Paper: CSIR-NET December 2011 · Subject: Physical Chemistry · Chapter: Thermodynamics · Topic: Laws State Functions · Marks: 2 · Difficulty: Medium
For the liquid $\rightleftharpoons$ vapour equilibrium of a substance $\frac{d P}{d T}$ at 1 bar and 400 K is $8 \times 10^{-3}$ bar $\mathrm{K}^{-1}$. If the molar volume in the vapour form is $200\,\mathrm{L\,mol}^{-1}$ and the molar volume in the liquid form is negligible, the molar enthalpy of vapourisation is (1.0 bar L = 100 J)
(a)$640\mathrm{kJmol}^{-1}$
(b)$100\mathrm{kJmol}^{-1}$
(c)$80\mathrm{kJmol}^{-1}$
(d)$64 \mathrm{kJmol}^{-1}$
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
Clapeyron: ΔH = TΔV(dP/dT) = 400 × 200 L × 8×10⁻³ bar K⁻¹ = 640 bar L = 64 kJ mol⁻¹ (d).