Question

A substance has a vapor pressure of 77.86 mm Hg at 318.3 K and a vapor pressure of161.3 mmHg at 340. 7 K. Calculate its heat of vaporization in kJ/mol.

Answer 1

Explanation:

It is given that,

Initial vapor pressure, P₁ = 77.86 mm

Initial temperature, T₁ = 318.3 K

Final vapor pressure, P₂ = 161.3 mm

Initial temperature, T₂ = 340.7 K

We need to find its heat of vaporization. It can be calculated by using Clausius-Clapeyron equation.

$ln(\dfrac{P_2}{P_1})=\dfrac{\Delta_{vap}H}{R}(\dfrac{1}{T_1}-\dfrac{1}{T_2})$

$\Delta _{vap} H=\dfrac{R\ ln(\dfrac{P_2}{P_1})}{(\dfrac{1}{T_1}-\dfrac{1}{T_2})}$

$\Delta _{vap} H=\dfrac{0.008 314\ ln(\dfrac{161.3}{77.86})}{(\dfrac{1}{318.3}-\dfrac{1}{340.7})}$

$\Delta _{vap} H=29.31\ kJ/mol$

So, the heat of vaporization of a substance is 29.31 kJ/mol. Hence, this is the required solution.