For water Δᵥₐₚ H = 41 kJ mol⁻¹ at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___________ kJ mol⁻¹
[Use : R = 8.3 J mol⁻¹ K⁻¹]

Solution

Chemistry ma’am · 4 steps

Evaporation makes one mole of gas from nothing, so subtract that gas's RT from Δ H to get Δ U.

1. Link H and U. Enthalpy and internal energy differ by the pressure–volume work, Δ H = Δ U + pΔ V. Liquid water has a tiny volume next to its vapour, so only the vapour counts. For an ideal gas, pΔ V = Δ n_g RT.

   Δ U = Δ H - Δ n_g RT

2. Find the gas change. In H₂O(l) → H₂O(g), the moles of gas go from 0 to 1, so Δ n_g = 1.
3. Put in numbers. Δ n_g RT = 1 × 8.3 × 373 = 3095.9 J mol⁻¹ ≈ 3.1 kJ mol⁻¹. Then Δ U = 41 - 3.1 = 37.9 kJ mol⁻¹.
4. So. The internal energy change is about 37.9 kJ mol⁻¹, which rounds to 38.

The common slip: 44. Adding Δ n_g RT to Δ H instead of subtracting it gives 44. The system spends energy pushing the vapour out, so Δ U must be smaller than Δ H.

The idea to keep. For a phase change or reaction that makes gas, Δ U = Δ H - Δ n_g RT. Count only gas moles, and turn RT into kJ before you subtract.

Answer from the NTA final answer key. Solution by PrepWiser.