Answer:
The vapor pressure is 748.77 torr
Explanation:
Using Clausius-Clapeyron equation:
[tex]ln(\frac{P_2}{P_1}) = \frac{\delta H_v_a_p}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where;
T₁ is the initial temperature = 85.0°F = 302.5 K
T₂ is the final temperature = 100 °C = 373 K
P₂ is the final pressure = 760 torr
P₁ is the initial pressure = vapor pressure = ?
R is gas constant = 8.314 J/K.mol
ΔHvap is the heat of vaporization of water = 40.7 kJ/mol
[tex]ln(\frac{P_2}{P_1}) = \frac{\delta H_v_a_p}{R}[\frac{1}{T_1}-\frac{1}{T_2}] = \frac{40.7}{8.314}[\frac{1}{302.5}-\frac{1}{373}][/tex]
[tex]ln(\frac{P_2}{P_1}) =[/tex] 4.895(0.00331 - 0.00268) = 0.01489
[tex]\frac{P_2}{P_1} = e^{0.01489}[/tex] = 1.015
P₁ = (760 torr)/(1.015) = 748.77 torr
Therefore, the vapor pressure is 748.77 torr
