# Question #ff3bb

Mar 7, 2015

The temperature will be 356 K.

So, you know that you must use the Clausius-Clapeyron equation. Now, you'll find this equation written is several equivalent forms, so I'll just choose one of these forms

$\ln \left({P}_{1} / {P}_{2}\right) = \frac{\Delta {H}_{\text{vap}}}{R} \cdot \left(\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right)$, where

${P}_{1}$ - the vapor pressure measured at ${T}_{1}$;
${P}_{2}$ - the vapor pressure measured at ${P}_{2}$;
$\Delta {H}_{\text{vap}}$ - the enthalpy of vaporization;
$R$ - the gas constant - expressed in Joules per mol K;

You have everything you need to solve for ${T}_{2}$. Since the pressure measured at this new temperature will be 5.00 times bigger than ${P}_{1}$, you can write it as ${P}_{2} = 5 \cdot {P}_{1}$ and use it in this form in the equation.

So, plug all in and you'll get

$\ln \left({P}_{1} / \left(5 \cdot {P}_{1}\right)\right) = \left(46340 \text{J"/"mol")/(8.31446"J"/("mol" * "K")) * (1/T_2 - 1/"323 K}\right)$

$\ln \left(\frac{1}{5}\right) = \text{5573.4" * 1/T_2 - "5573.4" * 1/"323}$

$- 1.6094 = \frac{\text{5573.4}}{T} _ 2 - 17.2252$

$15.646 = \text{5573.4"/T_2 => T_2 = 5573.4/15.646 = "356.2 K}$

Rounded to three sig figs, the answer will be

${T}_{2} = \text{356 K}$