# Question #f7a41

Apr 28, 2016

The temperature dependence of vapor pressure, or any other equilibrium constant, depends on the $\Delta H$ for that process, but we also need a reference point, which is the normal boiling point of water.

#### Explanation:

The ratio of vapor pressures at two different temperatures, ${T}_{1}$ and ${T}_{2}$ can be calculated approximately from the relation

$\ln \left(\frac{{P}_{2}}{{P}_{1}}\right) = \frac{\Delta {H}_{v a p}}{R} \left(\frac{1}{{T}_{1}} - \frac{1}{{T}_{2}}\right)$

Using ${T}_{1} = 373.15 K$ (100 C), ${P}_{1} = 1$ atm,
and ${T}_{2} = 310.15 K$ (37 C)

$\ln \left(\frac{{P}_{2}}{1}\right) = \frac{40700 \frac{J}{m o l}}{8.314 \frac{J}{m o l - K}} \left(\frac{1}{373.15 K} - \frac{1}{310.15 K}\right) = - 2.66$

${P}_{2} = {e}^{- 2.66}$ atm = 0.0696 atm = 53.2 torr

The first equation ignores the (slight) temperature dependence of $\Delta {H}_{v a p}$, but the true vapor pressure is only slightly less: 47.1 torr.