Assuming that water vapour is an ideal gas,th | vedclass

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(D) The process is: $H_2O(l) \rightarrow H_2O(g)$.The change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.Given: $\Delta H = 41\, kJ\, mo

Vedclass · Accepted Answer

$37.904$

Vedclass · Answer

(D) The process is: $H_2O(l) ightarrow H_2O(g)$.The change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.Given: $\Delta H = 41\, kJ\, mol^{-1} = 41000\, J\, mol^{-1}$,$T = 373\, K$,and $R = 8.3\, J\, mol^{-1}\, K^{-1}$.Using the relation $\Delta H = \Delta U + \Delta n_g RT$,we get $\Delta U = \Delta H - \Delta n_g RT$.$\Delta U = 41000\, J\, mol^{-1} - (1 	imes 8.3\, J\, mol^{-1}\, K^{-1} 	imes 373\, K)$.$\Delta U = 41000 - 3095.9 = 37904.1\, J\, mol^{-1}$.Converting to $kJ\, mol^{-1}$,we get $\Delta U = 37.9041\, kJ\, mol^{-1}$.

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