Assuming that water vapour is an ideal gas, the internal energy change (∆U) when 1 mol of water is vapourised at 1 bar pressure and 100 $^{\circ}$ C, (given : molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol $^{- 1}$ and R = 8.3 J mol $^{- 1}$ K $^{- 1}$ ) will be

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If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourization of 1 mol of water at 1 bar and 1000C is 41 kJ.mol−1. Calculate the internal energy change, when 1 mol of water is vaporized at 1 bar pressure and 1000C.

i) The reaction is: H2O(l)→H2O(g)We know that,ΔH=ΔU+ΔngRT⟹ΔU=ΔH−ΔngRTSo on substituting the given values we get,ΔU=41kJmol−1−8.3×3731000⟹ΔU=37.904 kJ.mol−1

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourization of 1 mol of water at 1 bar and $100^{0} C$ is $41 k J . {m o l}^{- 1}$ . Calculate the internal energy change, when 1 mol of water is vaporized at 1 bar pressure and $100^{0} C$ .

i) The reaction is: $H_{2} O_{(l)} \to H_{2} O_{(g)}$
We know that,
$Δ H = Δ U + Δ n_{g} R T ⟹ Δ U = Δ H - Δ n_{g} R T$
So on substituting the given values we get,
$Δ U = 41 k J m o l^{- 1} - \frac{8.3 \times 373}{1000}$
$⟹ Δ U = 37.904 k J . m o l^{- 1}$