The internal energy change (in J) When 90 g of water undergoes complete evaporation at 100ᵒC is ____________. (Round off to the nearest integer)
(Given : ΔHᵥₐₚ for water at 373 K = 41 kJ/mol,
R = 8.314 JK –1 mol –1 )

Solution

Chemistry ma’am · 3 steps

Evaporation makes gas, so subtract the work of pushing back the atmosphere from Δ H to get Δ U.

1. Moles of water. 90 g ÷ 18 g mol⁻¹ = 5 mol. All of it turns from liquid into vapour, so Δ n_g = 5 - 0 = 5 mol. The liquid's volume is negligible.

   Δ U = Δ H - Δ n_g RT

2. Put in numbers. Δ H = 5 × 41000 = 205000 J. The work term is Δ n_g RT = 5 × 8.314 × 373 = 15505.61 J.
3. So. Δ U = 205000 - 15505.61 = 189494.39 J, which rounds to 189494.

The common slip: 220506. Adding Δ n_g RT instead of subtracting it gives 220506 J. Making gas costs expansion work, so Δ U must come out smaller than Δ H.

The idea to keep. Use Δ H = Δ U + Δ n_g RT and count only gaseous moles in Δ n_g. When gas is produced, Δ U is smaller than Δ H. Keep everything in joules.

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