CONTINUITY: Why Do Crowned Pulleys Work? The Second Law of Thermodynamics!

 Posted at June 4, 2021 14:22

CONTEXT â¢ THE HAPPENING WORLD â¢ CONTINUITY â¢ TRACKING WITH CLOSEUPS
 
« NASA: âWe're Going Back to Venusâ, the DAVINCI+ and VERITAS Missions |
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| Evidence for Active Volcanoes on Mars? »
Friday, June 4, 2021
CONTINUITY: Why Do Crowned Pulleys Work? The Second Law of Thermodynamics!
Posted at June 4, 2021 14:22
SCANALYZER
https://www.fourmilab.ch/scanalyzer/archives/2021/06/why-do-crowned-pulleys-work-the-second-law-of-thermodynamics.html
https://arxiv.org/abs/2201.06956
General formalism for calculating the thermal efficiency of thermodynamic cycles defined in a $p-V$ diagram
2022-01-10
Könye; Viktor; Cserti; József
<div><div>
 
 
 
 
 
 
 <p><a href="/pdf/2201.06956">Download PDF</a></p><blockquote>
 Abstract: We develop a general method for calculating the thermal efficiency of
arbitrary thermodynamic cycles defined in the pressure-volume ($p-V$) diagram.
To demonstrate how effective our approach is, we calculate the thermal
efficiency of ideal gas engines for a few non-trivial cycles in the $p-V$
diagram, including a circular shape, a heart shape, a cycloid of Ceva, and a
star-shaped curve. We determine the segments along the cycle where heat is
absorbed or released from the heat engine. Our method can be applied to any gas
model, and, as an example, we present the results for the van der Waals gas.
 </blockquote>
 
 
 </div><div>
 <h2>Submission history</h2><p> From: Jozsef Cserti [<a href="/show-email/1f104c13/2201.06956">view email</a>]
 <br /><strong>[v1]</strong>
Mon, 10 Jan 2022 12:46:12 UTC (212 KB)<br /></p></div>||||I|||| Skip to main content
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 arXiv:2201.06956 (cond-mat)
 [Submitted on 10 Jan 2022]

 Title: General formalism for calculating the thermal efficiency of thermodynamic cycles defined in a $p-V$ diagram

 Authors: Viktor Könye, József Cserti
 Download PDF
 Abstract: We develop a general method for calculating the thermal efficiency of arbitrary thermodynamic cycles defined in the pressure-volume ($p-V$) diagram. To demonstrate how effective our approach is, we calculate the thermal efficiency of ideal gas engines for a few non-trivial cycles in the $p-V$ diagram, including a circular shape, a heart shape, a cycloid of Ceva, and a star-shaped curve. We determine the segments along the cycle where heat is absorbed or released from the heat engine. Our method can be applied to any gas model, and, as an example, we present the results for the van der Waals gas.
 Comments: Accepted in American Journal of Physics, 7 pages, 4 figures 
 Subjects: Statistical Mechanics (cond-mat.stat-mech) 
 Cite as: arXiv:2201.06956 [cond-mat.stat-mech] 
 (or arXiv:2201.06956v1 [cond-mat.stat-mech] for this version)
 https://doi.org/10.48550/arXiv.2201.06956 
 Focus to learn more 
 arXiv-issued DOI via DataCite 
 https://doi.org/10.1119/5.0064010 
 Related DOI: Focus to learn more 
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General formalism for calculating the thermal efficiency of thermodynamic cycles defined in a p − V diagram
Viktor Könye 
Institute for Theoretical Solid State Physics, IFW Dresden and Würzburg-Dresden Cluster of Excellence ct.qmat
Department of Physics of Complex Systems
Eötvös Loránd University
Helmholtzstr. 20, Pázmány Péter sétány 101069, H-1117Dresden, BudapestGermany, Hungary
József Cserti 
Institute for Theoretical Solid State Physics, IFW Dresden and Würzburg-Dresden Cluster of Excellence ct.qmat
Department of Physics of Complex Systems
Eötvös Loránd University
Helmholtzstr. 20, Pázmány Péter sétány 101069, H-1117Dresden, BudapestGermany, Hungary
General formalism for calculating the thermal efficiency of thermodynamic cycles defined in a p − V diagram
(Dated: January 19, 2022)
We develop a general method for calculating the thermal efficiency of arbitrary thermodynamic cycles defined in the pressure-volume (p − V ) diagram. To demonstrate how effective our approach is, we calculate the thermal efficiency of ideal gas engines for a few non-trivial cycles in the p − V diagram, including a circular shape, a heart shape, a cycloid of Ceva, and a star-shaped curve. We determine the segments along the cycle where heat is absorbed or released from the heat engine. Our method can be applied to any gas model, and, as an example, we present the results for the van der Waals gas. arXiv:2201.06956v1 [cond-mat.stat-mech]
I. INTRODUCTION
Calculating the efficiency of thermodynamic cycles is a central issue in practical applications. Moreover, when teaching, the basic concepts of thermodynamics can be introduced very efficiently through calculations of energy efficiency. In most textbook examples [1][2][3], the thermodynamic cycles in the pressure-volume (p − V ) diagram include only four types of processes: isothermal (constant temperature), isochoric (constant volume), isobaric (constant pressure), and adiabatic (heat-insulated system). Engines including such cycles, like the Carnot, Otto, Brayton (or Joule), Diesel, Stirling, Rankine, and Sargent cycles, are well known in the literature. To calculate the exchanged heat and work done for these systems, it is assumed that all processes are idealized; i.e., they are quasi-static and reversible, and there is no friction, turbulence, or heat loss. The common feature in the four above-mentioned types of idealized processes is that the exchanged heat and the work done can be calculated analytically for a gas characterized by its equation of state and its internal energy.
The calculation of the specific heat and of the thermal efficiency for unconventional thermodynamic processes (which deviate from the four special ones mentioned above) has been a frequently studied issue in the literature. For example, the specific heat has been calculated for gases in an arbitrary process in Ref. [4] and for the ideal gas along an elliptical p − V cycle in Ref. [5]. Similarly, to calculate the thermal efficiency, the use of logarithmic plots of the p−V diagram has been suggested in Ref. [6]. The thermal efficiency has been calculated for cycles in the p − V diagram involving diagonal processes with negative slope [7][8][9][10][11], parabolic processes [10], circular processes [12], an unconventional lobe [13], and an alternative thermodynamic cycle for the Stirling machine [14].
Extending the calculation of the thermal efficiency to general cycles then is a natural development. Indeed, for unconventional processes, the calculation of the efficiency is not obvious at all since it is rather difficult to calculate the absorbed and released heat during the cycle. In principle, one needs to locate the segments of the cycle where heat is absorbed or released by the fluid. These segments are separated by points at which the adiabatic curves are tangent to the curve representing the given cycle [11]. In most cases, such calculations cannot be performed analytically. This problem can be complicated even in the simple case of a circular process [12].
To avoid this problem, in this work, we present a general formalism for calculating the thermal efficiency for arbitrary cycles parametrized in the p − V diagram. Figure 1 shows such an example in which the cycle is a heart-shaped curve. Although such a heat engine is probably not realistic, our general approach could be useful in practice and may provide a didactically useful example for teaching the calculation of thermal efficiency as well.
Our formalism is quite versatile; it is valid for all gases provided their equation of state and their internal energy are known. We shall calculate the thermal efficiency both for ideal gas and for the van der Waals gas. From the calculation we shall identify the segments of the p − V diagram where heat is absorbed or released from the heat engine.
The paper is organized as follows. In Sec. II, we present our general formalism to obtain the thermodynamic efficiency. In Sec. III, we demonstrate the universality of our method by calculating the thermal efficiency of ideal gas engines for a few non-trivial processes defined in the p − V diagram. In Sec. IV, we extend our formalism to the van der Waals gas. Finally, in Sec. V, we summarize our main conclusions.  16)). The red lines and the blue dotted lines correspond to segments where heat is absorbed and released by the system, respectively.
II. GENERAL FORMALISM
First, we consider an arbitrary gas, assuming that its equation of state and its internal energy U (T, V, n) are known. Here, T is the temperature and n is the number of moles. From the differential form of the first law of thermodynamics, the exchanged heat for an infinitesimal quasi-static process depending only on the thermodynamic coordinates T and V is given bȳ
dQ = dU + p dV = ∂U ∂T V dT + ∂U ∂V T + p dV . (1)
The notationd signals thatdQ is not an exact differential of the state variables. The partial derivative of the internal energy U in the first term is the heat capacity at constant volume: C V = [∂U/∂T ] V (in general, it is known for a given gas), while the second term can be rewritten using the relation 
∂U ∂V T + p = T ∂S ∂V T = T ∂p ∂T V ,(2)dQ = C V ∂T ∂p V dp + C V ∂T ∂V p + T ∂p ∂T V dV . (3)
One can see that the exchanged heatdQ can be obtained by knowing only the equation of state and the heat capacity at constant volume C V . Note that Eq. (3) is valid for all gases, and that the temperature dependence of the three partial derivatives can be eliminated using the equation of state, so that they depend only on p and V . Equation (3) is one of the central results in this work and a starting point for further calculations.
Let us now parameterize the cycle in the p−V diagram. For a general cycle, we define the dimensionless pressure P(ϕ) and volume V(ϕ) as a function of a parameter ϕ in the following way:
p(ϕ) = p 0 P(ϕ), and V (ϕ) = V 0 V(ϕ),(4)
where p 0 and V 0 are constants characterizing the dimensions of p and V , and the parameter ϕ is a real number varying in a closed interval
[ϕ i , ϕ f ]. For a closed cycle, p(ϕ i ) = p(ϕ f ) and V (ϕ i ) = V (ϕ f ).
In this work, we consider only heat engines in which the cycle runs clockwise in the p − V diagram, i.e., the point (V(ϕ), P(ϕ)) on the curve of the cycle moves in the clockwise direction as ϕ increases. Note that a similar approach has been used in Ref. [13] for an unconventional lobe in the p − V diagram with an ideal gas and Ref. [14] for a Stirling engine where the volume of the gas is parameterized by the camshaft angle θ. However, the formalism presented below is valid for arbitrary cycles and gases. Since dp = p 0 (dP (ϕ)/ dϕ) dϕ and dV = V 0 (dV (ϕ)/ dϕ) dϕ, the exchanged heat in Eq. (3) along the curve is given bȳ
dQ = p 0 V 0 q(ϕ) dϕ , where (5a) q(ϕ) = C V V 0 ∂T ∂p V dP (ϕ) dϕ + + C V p 0 ∂T ∂V p + T p 0 ∂p ∂T V dV (ϕ) dϕ . (5b)
Note that to obtain the dimensionless heat function q(ϕ), it is enough to know the equation of state of the gas, the heat capacity C V and the parametrized curve of the cycle in the p−V diagram. When q(ϕ) > 0 the system absorbs heat, and when q(ϕ)  0 then q + = q and q − = 0, while for q  0. In a closed cycle, it follows from the first law of thermodynamics that Q + − Q − + W = 0, where Q + and Q − (both are positive numbers) are the total heat absorbed and released from the system, respectively, and −W is the work done by the engine (for clockwise direction in the p − V diagram, W  0 and b > 0 are constants characterizing the interactions and the size of the particles of a given gas. Introducing the dimensionless variablesV = V /V c ,p = p/p c , andT = T /T c the equation of state (19a) can be written in a universal form independent of the parameters of the gas:
p + 3 V 2 3V − 1 = 8T ,(20)
where V c = 3bn, p c = 1 27 a b 2 and RT c = 8 27 a b are the volume, pressure and the temperature at the critical point of the van der Waals gas. This is the law of corresponding states.
If we now choose the previously defined parameters in Eq. (4) as p 0 = p c and V 0 = V c we get P ≡p and V ≡V . Now, using Eqs. (5) and (19) we find that the heat function is
q(ϕ) = 1 Γ − 1 Γ P + 3 V 2 − 2 3V − 1 V 3 dV dϕ + + 3V − 1 3 dP dϕ ,(21)
where Γ = 1 + nR Cv . Note that for the ideal gas Γ = γ = C p /C V .
We now calculate the thermal efficiency and the heat function q(ϕ) of the heart-shaped curve defined in Eq. (16) using the van der Waals gas (see Fig. 4). The curve is chosen so that T > T c is always satisfied throughout the cycle. Using Γ ≈ 5/3 calculated from the heat capacity of argon, we get η = 0.1396. As it is expected the difference between the monoatomic ideal gas and a more realistic model for argon is very small above the critical temperature. 
V. CONCLUSIONS
We developed a general formalism to calculate the thermal efficiency of heat engines for arbitrary cycles defined in the p−V diagram. Our method can be applied to all gases provided their equation of state and their heat capacity are known. We derived a general and rather simple expression for the thermal efficiency. To demonstrate how versatile our formalism is, we calculated the thermal efficiency for a few rather non-trivial cycles. Finally, we derived the heat function for the van der Waals gas. A python code for calculating the thermal efficiency of the cycles presented in this work is available in the online supplemental information.
We believe that our general formalism provides a convenient and practical method to calculate the thermal efficiency of thermodynamic cycles defined on a p − V diagram, without the need to locate the adiabatic points. Moreover, from a practical point of view, our method can also be applied to model more general cycles, different from the ones usually described in textbooks. We hope that our work can be useful for physics teachers and students to get more insight into the calculation of thermal efficiency.
FIG. 1 .
1Heart-shaped curve in the p − V diagram. The parametric curve is given in the text (Eq. (
FIG. 2 .
2Circle-shaped cycle in the p − V diagram given by Eq. (11) with a = b = 1 (left panel) and the heat function q(ϕ) (right panel). Solid red lines are used where q(ϕ) > 0 and dotted blue lines where q(ϕ)
General formalism for calculating the thermal efficiency of thermodynamic
 cycles defined in a p–V diagram
https://arxiv.org/abs/2201.06956
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BSI-021: Thermodynamics: It's Not Rocket Science
2009-05-15
Joseph Lstiburek
<div><div><p>Of course the crazy way we speak about the Second Law does not help: “In an isolated system, a process can occur only if it increases the total entropy of the system.” Huh? It makes you want to hate Rudolf Clausius. Couldn’t he just say that heat goes from warm to cold?<a href="#Foot_01"><sup>1</sup></a> Most of us get the heat goes from warm to cold thing. It’s the other simple applications of the Second Law that we miss:</p><ul><li><p>moisture goes from warm to cold</p></li><li><p>moisture goes from more to less<a href="#Foot_02"><sup>2</sup></a></p></li><li><p>air goes from a higher pressure to a lower pressure</p></li><li><p>gravity acts down</p></li></ul><p>The two moisture “rules”—“warm to cold” and “more to less” are nothing more than a “thermal gradient” and a “concentration gradient.” If you combine the thermal gradient and the concentration gradient you get the “thermodynamic potential” and according to the Second Law water always follows the thermodynamic potential. The psychrometric chart is a graphical representation of “warm to cold” and “more to less”—thank you Willis Carrier.</p><p><a href="/sites/default/files/migrate/jpg/BSI021_Photo_01.jpg"></a><br /><strong>Photograph 1:</strong> <strong> Space Shuttle</strong>—One of the fundamental problems with the Space Shuttle is the shedding of foam from the external tank.</p><p>This “warm to cold” and “more to less” thing should be pretty easy. But seemingly smart folks get it wrong all the time. Take the Space Shuttle. The problem with the Space Shuttle is that foam keeps coming off of the external tank of the orbiter (<a href="#Photo_01"><strong>Photograph 1</strong></a>). The problem is best described thusly: we have an aluminum pressure vessel filled with a cryogenic liquid on a beach in Florida in July. The best analogy for this is <a href="#Photo_02"><strong>Photograph 2</strong></a>. The author does not have the research budget of NASA so a few liberties were taken with the authors’ experimental design defining the problem (<a href="#Photo_03"><strong>Photograph 3</strong></a> and <a href="#Photo_04"><strong>Photograph 4</strong></a>).</p><p><a href="/sites/default/files/migrate/jpg/BSI021_Photo_02.jpg"></a><strong><br />Photograph 2:  Engineering Analogy</strong>—“Aluminum pressure vessel filled with a cryogenic liquid on a beach in Florida in July.”</p><p><a href="/sites/default/files/migrate/jpg/BSI021_Photo_03.jpg"></a><strong><br />Photograph 3:  Limited Research Budget</strong>—“Richard Feynman eat your heart out…you only had a glass of ice water at the Challenger enquiry.”</p><p><a href="/sites/default/files/migrate/jpg/BSI021_Photo_04.jpg"></a><strong><br />Photograph 4:  Experimental Design</strong>—“Authors view of the Space Shuttle foam shedding problem…”</p><p><br />Most of us know intuitively that the Bud will sweat. And most of us know we need to insulate the Bud. But how many of us know that the insulation needs to control the vapor drive?  And what is the magnitude of the vapor drive anyway?  <a href="#Figure_01"><strong>Figure 1</strong></a> shows the boundary conditions. The surface of the foam goes to 150 degrees F. on a sunny day—it is brown in color after all.  In the summer it rains in Florida every day at 1 pm. So let us add a film of liquid phase water to the surface of the foam and wait for the sun to come back out. Now try to find where the saturation line (i.e. 100 percent RH) crosses the 150 degree F line on a pyschrometric chart. Nope. Not there. Can’t find it. Off the chart, so far off the chart we are in uncharted territory…Try the steam tables. Holy vapor pressure Batman, 13 kPa. Now, how about finding the vapor pressure at minus 420 degrees F?  This one is easy—you don’t need no stinkin’ chart for that…can we say “zero’?  We are talking some kind of serious inward vapor drive—13 to 0 kpa.</p><p><a href="/sites/default/files/migrate/jpg/BSI021_Figure_01.jpg"></a><br /><strong>Figure 1:  Boundary Conditions</strong>—A film of rainwater on the surface of solar heated foam insulation leads to enormous vapor drives.</p><p>Now ask the question that certain rocket scientists apparently have not asked: how resistant is the foam insulation on the external tank to vapor flow? The foam has a permeance of 3 perms per inch. The foam is 2 inches thick. Yup, that would make the resistance to flow 1.5 perms. I am no rocket scientist<a href="#Foot_03"><sup>3</sup></a> but that sounds kind of high permeance wise to me. Seems to me that water vapor will flow through the foam under that magnitude of vapor drive like brown organic matter through a goose. </p><p>I wonder what happens when the vapor hits the aluminum lining of the tank? It’s not just foam coming off the tank, but hunks of ice. A way of thinking of this is that we have a vapor barrier on the wrong side of the assembly. Seems to me that we need a vapor barrier on the outside of the assembly. Just for laughs go and “Google” the first shuttle launches and check out the pictures and ask yourself what looks different? Notice the “white” stuff on the external tank of these early launches—looks like a low perm aluminum skin to me. Apparently, the vapor control got “value engineered” out of the assembly after the first few launches to save weight. Violate the Second Law at your peril.</p><p>Let’s move on to FEMA and New Orleans and Katrina. It is hard to believe this if you have not seen this with your own eyes. Trailer after trailer after trailer (<a href="#Photo_05"><strong>Photograph 5</strong></a>). Where is New Orleans on the map? It is smack dab in the ASHRAE hot-humid climate region. Most of us intuitively know that this means it is pretty much hot and humid outside almost all the time in New Orleans. Here is the tricky part—we air condition the inside in New Orleans. Yes, it is true. We make the inside colder and drier than the outside in New Orleans. So, work with me on this…the outside is hot and humid and the inside is cold and dry. It is that pesky Second Law again.  What direction is the vapor drive? We can actually use a pyschrometric chart for this example and not have to go to the steam tables—so it is not as hard as NASA’s problem.</p><p><a href="/sites/default/files/migrate/jpg/BSI021_Photo_05.jpg"></a><strong><br />Photograph 5:  FEMA Trailers</strong>—Thousands and thousands of these are located in hot-humid climates.</p><p>With a dominant inward vapor drive (there is not much of a winter in New Orleans) what is the last thing you would want to see on the inside of an assembly in New Orleans? Yes, that is right, a vapor barrier. You absolutely positively would not ever want to see a vapor barrier on the inside of an air-conditioned assembly in New Orleans. Yes, you guessed it. All FEMA trailers have an interior vapor barrier (<a href="#Photo_06"><strong>Photograph 6</strong></a>). The FEMA trailers are lined with gypsum board wrapped with vinyl facings. And they are turning the trailers into moldy stinky messes. And it is not like FEMA wasn’t warned. They and HUD and the mobile home/manufactured home industry all blindly continue to believe that this is not a problem. But hey, it is poor people in these trailers, who is going to notice?</p><p><a href="/sites/default/files/migrate/jpg/BSI021_Photo_06.jpg"></a><strong><br />Photograph 6:  Vinyl Covered Gypsum Board</strong>—FEMA trailers are lined with vinyl-covered gypsum board as are HUD manufactured houses. This leads to mold problems. The only solution is to not install vinyl covered gypsum board.</p><p>But this is America, we can abuse rich people too—especially the ones that stay in Michelin Five Star rated hotels. It is not well known, but in order to get a Michelin Five Star rating a hotel has to have interior vinyl wall coverings. I’m not kidding, this is true. For real. The vinyl wall coverings are associated with “higher quality.” Now for the good part. Vinyl is a vapor barrier. Installing vinyl wall coverings on the inside of air-conditioned hotels in hot humid climates results in mold. See FEMA trailers above.</p><p>One of the first signs are “pink spots” on the surface of the vinyl wall coverings (<a href="#Photo_07"><strong>Photograph 7</strong></a>). Molds growing on the back of the vinyl (<a href="#Photo_08"><strong>Photograph 8</strong></a>) exude digestive enzymes that react with the plasticizers in the vinyl leading to the discoloration. These spots are typically referred to as the Marriott measles, the Radisson rash, and the Hyatt hives. Once again the Second Law asserts itself. When I am asked to work on fixing these expensive hotel properties I never stay in them because of the mold, I always stay in La Quinta Inns<a href="#Foot_04"><sup>4</sup></a> since their “brand look” is “mission style” and has no interior vinyl. </p><p><a href="/sites/default/files/migrate/jpg/BSI021_Photo_07.jpg"></a><strong><br />Photograph 7:  Vinyl Wall Coverings and Hotels</strong>—Marriott measles, the Radisson rash, and the Hyatt hives.</p><p><a href="/sites/default/files/migrate/jpg/BSI021_Photo_08.jpg"></a><strong><br />Photograph 8: Hotel Mold</strong>—Vinyl wall coverings are low perm materials that retard inward vapor movement and lead<br />to mold growth.</p><p>One of the neat things about the Second Law is that you can magnify the effects with sun and storage and turn what most folks consider a hot humid climate problem into a problem everywhere in the Lower 48 States. Problems with reservoir claddings are making homebuilders forget about sub-prime mortgages. Brick, stucco and fiber-cement siding are all claddings that store water after rain exposure—they act as “reservoirs” that are “charged” during a rain event. It is pretty obvious that reservoir claddings can store a significant amount of water at the exterior of a wall assembly. </p><p><a href="/sites/default/files/migrate/jpg/BSI021_Figure_02.jpg"></a></p><p><strong><br />Figure 2:</strong> Brick “reservoir” is charged during a rain event and “discharges” both inward and outward magnifying the vapor drives.</p><p>Incident solar radiation heats the water stored in the cladding creating a huge vapor drive that is both inward and outward (<a href="#Figure_02"><strong>Figure 2</strong></a>). The vapor drive inward out of a solar heated rain wetted reservoir cladding located in a building in Chicago in July dwarfs by several hundred percent the outward vapor drive in a humidified building located in Barrow, Alaska in January when it is minus 40.<a href="#Foot_05"><sup>5</sup></a> This leads to summer condensation on the cavity side of polyethylene vapor barriers in places like Chicago (<a href="#Photo_09"><strong>Photograph 9</strong></a>) and Minneapolis (<a href="#Photo_10"><strong>Photograph 10</strong></a>). </p><p><a href="/sites/default/files/migrate/jpg/BSI021_Photo_09.jpg"></a><strong><br />Photograph 9:  Condensation on Polyethylene Vapor Barriers</strong>—Solar driven moisture out of reservoir claddings can lead to cavity side condensation on polyethylene vapor barriers in cold climates such as Chicago.</p><p><a href="/sites/default/files/migrate/jpg/BSI021_Photo_10.jpg"></a><strong><br />Photograph 10: Minneapolis Misery</strong>—Pat Huelman of the University of Minnesota gets a whiff of mold behind a plastic vapor barrier in a Minneapolis house with a plastic vapor barrier.</p><p>Plastic vapor barriers are a “no-no” on the inside of wall assemblies with reservoir claddings everywhere in the Lower 48 States. Many old timers intuitively knew this and were passionate about using kraft-faced batts rather than polyethylene. Kraft-faced batts act as vapor “retarders” rather than vapor “barriers” in cold climate wall assemblies (<a href="#Photo_11"><strong>Photograph 11</strong></a>). The kraft-faced batts allowed the wall assemblies to dry inwards during the summer especially when the assembly was “stressed” by inward vapor drive out of a reservoir cladding. It appears that mostly the young got “sucked in” to the plastic vapor barrier craze. </p><p><a href="#Photo_11"></a><strong><br />Photograph 11: Kraft-Faced Batts</strong>—Bless old timers and common sense—the kraft facing breathes during the summer allowing inward drying—polyethylene vapor barriers do not.</p><p>If you think about it, if NASA couldn’t get the moisture goes from warm to cold and more to less thing right, then it should not be surprising that FEMA, HUD, the hotel industry and homebuilders couldn’t either. We seem to forget that there are two sides to a wall and that there are several seasons—two of which are summer and winter and that means that vapor drives go both ways. Once again the Second Law strikes.</p><hr /><h3>Footnotes:</h3><ol><li><p><a></a>Actually he kind of said it—Clausius said “Heat cannot of itself pass from a colder to a hotter body.”  I am not really picking on him—Clausius’ work on the Second Law is probably one of the greatest gifts of any one individual to the human race….right up there with Newton and Einstein.  But how many people get it?  Probably about as many people as get Einstein’s Theory of Relativity.  A few more folks do understand Newton—that calculus thing he did is pretty cool.  Yes, yes, my German friends, I know about Leibniz, but you already have two of the three folks on my list of important dudes in this footnote…</p></li><li><p><a></a>Thermal diffusion and molecular diffusion are distinct driving forces—so both of these statements are actually true—but separating them is typically not of any practical significance except for reasons that follow.  In almost all practical applications in building science “warm to cold” is also in the same direction of “more to less.” It is typically easier to sense or measure temperature so that as a first approximation you can use temperature as your guide to understanding the problem rather than the concentration gradient. Every now and then “more to less” is in the opposite direction as “warm to cold”—under such conditions “more to less” usually wins…rock beats scissors, paper beats rock…more to less usually beats warm to cold.” Once again the “engineer” in me is “approximately correct.” Or just go look at a psychrometric chart—but that is not strictly true either…</p></li><li><p><a></a>Well, not any more—my undergraduate degree was in aerospace engineering—then I got a personality and became a civil engineer.</p></li><li><p><a></a>La Quinta means “near Denny’s” in Spanish.</p></li><li><p><a></a>At this temperature it does not matter whether it is Fahrenheit or Celsius because at minus 40 both scales cross—minus 40 F equals minus 40 C—we refer to that as minus 40 FC for “freaking cold.”</p></li></ol></div>||||I||||Skip to main content

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 BSI-021: Thermodynamics: It's Not Rocket Science

 Joseph Lstiburek
 Effective Date
 May 15, 2009
 Text

 Of course the crazy way we speak about the Second Law does not help: “In an isolated system, a process can occur only if it increases the total entropy of the system.” Huh? It makes you want to hate Rudolf Clausius. Couldn’t he just say that heat goes from warm to cold?1 Most of us get the heat goes from warm to cold thing. It’s the other simple applications of the Second Law that we miss:

 * moisture goes from warm to cold

 * moisture goes from more to less2

 * air goes from a higher pressure to a lower pressure

 * gravity acts down

 The two moisture “rules”—“warm to cold” and “more to less” are nothing more than a “thermal gradient” and a “concentration gradient.” If you combine the thermal gradient and the concentration gradient you get the “thermodynamic potential” and according to the Second Law water always follows the thermodynamic potential. The psychrometric chart is a graphical representation of “warm to cold” and “more to less”—thank you Willis Carrier.


 Photograph 1: Space Shuttle—One of the fundamental problems with the Space Shuttle is the shedding of foam from the external tank.

 This “warm to cold” and “more to less” thing should be pretty easy. But seemingly smart folks get it wrong all the time. Take the Space Shuttle. The problem with the Space Shuttle is that foam keeps coming off of the external tank of the orbiter (Photograph 1). The problem is best described thusly: we have an aluminum pressure vessel filled with a cryogenic liquid on a beach in Florida in July. The best analogy for this is Photograph 2. The author does not have the research budget of NASA so a few liberties were taken with the authors’ experimental design defining the problem (Photograph 3 and Photograph 4).


 Photograph 2: Engineering Analogy—“Aluminum pressure vessel filled with a cryogenic liquid on a beach in Florida in July.”


 Photograph 3: Limited Research Budget—“Richard Feynman eat your heart out…you only had a glass of ice water at the Challenger enquiry.”


 Photograph 4: Experimental Design—“Authors view of the Space Shuttle foam shedding problem…”


 Most of us know intuitively that the Bud will sweat. And most of us know we need to insulate the Bud. But how many of us know that the insulation needs to control the vapor drive? And what is the magnitude of the vapor drive anyway? Figure 1 shows the boundary conditions. The surface of the foam goes to 150 degrees F. on a sunny day—it is brown in color after all. In the summer it rains in Florida every day at 1 pm. So let us add a film of liquid phase water to the surface of the foam and wait for the sun to come back out. Now try to find where the saturation line (i.e. 100 percent RH) crosses the 150 degree F line on a pyschrometric chart. Nope. Not there. Can’t find it. Off the chart, so far off the chart we are in uncharted territory…Try the steam tables. Holy vapor pressure Batman, 13 kPa. Now, how about finding the vapor pressure at minus 420 degrees F? This one is easy—you don’t need no stinkin’ chart for that…can we say “zero’? We are talking some kind of serious inward vapor drive—13 to 0 kpa.


 Figure 1: Boundary Conditions—A film of rainwater on the surface of solar heated foam insulation leads to enormous vapor drives.

 Now ask the question that certain rocket scientists apparently have not asked: how resistant is the foam insulation on the external tank to vapor flow? The foam has a permeance of 3 perms per inch. The foam is 2 inches thick. Yup, that would make the resistance to flow 1.5 perms. I am no rocket scientist3 but that sounds kind of high permeance wise to me. Seems to me that water vapor will flow through the foam under that magnitude of vapor drive like brown organic matter through a goose.

 I wonder what happens when the vapor hits the aluminum lining of the tank? It’s not just foam coming off the tank, but hunks of ice. A way of thinking of this is that we have a vapor barrier on the wrong side of the assembly. Seems to me that we need a vapor barrier on the outside of the assembly. Just for laughs go and “Google” the first shuttle launches and check out the pictures and ask yourself what looks different? Notice the “white” stuff on the external tank of these early launches—looks like a low perm aluminum skin to me. Apparently, the vapor control got “value engineered” out of the assembly after the first few launches to save weight. Violate the Second Law at your peril.

 Let’s move on to FEMA and New Orleans and Katrina. It is hard to believe this if you have not seen this with your own eyes. Trailer after trailer after trailer (Photograph 5). Where is New Orleans on the map? It is smack dab in the ASHRAE hot-humid climate region. Most of us intuitively know that this means it is pretty much hot and humid outside almost all the time in New Orleans. Here is the tricky part—we air condition the inside in New Orleans. Yes, it is true. We make the inside colder and drier than the outside in New Orleans. So, work with me on this…the outside is hot and humid and the inside is cold and dry. It is that pesky Second Law again. What direction is the vapor drive? We can actually use a pyschrometric chart for this example and not have to go to the steam tables—so it is not as hard as NASA’s problem.


 Photograph 5: FEMA Trailers—Thousands and thousands of these are located in hot-humid climates.

 With a dominant inward vapor drive (there is not much of a winter in New Orleans) what is the last thing you would want to see on the inside of an assembly in New Orleans? Yes, that is right, a vapor barrier. You absolutely positively would not ever want to see a vapor barrier on the inside of an air-conditioned assembly in New Orleans. Yes, you guessed it. All FEMA trailers have an interior vapor barrier (Photograph 6). The FEMA trailers are lined with gypsum board wrapped with vinyl facings. And they are turning the trailers into moldy stinky messes. And it is not like FEMA wasn’t warned. They and HUD and the mobile home/manufactured home industry all blindly continue to believe that this is not a problem. But hey, it is poor people in these trailers, who is going to notice?


 Photograph 6: Vinyl Covered Gypsum Board—FEMA trailers are lined with vinyl-covered gypsum board as are HUD manufactured houses. This leads to mold problems. The only solution is to not install vinyl covered gypsum board.

 But this is America, we can abuse rich people too—especially the ones that stay in Michelin Five Star rated hotels. It is not well known, but in order to get a Michelin Five Star rating a hotel has to have interior vinyl wall coverings. I’m not kidding, this is true. For real. The vinyl wall coverings are associated with “higher quality.” Now for the good part. Vinyl is a vapor barrier. Installing vinyl wall coverings on the inside of air-conditioned hotels in hot humid climates results in mold. See FEMA trailers above.

 One of the first signs are “pink spots” on the surface of the vinyl wall coverings (Photograph 7). Molds growing on the back of the vinyl (Photograph 8) exude digestive enzymes that react with the plasticizers in the vinyl leading to the discoloration. These spots are typically referred to as the Marriott measles, the Radisson rash, and the Hyatt hives. Once again the Second Law asserts itself. When I am asked to work on fixing these expensive hotel properties I never stay in them because of the mold, I always stay in La Quinta Inns4 since their “brand look” is “mission style” and has no interior vinyl.


 Photograph 7: Vinyl Wall Coverings and Hotels—Marriott measles, the Radisson rash, and the Hyatt hives.


 Photograph 8: Hotel Mold—Vinyl wall coverings are low perm materials that retard inward vapor movement and lead
 to mold growth.

 One of the neat things about the Second Law is that you can magnify the effects with sun and storage and turn what most folks consider a hot humid climate problem into a problem everywhere in the Lower 48 States. Problems with reservoir claddings are making homebuilders forget about sub-prime mortgages. Brick, stucco and fiber-cement siding are all claddings that store water after rain exposure—they act as “reservoirs” that are “charged” during a rain event. It is pretty obvious that reservoir claddings can store a significant amount of water at the exterior of a wall assembly.


 Figure 2: Brick “reservoir” is charged during a rain event and “discharges” both inward and outward magnifying the vapor drives.

 Incident solar radiation heats the water stored in the cladding creating a huge vapor drive that is both inward and outward (Figure 2). The vapor drive inward out of a solar heated rain wetted reservoir cladding located in a building in Chicago in July dwarfs by several hundred percent the outward vapor drive in a humidified building located in Barrow, Alaska in January when it is minus 40.5 This leads to summer condensation on the cavity side of polyethylene vapor barriers in places like Chicago (Photograph 9) and Minneapolis (Photograph 10).


 Photograph 9: Condensation on Polyethylene Vapor Barriers—Solar driven moisture out of reservoir claddings can lead to cavity side condensation on polyethylene vapor barriers in cold climates such as Chicago.


 Photograph 10: Minneapolis Misery—Pat Huelman of the University of Minnesota gets a whiff of mold behind a plastic vapor barrier in a Minneapolis house with a plastic vapor barrier.

 Plastic vapor barriers are a “no-no” on the inside of wall assemblies with reservoir claddings everywhere in the Lower 48 States. Many old timers intuitively knew this and were passionate about using kraft-faced batts rather than polyethylene. Kraft-faced batts act as vapor “retarders” rather than vapor “barriers” in cold climate wall assemblies (Photograph 11). The kraft-faced batts allowed the wall assemblies to dry inwards during the summer especially when the assembly was “stressed” by inward vapor drive out of a reservoir cladding. It appears that mostly the young got “sucked in” to the plastic vapor barrier craze.


 Photograph 11: Kraft-Faced Batts—Bless old timers and common sense—the kraft facing breathes during the summer allowing inward drying—polyethylene vapor barriers do not.

 If you think about it, if NASA couldn’t get the moisture goes from warm to cold and more to less thing right, then it should not be surprising that FEMA, HUD, the hotel industry and homebuilders couldn’t either. We seem to forget that there are two sides to a wall and that there are several seasons—two of which are summer and winter and that means that vapor drives go both ways. Once again the Second Law strikes.

 Footnotes:

 1. Actually he kind of said it—Clausius said “Heat cannot of itself pass from a colder to a hotter body.” I am not really picking on him—Clausius’ work on the Second Law is probably one of the greatest gifts of any one individual to the human race….right up there with Newton and Einstein. But how many people get it? Probably about as many people as get Einstein’s Theory of Relativity. A few more folks do understand Newton—that calculus thing he did is pretty cool. Yes, yes, my German friends, I know about Leibniz, but you already have two of the three folks on my list of important dudes in this footnote…

 2. Thermal diffusion and molecular diffusion are distinct driving forces—so both of these statements are actually true—but separating them is typically not of any practical significance except for reasons that follow. In almost all practical applications in building science “warm to cold” is also in the same direction of “more to less.” It is typically easier to sense or measure temperature so that as a first approximation you can use temperature as your guide to understanding the problem rather than the concentration gradient. Every now and then “more to less” is in the opposite direction as “warm to cold”—under such conditions “more to less” usually wins…rock beats scissors, paper beats rock…more to less usually beats warm to cold.” Once again the “engineer” in me is “approximately correct.” Or just go look at a psychrometric chart—but that is not strictly true either…

 3. Well, not any more—my undergraduate degree was in aerospace engineering—then I got a personality and became a civil engineer.

 4. La Quinta means “near Denny’s” in Spanish.

 5. At this temperature it does not matter whether it is Fahrenheit or Celsius because at minus 40 both scales cross—minus 40 F equals minus 40 C—we refer to that as minus 40 FC for “freaking cold.”

 Related Documents

 * BSD-109: Pressures in Buildings
 * BSD-138: Moisture and Materials
 * BSI-005: A Bridge Too Far
 * RR-0002: The Thermal Metric Project
 * BA-1204: External Insulation of Masonry Walls and Wood Framed Walls
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Of course the crazy way we speak about the Second Law does not help: “In an isolated system, a process can occur only if it increases the total entropy of the system.” Huh? It makes you want to hate Rudolf Clausius. Couldn’t he just say that heat goes from warm to cold? 1  Most of us get the heat goes from warm to cold thing. It’s the other simple applications of the Second Law that we miss:   moisture goes from warm to cold   moisture goes from more to less 2     air goes from a higher pressure to a lower pressure   gravity acts down   The two moisture “rules”—“warm to cold” and “more to less” are nothing more than a “thermal gradient” and a “concentration gradient.” If you combine the thermal gradient and the concentration gradient you get the “thermodynamic potential” and according to the Second Law water always follows the thermodynamic potential. The psychrometric chart is a graphical representation of “warm to cold” and “more to less”—thank you Willis Carrier.     Photograph 1:  Space Shuttle—One of the fundamental problems with the Space Shuttle is the shedding of foam from the external tank. This “warm to cold” and “more to less” thing should be pretty easy. But seemingly smart folks get it wrong all the time. Take the Space Shuttle. The problem with the Space Shuttle is that foam keeps coming off of the external tank of the orbiter ( Photograph 1 ). The problem is best described thusly: we have an aluminum pressure vessel filled with a cryogenic liquid on a beach in Florida in July. The best analogy for this is  Photograph 2 . The author does not have the research budget of NASA so a few liberties were taken with the authors’ experimental design defining the problem ( Photograph 3  and  Photograph 4 ).     Photograph 2:  Engineering Analogy—“Aluminum pressure vessel filled with a cryogenic liquid on a beach in Florida in July.”     Photograph 3:  Limited Research Budget—“Richard Feynman eat your heart out…you only had a glass of ice water at the Challenger enquiry.”     Photograph 4:  Experimental Design—“Authors view of the Space Shuttle foam shedding problem…”  Most of us know intuitively that the Bud will sweat. And most of us know we need to insulate the Bud. But how many of us know that the insulation needs to control the vapor drive?  And what is the magnitude of the vapor drive anyway?   Figure 1  shows the boundary conditions. The surface of the foam goes to 150 degrees F. on a sunny day—it is brown in color after all.  In the summer it rains in Florida every day at 1 pm. So let us add a film of liquid phase water to the surface of the foam and wait for the sun to come back out. Now try to find where the saturation line (i.e. 100 percent RH) crosses the 150 degree F line on a pyschrometric chart. Nope. Not there. Can’t find it. Off the chart, so far off the chart we are in uncharted territory…Try the steam tables. Holy vapor pressure Batman, 13 kPa. Now, how about finding the vapor pressure at minus 420 degrees F?  This one is easy—you don’t need no stinkin’ chart for that…can we say “zero’?  We are talking some kind of serious inward vapor drive—13 to 0 kpa.     Figure 1:  Boundary Conditions—A film of rainwater on the surface of solar heated foam insulation leads to enormous vapor drives. Now ask the question that certain rocket scientists apparently have not asked: how resistant is the foam insulation on the external tank to vapor flow? The foam has a permeance of 3 perms per inch. The foam is 2 inches thick. Yup, that would make the resistance to flow 1.5 perms. I am no rocket scientist 3  but that sounds kind of high permeance wise to me. Seems to me that water vapor will flow through the foam under that magnitude of vapor drive like brown organic matter through a goose.  I wonder what happens when the vapor hits the aluminum lining of the tank? It’s not just foam coming off the tank, but hunks of ice. A way of thinking of this is that we have a vapor barrier on the wrong side of the assembly. Seems to me that we need a vapor barrier on the outside of the assembly. Just for laughs go and “Google” the first shuttle launches and check out the pictures and ask yourself what looks different? Notice the “white” stuff on the external tank of these early launches—looks like a low perm aluminum skin to me. Apparently, the vapor control got “value engineered” out of the assembly after the first few launches to save weight. Violate the Second Law at your peril. Let’s move on to FEMA and New Orleans and Katrina. It is hard to believe this if you have not seen this with your own eyes. Trailer after trailer after trailer ( Photograph 5 ). Where is New Orleans on the map? It is smack dab in the ASHRAE hot-humid climate region. Most of us intuitively know that this means it is pretty much hot and humid outside almost all the time in New Orleans. Here is the tricky part—we air condition the inside in New Orleans. Yes, it is true. We make the inside colder and drier than the outside in New Orleans. So, work with me on this…the outside is hot and humid and the inside is cold and dry. It is that pesky Second Law again.  What direction is the vapor drive? We can actually use a pyschrometric chart for this example and not have to go to the steam tables—so it is not as hard as NASA’s problem.     Photograph 5:  FEMA Trailers—Thousands and thousands of these are located in hot-humid climates. With a dominant inward vapor drive (there is not much of a winter in New Orleans) what is the last thing you would want to see on the inside of an assembly in New Orleans? Yes, that is right, a vapor barrier. You absolutely positively would not ever want to see a vapor barrier on the inside of an air-conditioned assembly in New Orleans. Yes, you guessed it. All FEMA trailers have an interior vapor barrier ( Photograph 6 ). The FEMA trailers are lined with gypsum board wrapped with vinyl facings. And they are turning the trailers into moldy stinky messes. And it is not like FEMA wasn’t warned. They and HUD and the mobile home/manufactured home industry all blindly continue to believe that this is not a problem. But hey, it is poor people in these trailers, who is going to notice?     Photograph 6:  Vinyl Covered Gypsum Board—FEMA trailers are lined with vinyl-covered gypsum board as are HUD manufactured houses. This leads to mold problems. The only solution is to not install vinyl covered gypsum board. But this is America, we can abuse rich people too—especially the ones that stay in Michelin Five Star rated hotels. It is not well known, but in order to get a Michelin Five Star rating a hotel has to have interior vinyl wall coverings. I’m not kidding, this is true. For real. The vinyl wall coverings are associated with “higher quality.” Now for the good part. Vinyl is a vapor barrier. Installing vinyl wall coverings on the inside of air-conditioned hotels in hot humid climates results in mold. See FEMA trailers above. One of the first signs are “pink spots” on the surface of the vinyl wall coverings ( Photograph 7 ). Molds growing on the back of the vinyl ( Photograph 8 ) exude digestive enzymes that react with the plasticizers in the vinyl leading to the discoloration. These spots are typically referred to as the Marriott measles, the Radisson rash, and the Hyatt hives. Once again the Second Law asserts itself. When I am asked to work on fixing these expensive hotel properties I never stay in them because of the mold, I always stay in La Quinta Inns 4  since their “brand look” is “mission style” and has no interior vinyl.      Photograph 7:  Vinyl Wall Coverings and Hotels—Marriott measles, the Radisson rash, and the Hyatt hives.     Photograph 8: Hotel Mold—Vinyl wall coverings are low perm materials that retard inward vapor movement and leadto mold growth. One of the neat things about the Second Law is that you can magnify the effects with sun and storage and turn what most folks consider a hot humid climate problem into a problem everywhere in the Lower 48 States. Problems with reservoir claddings are making homebuilders forget about sub-prime mortgages. Brick, stucco and fiber-cement siding are all claddings that store water after rain exposure—they act as “reservoirs” that are “charged” during a rain event. It is pretty obvious that reservoir claddings can store a significant amount of water at the exterior of a wall assembly.        Figure 2: Brick “reservoir” is charged during a rain event and “discharges” both inward and outward magnifying the vapor drives. Incident solar radiation heats the water stored in the cladding creating a huge vapor drive that is both inward and outward ( Figure 2 ). The vapor drive inward out of a solar heated rain wetted reservoir cladding located in a building in Chicago in July dwarfs by several hundred percent the outward vapor drive in a humidified building located in Barrow, Alaska in January when it is minus 40. 5  This leads to summer condensation on the cavity side of polyethylene vapor barriers in places like Chicago ( Photograph 9 ) and Minneapolis ( Photograph 10 ).      Photograph 9:  Condensation on Polyethylene Vapor Barriers—Solar driven moisture out of reservoir claddings can lead to cavity side condensation on polyethylene vapor barriers in cold climates such as Chicago.     Photograph 10: Minneapolis Misery—Pat Huelman of the University of Minnesota gets a whiff of mold behind a plastic vapor barrier in a Minneapolis house with a plastic vapor barrier. Plastic vapor barriers are a “no-no” on the inside of wall assemblies with reservoir claddings everywhere in the Lower 48 States. Many old timers intuitively knew this and were passionate about using kraft-faced batts rather than polyethylene. Kraft-faced batts act as vapor “retarders” rather than vapor “barriers” in cold climate wall assemblies ( Photograph 11 ). The kraft-faced batts allowed the wall assemblies to dry inwards during the summer especially when the assembly was “stressed” by inward vapor drive out of a reservoir cladding. It appears that mostly the young got “sucked in” to the plastic vapor barrier craze.      Photograph 11: Kraft-Faced Batts—Bless old timers and common sense—the kraft facing breathes during the summer allowing inward drying—polyethylene vapor barriers do not. If you think about it, if NASA couldn’t get the moisture goes from warm to cold and more to less thing right, then it should not be surprising that FEMA, HUD, the hotel industry and homebuilders couldn’t either. We seem to forget that there are two sides to a wall and that there are several seasons—two of which are summer and winter and that means that vapor drives go both ways. Once again the Second Law strikes.  Footnotes:     Actually he kind of said it—Clausius said “Heat cannot of itself pass from a colder to a hotter body.”  I am not really picking on him—Clausius’ work on the Second Law is probably one of the greatest gifts of any one individual to the human race….right up there with Newton and Einstein.  But how many people get it?  Probably about as many people as get Einstein’s Theory of Relativity.  A few more folks do understand Newton—that calculus thing he did is pretty cool.  Yes, yes, my German friends, I know about Leibniz, but you already have two of the three folks on my list of important dudes in this footnote…     Thermal diffusion and molecular diffusion are distinct driving forces—so both of these statements are actually true—but separating them is typically not of any practical significance except for reasons that follow.  In almost all practical applications in building science “warm to cold” is also in the same direction of “more to less.” It is typically easier to sense or measure temperature so that as a first approximation you can use temperature as your guide to understanding the problem rather than the concentration gradient. Every now and then “more to less” is in the opposite direction as “warm to cold”—under such conditions “more to less” usually wins…rock beats scissors, paper beats rock…more to less usually beats warm to cold.” Once again the “engineer” in me is “approximately correct.” Or just go look at a psychrometric chart—but that is not strictly true either…     Well, not any more—my undergraduate degree was in aerospace engineering—then I got a personality and became a civil engineer.     La Quinta means “near Denny’s” in Spanish.     At this temperature it does not matter whether it is Fahrenheit or Celsius because at minus 40 both scales cross—minus 40 F equals minus 40 C—we refer to that as minus 40 FC for “freaking cold.”
BSI-021: Thermodynamics: It's Not Rocket Science
https://www.buildingscience.com/documents/insights/bsi-021-thermodynamics-its-not-rocket-science