Question

The vapor pressure of benzene (C

6

H

6

) is 40.1 mmHg at 7.6°C. What is its vapor pressure at

60.6°C? The molar heat of vaporization of benzene is 31.0 kJ/mol

Answer 1

Clausius-Clapeyron equation:

$ln(\frac{p_2}{p_1}) = (\frac{∆H_{vap}}{R})\times(\frac{1}{T_1} – \frac{1}{T_2})$

_R — the gas constant (8.3145 J/mol K)

p₁ – the vapor pressure of benzene at T₁;

p₂ – the vapor pressure of benzene at T₂;

∆H_vap – the enthalpy of vaporization;

p₁ = 40.1 mmHg

T₁ = 7.6 °C + 273.15 = 280.75 K

T₂ = 60.6 °C + 273.15 = 333.75 K

∆H_vap = 31000 J/mol

$ln(\frac{p_2}{40.1}) = \frac{31000}{8.3145}\times(\frac{1}{280.75} – \frac{1}{333.75})$

$ln(\frac{p_2}{40.1}) = 3728.42×(0.00356 – 0.00299)$

$ln(\frac{p_2}{40.1}) = 3728.42×0.00057$

$ln(\frac{p_2}{40.1}) = 2.12$

$\frac{p_2}{40.1} = e^{2.12} = 8.33$

$p_2 = 40.1\times8.33 = 334.03 mmHg$

Answer: 334.03 mmHg