Answer :
Answer:
[tex]\Delta \:H_{vap}=40383.88\ J/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] = 13.95 torr
[tex]P_2[/tex] = 144.78 torr
[tex]T_1[/tex] = 25°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
[tex]T_1[/tex] = 298.15 K
[tex]T_2[/tex] = 75°C = 348.15 K
So,
[tex]\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)[/tex]
[tex]\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}[/tex]
[tex]\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)[/tex]
[tex]\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)[/tex]
[tex]\Delta \:H_{vap}=40383.88\ J/mol[/tex]