Question
The value for $$\Delta U$$ for the reversible isothermal evaporation of $$90\,g$$ water at $$100{\,^ \circ }C$$ will be $$\left( {\Delta {H_{{\text{evap}}}}} \right.$$ of water $$ = 40.8\,kJ\,mo{l^{ - 1}},$$ $$\left. {R = 8.314\,J\,{K^{ - 1}}\,mo{l^{ - 1}}} \right)$$
A.
$$4800\,kJ$$
B.
$$188.494\,kJ$$
C.
$$40.8\,kJ$$
D.
$$125.03\,kJ$$
Answer :
$$188.494\,kJ$$
Solution :
$$\eqalign{
& \Delta H\,\,{\text{for}}\,\,{\text{18}}\,{\text{g}}\,\,{\text{water}} = 40.8\,kJ \cr
& {\text{For}}\,\,90\,g\,\,{\text{water}} = \frac{{40.8}}{{18}} \times 90 = 204\,kJ \cr
& n\,\,{\text{for}}\,\,90\,g\,\,{\text{water}} = \frac{{90}}{{18}} = 5 \cr
& \Delta {n_g} = 5 - 0 = 5;\,\,\Delta H = \Delta U + \Delta {n_g}\,RT \cr
& \Delta U = 204000 - \left( {5 \times 8.314 \times 373} \right) \cr
& \,\,\,\,\,\,\,\,\,\, = 188494\,J\,\,{\text{or}}\,\,188.494\,kJ \cr} $$