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Standard enthalpy of vaporisation for water at 100°C is 40.66 kJ mol^–1. The internal energy of vaporisation of water at 100°C (in kJ mol^–1) is (Assume water vapour to behave like an ideal gas).

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+37.56

–43.76

+43.76

+40.66

answer is A.

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Detailed Solution

H₂O_(l) $⇌$ H₂O_(g) ; $∆_{vap} H$ = 40.66kJ/mole

$∆ n_{g}$ = 1

$∆ H = ∆ E + ∆ n_{g} RT$

$∆ E =$ 40.66 - (1 $\times$ 8.314 $\times$ 10^-3 $\times$ 373) = 37.56 kJ/mole

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