1. Tardigrade
  2. Question
  3. Chemistry
  4. For water Δ text vap H =41 kJ mol -1 at 373 K 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 evaporation of water is kJ mol -1 [Use: R =8.3 J mol -1 K -1 ]

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Q. For water $\Delta_{\text {vap }} H =41\, kJ\, mol ^{-1}$ at $373\, K$ 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 evaporation of water is ___$kJ \,mol ^{-1}$
[Use : $R =8.3 \,J \,mol ^{-1} \,K ^{-1}$ ]

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Solution:

Given equation is
$H _{2} O (\ell) \longrightarrow H _{2} O ( g ): \Delta H =41 \frac{ kJ }{ mol }$
$\Rightarrow $ From the relation : $\Delta H =\Delta U +\Delta n _{ g } RT$
$\Rightarrow 41 \frac{ kJ }{ mol }=\Delta U +(1) \times \frac{8.3}{1000} \times 373$
$\Rightarrow DU =41-3.0959$
$=38 \,kJ / mol$