Answer:
The vapor pressure at 60.21°C is 327 mmHg.
Explanation:
Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.
We need to find vapor pressure at 60.21°C.
The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.
[tex]ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]
We have given in the question
[tex]P_1=102\ mmHg[/tex]
[tex]T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol[/tex]
And [tex]R[/tex] is the Universal Gas Constant.
[tex]R=0.008 314 kJ/Kmol[/tex]
[tex]ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165[/tex]
Taking inverse log both side we get,
[tex]\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg[/tex]
