A fish swimming in water body when taken out from the water body is covered with a film of water of weight 36 g. When it is subjected to cooking at 100°C, then the internal energy for vaporization in kJ mol⁻¹ is ___________. [nearest integer]
[Assume steam to be an ideal gas. Given ΔᵥₐₚH^Θ for water at 373 K and 1 bar is 41.1 kJ mol⁻¹ ; R = 8.31 J K⁻¹ mol⁻¹]

Solution

Chemistry ma’am · 4 steps

The 36 g is a decoy: per mole, take the expansion work RT away from Δ H and you have Δ U.

1. Link H and U. For the change H₂O(l) → H₂O(g), only the steam counts as gas. So Δ n_g = 1 - 0 = 1 for every mole vaporised.

   ΔᵥₐₚU = ΔᵥₐₚH - Δ n_g RT

2. Find RT. RT = 8.31 × 373 = 3099.6 J mol⁻¹ = 3.10 kJ mol⁻¹. This is the work the steam does pushing back the atmosphere.
3. Subtract. ΔᵥₐₚU = 41.1 - 1 × 3.10 = 38.0 kJ mol⁻¹. The 36 g (2 mol) does not enter, because the answer is asked per mole.
4. So. The internal energy of vaporisation is about 38.0 kJ mol⁻¹, which rounds to 38.

The common slip: 76. Multiplying by 2 mol for the 36 g of water gives the total, about 76 kJ, but the question asks for kJ per mole.

The idea to keep. Use Δ U = Δ H - Δ n_g RT, counting only gaseous moles in Δ n_g. Check the units asked: per mole means the sample mass does not matter.

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