Answer :
Answer:
[tex]\Delta \:H_{vap}=23.3054\ kJ/mol[/tex]
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
[tex]\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c [/tex]
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:
[tex]\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)[/tex]
Given:
[tex]P_1[/tex] = 234.2 mmHg
[tex]P_2[/tex] = 522.6 mmHg
[tex]T_1[/tex] = 217.9 K
[tex]T_2[/tex] = 232.4 K
So,
[tex]\ln \:\left(\:\frac{234.2\ mmHg}{522.6\ mmHg}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314\times 10^{-3}\ kJ /mol K}\:\left(\:\frac{1}{232.4\ K}-\:\frac{1}{217.9\ K}\:\right)[/tex]
[tex]\Delta \:H_{vap}=\frac{\left\{\ln \left(\:\frac{234.2}{522.6}\right)\:\times \:8.314\times 10^{-3}\right\}}{\left(\:\frac{1}{232.4}-\:\frac{1}{217.9}\:\right)}\ kJ/mol[/tex]
[tex]\Delta \:H_{vap}=-\frac{10^{-3}\times \:8.314\ln \left(\frac{234.2}{522.6}\right)}{\frac{14.5}{50639.96}}\ kJ/mol[/tex]
[tex]\Delta \:H_{vap}=-\frac{421020.62744\ln \left(\frac{234.2}{522.6}\right)}{14500}\ kJ/mol[/tex]
[tex]\Delta \:H_{vap}=23.3054\ kJ/mol[/tex]