For water Delta_{vap} H = 41 kJ mol^{-1} at | vedclass

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Question

(B) The evaporation reaction is: $H_2O(\ell) \longrightarrow H_2O(g)$.Given $\Delta H = 41 \ kJ \ mol^{-1}$,$T = 373 \ K$,and $R = 8.3 \ J \ mol^{-1}

Vedclass · Accepted Answer

$38$

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(B) The evaporation reaction is: $H_2O(\ell) \longrightarrow H_2O(g)$.Given $\Delta H = 41 \ kJ \ mol^{-1}$,$T = 373 \ K$,and $R = 8.3 \ J \ mol^{-1} \ K^{-1} = 8.3 	imes 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$.The change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.Using the relation $\Delta H = \Delta U + \Delta n_g RT$:$\Delta U = \Delta H - \Delta n_g RT$$\Delta U = 41 \ kJ \ mol^{-1} - (1 \ mol) 	imes (8.3 	imes 10^{-3} \ kJ \ mol^{-1} \ K^{-1}) 	imes (373 \ K)$$\Delta U = 41 - 3.0959 = 37.9041 \ kJ \ mol^{-1} \approx 38 \ kJ \ mol^{-1}$.

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