Answer:
H.vap = 6.544 kJ/mol
Explanation:
Clausius-Clapeyron Equation
In P = H.vap / RT where T = 20 + 273 = 293K
H.vap = In P (RT) = In 11 (8.314 x 293) = 6543.861 J/mol = 6.544 kJ/mol
The molar heat of vaporization of octane if its vapor pressure at 20° is 11.0 torr is 38.92 kJ/mol.
We can find the molar heat of vaporization of any compound with the Clausius-Clapeyron equation:
[tex] ln(\frac{P_{2}}{P_{1}}) = -\frac{\Delta H_{vap}}{R}*(\frac{1}{T_{2}} - \frac{1}{T_{1}}) [/tex] (1)
Where:
[tex] \Delta H_{vap}[/tex]: is the molar heat of vaporization =?
P₁: is the vapor pressure at T₁ = 11.0 torr
P₂: is the vapor pressure at T₂
T₁ = 20 °C = 293 K
R: is the gas constant = 8.314 J/K*mol
Assuming that the compound is octane, which boils at 125.7 °C, and remembering that any compound boils at a pressure of 1 atm (760 torr), we have:
T₂ = 125.7 °C = 398.7 K
P₂ = 1 atm = 760 torr
Hence, the molar heat of vaporization is (eq 1):
[tex] ln(\frac{760 torr}{11.0 torr}) = -\frac{\Delta H_{vap}}{8.314 J/K*mol}*(\frac{1}{398.7 K} - \frac{1}{293 K}) [/tex]
[tex] \Delta H_{vap} = -\frac{ln(\frac{760 torr}{11.0 torr})*8.314 J/K*mol}{(\frac{1}{398.7 K} - \frac{1}{293 K})} = 38917.6 J/mol = 38.92 kJ/mol [/tex]
Therefore, the molar heat of vaporization is 38.92 kJ/mol.
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