Question

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 kJ/mol.)

Answer 1

The vapor pressure at 60.21°C is 327 mmHg.

Step-by-step 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.

`ln((P_2)/(P_1))=(\Delta_(vap)H)/(R)((1)/(T_1)-(1)/(T_2))`

We have given in the question

`P_1=102\ mmHg`

`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`

And
`R` is the Universal Gas Constant.

`R=0.008 314 kJ/Kmol`

`ln((P_2)/(102))=(39.3)/(0.008314)((1)/(308.05)-(1)/(333.36))\\\\ln((P_2)/(102))=4726.967((333.36-308.05)/(333.36*308.05))\\\\ln((P_2)/(102))=4726.967((25.31)/(333.36*308.05))\\\\ln((P_2)/(102))=4726.967((25.31)/(102691.548))\\\\ln((P_2)/(102))=1.165`

Taking inverse log both side we get,

`(P_2)/(102)=e^(1.165)\\\\P_2=102* 3.20\ mmHg\\P_2=327\ mmHg`