Question

Octane has a vapor pressure of 40 torr at 45.1°C and 400. torr at 104.0°

c. What is its heat of vaporization (R=8.314J/K mol)? (SHOW YOUR WORK!!!!)

Answer 1

If I am not wrong, you have to use the Clausius-Clapeyron equation:

`\log{(P_(1))/(P_(2)) = -(L_(vap))/(R)((1)/(T_(1))-(1)/(T_(2)))`

Substituting the values (note temperature should be converted to Kelvins!)

`L_(vap)=39000J/mol`

Answer 2

39.018 kJ/mol is heat of vaporization of octane.

Step-by-step explanation:

The Clausius-Clapeyron equation is given as:

`log((P_(2))/(P_(1))) = (\Delta H_(vap))/(2.303R)[(1)/(T_(1)) - (1)/(T_(2))]`

`P_1`=Vapor pressure at
`T_1` temperature.

`P_2`=Vapor pressure at
`T_2` temperature.

`\Delta H_(vap)` = Heat of vaporization

Octane has a vapor pressure of 40 torr at 45.1°C.

`P_1=40 Torr,T_1=45.1^oC=318.25 K`

Octane has a vapor pressure of 400. torr at 104.0°C

`P_2=400 Torr,T_2=104.0^oC=377.15 K`

`log((400.0Torr)/(40.0 Torr)) = (\Delta H_(vap))/(2.303* 8.314 J/K mol)[(1)/(318.25 K) - (1)/(377.15 K)]`

`1=(\Delta H_(vap))/(2.303* 8.314 J/K mol)[(1)/(318.25 K) - (1)/(377.15 K)]`

`\Delta H_(vap)=(2.303* 8.314 J/K mol)/(377.15K - 318.25 K)* (377.15* 318.25 K)`

`\Delta H_(vap)=39,018.55 J/mol=39.018 kJ/mol`