Answer:
The final vapor pressure is 687.24mmHg
Explanation:
The clausius-clapeyron equation below will be used to solve this problem:
[tex]ln\frac{P_2}{P_1} =\frac{\delta H}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]
Where;
ΔH is heat of vaporization = 450 kJ/mol = 450000J/mol
initial vapor pressure P₁ = 400mmHg
initial temperature T₁ = 3030K
Final temperature T₂ = 3070K
R is ideal gas constant = 8.314 J/molK
final vapor pressure P₂ = ?
[tex]ln\frac{P_2}{400} =\frac{450000}{8.314}(\frac{1}{3030}-\frac{1}{3070})[/tex]
[tex]ln\frac{P_2}{400} =(54125.57)(0.00001)[/tex]
[tex]ln\frac{P_2}{400} =(0.5412557)[/tex]
[tex]\frac{P_2}{400} = e^{(0.5412557)}[/tex]
P₂/400 = 1.7181
P₂ = (400*1.7181)mmHg
P₂ = 687.24mmHg
The final vapor pressure is 687.24mmHg
