Question

The molar heat of vaporization of toluene is $\Delta H_v$. If its vapour pressure at 315K is 60 torr & that at 355 K is 300 torr then $\Delta H_v=?$ (log 2 =0.3).

• A. $37.4$ kJ/mole
• B. $3.74$ kJ/mole
• C. $37.4$ J/mol
• D. $3.74$ J/mole

Answer 1

The correct option is A $37.4$ kJ/mole

The Clausius - Clapeyron equation is as given below.

$log\frac{P_2}{P_1}=\frac{\Delta H_v}{2.303R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

Substitute values in the above expression.

$log\frac{300}{60}=\frac{\Delta H_v}{2.303\times 8.314}(\frac{1}{315}-\frac{1}{355})$

Hence, $\Delta H_v=37414J/mol=37.41kJ/mol$

Hence, the correct option is $A$