Question
Assuming that water vapour is an ideal gas, the internal energy change $$\left( {\Delta U} \right)$$ when $$1\,mol$$ of water is vapourised at 1 bar pressure and $${100^ \circ }C,$$ ( given : molar enthalpy of vapourisation of water at $$1\,bar$$ and $$373\,K = 41\,kJ\,mo{l^{ - 1}}$$ and $$R = 8.3\,J\,mo{l^{ - 1}}\left. {{K^{ - 1}}} \right)$$ will be
A.
$$41.00\,kJ\,mo{l^{ - 1}}$$
B.
$$4.100\,kJ\,mo{l^{ - 1}}$$
C.
$$3.7904\,kJ\,mo{l^{ - 1}}$$
D.
$$37.904\,kJ\,mo{l^{ - 1}}$$
Answer :
$$37.904\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{Given}}\,\,\Delta H = 41\,kJ\,mo{l^{ - 1}} = 41000\,J\,mo{l^{ - 1}} \cr
& T = {100^ \circ }C = 273 + 100 = 373\,K \cr
& \Delta n = 1 \cr
& \Delta U = \Delta H - \Delta nRT = 41000 - \left( {1 \times 8.314 \times 373} \right) \cr
& = 37898.88\,J\,mo{l^{ - 1}} \simeq 37.9\,kJ\,mo{l^{ - 1}} \cr} $$