Respuesta :
Answer:
The vapor pressure of water at 50 °C is 93.7 torr.
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] = 23.8 torr
[tex]P_2[/tex] = ?
[tex]T_1[/tex] = 25°C
[tex]T_2[/tex] = 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
[tex]T_1[/tex] = 298.15 K
[tex]T_2[/tex] = 323.15 K
So, applying in the above equation as:-
[tex]\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)[/tex]
[tex]\frac{23.8}{P_2}=e^{\frac{43.9}{8.314\times \:10^{-3}}\left(\frac{1}{323.15}-\frac{1}{298.15}\right)}[/tex]
[tex]23.8=\frac{1}{e^{\frac{1097500}{801030.39216}}}P_2[/tex]
[tex]P_2=23.8e^{\frac{1097500}{801030.39216}}=93.7\ torr[/tex]
The vapor pressure of water at 50 °C is 93.7 torr.
