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JEE Main levelSingle correct optionDifficulty 2 of 3td009

The enthalpy of vaporisation of water at $373 K$ and $1 atm$ is $40.7 kJ mo l^{- 1}$ . Treating the vapour as ideal and neglecting the volume of the liquid ( $R = 8.314 J mo l^{- 1} K^{- 1}$ ), the internal energy of vaporisation is

A $43.8 kJ mo l^{- 1}$
B $40.7 kJ mo l^{- 1}$
C $3.1 kJ mo l^{- 1}$
D $37.6 kJ mo l^{- 1}$

Answer: D

Δ H = Δ U + Δ n_{g} RT

with

Δ n_{g} = + 1

.

Δ U = 40.7 - (8.314) (373) /1000 = 40.7 - 3.10 = 37.6 kJ mo l^{- 1}

.

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