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:

A
$4.100 kJ {mol}^{- 1}$
B
$3.7904 kJ {mol}^{- 1}$
C
$37.904 kJ {mol}^{- 1}$
D
$41.00 kJ {mol}^{- 1}$

Solution:

The vapourisation process is $H_{2} O (l) \to H_{2} O (g)$

For this process, $Δ v_{g} = + 1$ . Hence

$\begin{aligned} Δ_{r} U = Δ_{r} H - (Δ v_{g}) R T = (41 \times 10^{3} {mol}^{- 1}) - (1) (8.3 {JK}^{- 1} {mol}^{- 1}) (373 K) \\ = 41 \times 10^{3} J {mol}^{- 1} - 3095.9 J {mol}^{- 1} = 37904 J {mol}^{- 1} = 37.904 kJ {mol}^{- 1} \end{aligned}$

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

Solution:

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