Question

A certain substance has a heat of vaporization of 47.70 kJ/mol. At what kelvin temperature will the vapor pressure be 3.50 times higher than it was at 293k?=______K

Answer 1

Answer:

$T_2 = 313\ K$

Explanation:

he expression for Clausius-Clapeyron Equation is shown below as:

$\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c$

Where,  

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

$\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)$

Given:

$P_2$ = 3.50 $P_1$

$T_1$ = 293 K

$\Delta \:H_{vap}=47.70\ kJ/mol$

So,  

$\ln \left( \dfrac{P_1}{3.50\times P_1} \right) = \dfrac{47.70}{8.314\times 10^{-3}} \left( \dfrac{1}{T_2}- \dfrac{1}{293} \right)$

$\ln \left( \dfrac{1}{3.50} \right) = \dfrac{47.70}{8.314\times 10^{-3}} \left( \dfrac{1}{T_2}- \dfrac{1}{293} \right)$

$-\ln \left(3.5\right)=\frac{47700}{8.314}\left(\frac{1}{T_2}-\frac{1}{293}\right)$

$-2436.002\ln \left(3.5\right)T_2=13976100-47700T_2$

$T_2 = 313\ K$