JEE MAIN - Chemistry (2007 - No. 14)
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 $$R = 8.3\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$$ )
and $$R = 8.3\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$$ )
$$41.00\,kJ\,mo{l^{ - 1}}$$
$$4.100\,kJ\,mo{l^{ - 1}}$$
$$3.7904\,kJ\,mo{l^{ - 1}}$$
$$37.904\,kJ\,mo{l^{ - 1}}$$
Explanation
Given
$$\Delta H = 41\,kJ\,mo{l^{ - 1}}$$
$$ = 41000\,J\,mo{l^{ - 1}}$$
$$T = {100^ \circ }C = 273 + 100$$
$$ = 373\,K,n = 1$$
$$\Delta U = \Delta H - \Delta nRT$$
$$ = 41000 - \left( {2 \times 8.314 \times 373} \right)$$
$$ = 37898.88\,J\,mo{l^{ - 1}}$$
$$ = 37.9\,kJ\,mo{l^{ - 1}}$$
$$\Delta H = 41\,kJ\,mo{l^{ - 1}}$$
$$ = 41000\,J\,mo{l^{ - 1}}$$
$$T = {100^ \circ }C = 273 + 100$$
$$ = 373\,K,n = 1$$
$$\Delta U = \Delta H - \Delta nRT$$
$$ = 41000 - \left( {2 \times 8.314 \times 373} \right)$$
$$ = 37898.88\,J\,mo{l^{ - 1}}$$
$$ = 37.9\,kJ\,mo{l^{ - 1}}$$
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