 The mass loss $\Delta m$ in an interval $\Delta t$ is related to the collision flux by
$$
\Delta m=Z_{\mathrm{W}} A_0 m \Delta t
$$
where $A_0$ is the area of the hole and $m$ is the mass of one atom. It follows that
$$
Z_{\mathrm{W}}=\frac{\Delta m}{A_0 m \Delta t}
$$
Because $Z_{\mathrm{W}}$ is related to the pressure by eqn 20.14 , we can write
$$
p=\left(\frac{2 \pi R T}{M}\right)^{1 / 2} \frac{\Delta m}{A_0 \Delta t}
$$
Because $M=132.9 \mathrm{~g} \mathrm{~mol}^{-1}$, substitution of the data gives $p=8.7 \mathrm{kPa}$ (using $1 \mathrm{~Pa}=$ $\left.1 \mathrm{~N} \mathrm{~m}^{-2}=1 \mathrm{~J} \mathrm{~m}^{-1}\right)$.
Caesium (m.p. $29^{\circ} \mathrm{C}$, b.p. $686^{\circ} \mathrm{C}$ ) was introduced into a container and heated to $500^{\circ} \mathrm{C}$. When a hole of diameter $0.50 \mathrm{~mm}$ was opened in the container for $100 \mathrm{~s}$, a mass loss of $385 \mathrm{mg}$ was measured. Calculate the vapour pressure of liquid caesium at $500 \mathrm{~K}$.