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Assuming that water vapour behaves as an ideal gas, the internal energy change (ΔU) when 1 mol of water is vapourised at 1 bar pressure and 100 °C, [given: molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol⁻¹ and R = 8.3 J mol⁻¹ K⁻¹] will be

1. 41.00 kJ mol⁻¹
2. 4.100 kJ mol⁻¹
3. 3.7904 kJ mol⁻¹
4. 37.904 kJ mol⁻¹

Correct answer: 37.904 kJ mol⁻¹

Solution

Vaporising 1 mol gives dn_g = +1, so dU = dH - dn_g*RT = 41000 - (1)(8.3)(373) = 37904 J = 37.904 kJ/mol.

Related JEE Main Chemistry questions

• Using the following thermochemical equations, determine the enthalpy of vaporisation of water: H₂(g) + 1/2 O₂(g) → H₂O(l), ΔH = -286 kJ H₂(g) + 1/2 O₂(g) → H₂O(g), ΔH = -245.5 kJ
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