# How do you find vapor pressure given boiling point and heat of vaporization?

Jan 17, 2015

You do that by using the Clausius-Clapeyron equation, which allows you to estimate the vapor pressure at another temperature if the vapor pressure for that substance is known at some temperature, provided that you also know the enthalpy of vaporization.

The most useful form of the Clausius-Clapeyron is

ln(P_2/P_1) = -(DeltaH_("vap"))/"R" * (1/T_2 - 1/T_1), where

${P}_{1}$ - the vapor pressure of that substance at ${T}_{1}$;
${P}_{2}$ - the vapor pressure of that substance at ${T}_{2}$;
$\Delta {H}_{\text{vap}}$ - the enthalpy of vaporization.
$\text{R}$ - the gas constant - usually expressed as $8.314$ $\text{J/K" * "mol}$.

Now, it is important to know that under normal conditions, everything boils at 1 atm. A liquid boils when its vapor pressure equals the exterior pressure, and since we're at 1 atm, that is what the vapor pressure for a substance at its boiling point will be.

Here's an example of how to apply this euation.

The vapor pressure for a substance at ${34.9}^{\circ} \text{C}$ is 115 torr. Its enthalpy of vaporization is $\text{40.5 kJ/mol}$. What is its boling temperature?

SInce no mention is made about pressure, we'll assume we're at 1 atm. This means that the vapor pressure of this substance at boiling point is 760 torr, the equivalent of 1 atm. So,

$\ln \left(\frac{760}{115}\right) = - \frac{40.5 \cdot {10}^{3} \frac{J}{m o l}}{8.314 \frac{J}{K \cdot m o l}} \cdot \left(\frac{1}{T} _ 2 - \frac{1}{34.9 + 273.15}\right)$

$1.89 = \frac{- 4.87 \cdot {10}^{3}}{T} _ 2 + 15.8 \implies {T}_{2} = \frac{- 4.87 \cdot {10}^{3}}{-} 13.91 = 350 K$

Under normal conditions, this substance boils at $350$ $\text{K}$.