Assuming that water vapour is an ideal gas, the internal energy change (ΔU) when 1 mol of water is vaporised at 1 bar pressure and 100 °C, (Given: Molar enthalpy ofvapourisation of water at 1 bar and 373K=41kJmol−1 and R=8.3Jmol−1K−1 ) will be:

# Assuming that water vapour is an ideal gas, the internal energy change $\left(\mathrm{\Delta }U\right)$ when 1 mol of water is vaporised at 1 bar pressure and 100 °C, (Given: Molar enthalpy ofvapourisation of water at 1 bar and $373\mathrm{K}=41\mathrm{kJ}{\mathrm{mol}}^{-1}$ and $R=8.3\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$ ) will be:

1. A

$4.100\mathrm{kJ}{\mathrm{mol}}^{-1}$

2. B

$3.7904\mathrm{kJ}{\mathrm{mol}}^{-1}$

3. C

$37.904\mathrm{kJ}{\mathrm{mol}}^{-1}$

4. D

$41.00\mathrm{kJ}{\mathrm{mol}}^{-1}$

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### Solution:

The vapourisation process is ${\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{g}\right)$

For this process, $\mathrm{\Delta }{v}_{\mathrm{g}}=+1$. Hence

$\begin{array}{r}{\mathrm{\Delta }}_{\mathrm{r}}U={\mathrm{\Delta }}_{\mathrm{r}}H-\left(\mathrm{\Delta }{v}_{\mathrm{g}}\right)RT=\left(41×{10}^{3}{\mathrm{mol}}^{-1}\right)-\left(1\right)\left(8.3{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\right)\left(373\mathrm{K}\right)\\ =41×{10}^{3}\mathrm{J}{\mathrm{mol}}^{-1}-3095.9\mathrm{J}{\mathrm{mol}}^{-1}=37904\mathrm{J}{\mathrm{mol}}^{-1}=37.904\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$  