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(A) The vaporization process is represented as: $H_2O_{(l)} \rightarrow H_2O_{(g)}$.The number of moles of water is $n = \frac{36 \ g}{18 \ g \ mol^{-

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(A) The vaporization process is represented as: $H_2O_{(l)} ightarrow H_2O_{(g)}$.The number of moles of water is $n = \frac{36 \ g}{18 \ g \ mol^{-1}} = 2 \ mol$.The relationship between enthalpy change and internal energy change is given by: $\Delta U = \Delta H - \Delta n_g RT$.For the vaporization of $1 \ mol$ of water,$\Delta n_g = 1$ (since $1 \ mol$ of gas is produced from $1 \ mol$ of liquid).$\Delta U = \Delta_{vap} H^{\ominus} - RT = 41.1 \ kJ \ mol^{-1} - (8.31 \ J \ K^{-1} \ mol^{-1} 	imes 373 \ K) / 1000$.$\Delta U = 41.1 - 3.09963 \approx 41.1 - 3.1 = 38.0 \ kJ \ mol^{-1}$.Thus,the internal energy change for vaporization is $38 \ kJ \ mol^{-1}$.

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