The enthalpy of vaporization of a liquid at its boiling point (Tb=200 K) is 15.3 kJ mol⁻¹. If the molar… · ABC26PH0442

Subject: Physical Chemistry · Chapter: Chemical Equilibrium · Topic: Equilibria – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium

The enthalpy of vaporization of a liquid at its boiling point $\left(T_{b}=200 \mathrm{K}\right)$ is $15.3 \mathrm{kJ} \mathrm{mol}^{-1}$. If the molar volumes of the liquid and the vapour at 200 K are 100 and $12000 \mathrm{cm}^{3} \mathrm{mol}^{-1}$, respectively, then the slope $\frac{\mathrm{dP}}{\mathrm{dT}}$ of the liquid boundary is ____ $\mathrm{kPa} \mathrm{K}^{-1}$ (Upto two decimal places. Note : $1 \mathrm{Pa}=1 \mathrm{J} \mathrm{m}^{-1}$ )
Answer
SELF-PRACTICE — the source book printed no answer.

Nothing is invented here, so this question has no answer on record.

Explanation
Answer: ≈ 6.43 kPa K⁻¹ — Clapeyron: dP/dT = ΔH/(TΔV) = 15300/(200 × 11 900 × 10⁻⁶) = 6429 Pa K⁻¹.

Study loop for Chemical Equilibrium

1. Practise the PYQsPrevious-year questions, with answers

· Browse this chapter in the app