For water at 100° C and 1 bar,
Δᵥₐₚ H - Δᵥₐₚ U = _____________ × 10² J mol⁻¹. (Round off to the Nearest Integer)
[Use : R = 8.31 J mol⁻¹ K⁻¹]
[Assume volume of H₂O(l) is much smaller than volume of H₂O(g). Assume H₂O(g) treated as an ideal gas]

Solution

Chemistry ma’am · 3 steps

Only the vapour pushes the atmosphere back, so the gap between ΔH and ΔU is just RT for one mole of gas.

1. Link H and U. At constant pressure, Δ H = Δ U + pΔ V. For H₂O(l) → H₂O(g), the liquid's volume is negligible, so Δ V ≈ V_(gas).
2. Ideal gas. For an ideal gas, pV_(gas) = n_g RT. One mole of liquid gives one mole of vapour, so Δ n_g = 1.

   ΔᵥₐₚH - ΔᵥₐₚU = Δ n_g RT = 1 × 8.31 × 373 = 3099.6 J mol⁻¹

3. So. 3099.6 J mol⁻¹ = 30.996 × 10² J mol⁻¹, which rounds to 31.

The common slip: 8. Using T = 100 instead of 373 K gives 831 J mol⁻¹, that is 8, not 31. Convert °C to kelvin first.

The idea to keep. For any phase change or reaction, Δ H - Δ U = Δ n_g RT. Count only the gas moles, and always put T in kelvin.

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