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For water $Δ_{v a p} H = 41 k J m o l^{- 1}$ at 373K 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 vaporation of water is _____ $k J m o l^{- 1}$ $[U s e : R = 8.3 J m o l^{- 1} K^{- 1}]$

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answer is 38.

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Detailed Solution

Given equation is
$H_{2} O (l) \to H_{2} O (g) : Δ H = 41$
$\Rightarrow$ From the relation : $Δ H = Δ U + Δ n_{g} R T$
$\Rightarrow 41 \frac{k J}{m o l} = Δ U + (1) \times \frac{8.3}{1000} \times 373$
$Δ U = 41 - 3.0959 = 38 k j / m o l$

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