Respuesta :

If I am not wrong, you have to use the Clausius-Clapeyron equation:
[tex]\log{\frac{P_{1}}{P_{2}} = -\frac{L_{vap}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})[/tex]

Substituting the values (note temperature should be converted to Kelvins!)
[tex]L_{vap}=39000J/mol[/tex]

Answer:

39.018 kJ/mol is heat of vaporization of octane.

Explanation:

The Clausius-Clapeyron equation is given as:

[tex]log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}][/tex]

[tex]P_1[/tex]=Vapor pressure at [tex]T_1[/tex] temperature.

[tex]P_2[/tex]=Vapor pressure at [tex]T_2[/tex] temperature.

[tex]\Delta H_{vap} [/tex] = Heat of vaporization

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

[tex]P_1=40 Torr,T_1=45.1^oC=318.25 K[/tex]

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

[tex]P_2=400 Torr,T_2=104.0^oC=377.15 K[/tex]

[tex]log(\frac{400.0Torr}{40.0 Torr}) = \frac{\Delta H_{vap}}{2.303\times 8.314 J/K mol}[\frac{1}{318.25 K} - \frac{1}{377.15 K}][/tex]

[tex]1=\frac{\Delta H_{vap}}{2.303\times 8.314 J/K mol}[\frac{1}{318.25 K} - \frac{1}{377.15 K}][/tex]

[tex]\Delta H_{vap}=\frac{2.303\times 8.314 J/K mol}{377.15K - 318.25 K}\times (377.15\times 318.25 K)[/tex]

[tex]\Delta H_{vap}=39,018.55 J/mol=39.018 kJ/mol[/tex]