Assuming that water vapour is an ideal gas, the internal energy change (ΔE) when 1mol of water is vapourised at 1 bar pressure and 100°C, (given : molar enthalpy of vapourisation of water at 1 bar and 373K = 41kJ mol–1and R = 8.3J mol–1K–1 ) will be: (a) 4.100 kJ mol-1 (b) 3.7904 kJ mol-1 (c) 37.904 kJ mol-1 (d) 41.00 kJ mol-1

Answer: (c) 37.904 kJ mol-1

H2O(l) ⇌ H2O(g)

Given, the internal energy change (ΔE)

Molar enthalpy of vapourisation of water at 1 bar and 373K = 41kJ mol–1and R = 8.3J mol–1K–1

Δng=1

So,

ΔH = ΔE + ΔngRT

ΔE = ΔH − ΔngRT

ΔE =41 − (1 × 8.314 × 10−3 × 373)

ΔE = 37.904 J/mol

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