Question

The vapor pressure of water at 25 degrees Celsius is 23.8 torr, and the heat of vaporization of water at 25 degrees Celsius is 43.9 kJ/mol. Calculate the vapor pressure of water at 50 degrees Celsius.

Answer 1

Answer:

The vapor pressure of water at 50 °C  is 93.7 torr.

Explanation:

The 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:

$P_1$ = 23.8 torr

$P_2$ = ?

$T_1$ = 25°C

$T_2$ = 50 °C  

ΔHvap = 43.9 kJ/mol

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  

T = (50 + 273.15) K = 323.15 K  

$T_1$ = 298.15 K  

$T_2$ = 323.15 K

So,  applying in the above equation as:-

$\ln \:\left(\:\frac{23.8}{P_2}\right)\:=\:\frac{43.9}{8.314\times 10^{-3}}\:\left(\:\frac{1}{323.15}-\:\frac{1}{298.15}\:\right)$

$\frac{23.8}{P_2}=e^{\frac{43.9}{8.314\times \:10^{-3}}\left(\frac{1}{323.15}-\frac{1}{298.15}\right)}$

$23.8=\frac{1}{e^{\frac{1097500}{801030.39216}}}P_2$

$P_2=23.8e^{\frac{1097500}{801030.39216}}=93.7\ torr$

The vapor pressure of water at 50 °C  is 93.7 torr.