Verified Solution • 2025–26 Session moderate

QJEEM21Chfd11: JEE Main 2021 Chemistry - Class 11 JEEM21Chfd11 Solution: For water vap H = 41 kJ mol 1 at 373 K and 1 bar pressure.…

For water Δ\Delta vap H = 41 kJ mol −- 1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___________ kJ mol −- 1

[Use : R = 8.3 J mol −- 1 K −- 1]

Step-by-Step Verified Solution

H2O(l) →\to H2O(g) : Δ\Delta H = 41 kJmol{{kJ} \over {mol}}

⇒\Rightarrow From the relation : Δ\Delta H = Δ\Delta U + Δ\Delta ngRT

⇒\Rightarrow 41 kJmol{{kJ} \over {mol}} = Δ\Delta U + (1) ×\times 8.31000{{8.3} \over {1000}} ×\times 373

Δ\Delta DU = 41 −- 3.0959 = 38 kJ/mol

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