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

Step-by-step 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((P_2)/(P_1))=(\Delta_(vap)H)/(R)((1)/(T_1)-(1)/(T_2))`

`\Delta _(vap) H=(R\ ln((P_2)/(P_1)))/(((1)/(T_1)-(1)/(T_2)))`

`\Delta _(vap) H=(0.008 314\ ln((161.3)/(77.86)))/(((1)/(318.3)-(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.