Answer:
The vapor pressure of ethanol at 60.61 °C is 327.56 mmHg
Explanation:
using Clausius-Clapeyron equation
[tex]ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}(\frac{1}{T_1} -\frac{1}{T_2})[/tex]
where;
ΔH is the enthalpy of vaporization of ethanol = 39.3 kJ/mol
R is ideal gas constant = 8.314 J/mol.K
P₁ is the initial pressure of ethanol at T₁ = 1.00 × 10² mmHg
P₂ is the final pressure of ethanol at T₂ = ?
T₁ is the initial temperature = 34.9°C = 307.9 K
T₂ is the final temperature = 60.61°C = 333.61 K
[tex]ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}(\frac{1}{T_1} -\frac{1}{T_2}) \\\\ln(\frac{P_2}{100}) = \frac{39300}{8.314}(\frac{1}{307.9} -\frac{1}{333.61})\\\\ln(\frac{P_2}{100}) = 4726.9666 (0.003248 -0.002997)\\\\ln(\frac{P_2}{100}) = 4726.9666 (0.000251) = 1.1865\\\\(\frac{P_2}{100}) = e^{(1.1865)} \\\\(\frac{P_2}{100}) = 3.2756\\\\P_2 = (100*3.2756)mmHg\\\\P_2 = 327.56 mmHg[/tex]
Therefore, the vapor pressure of ethanol at 60.61 °C is 327.56 mmHg
