Question

Assuming that water vapour is an ideal gas, the internal energy $(ΔU)$ when 1 mol of water is vapourised at 1 bar pressure and $100^°C$, (Given: Molar enthalpy of vapourization of water at 1 bar and $373 \,K = 41\, kJ \,mol^{−1}$ and $R = 8.3\, J \,mol^{−1}K^{−1}$) will be

• A. $4.100\, kJ\, mol^{−1}$
• B. $3.7904\, kJ\, mol^{−1}$
• C. $37.904\, kJ\, mol^{−1}$
• D. $41.00\, kJ\, mol^{−1}$

Answer 1

Answer: (C)

$H_{2}O(\ell) \xrightarrow {\text{vaporisation}} H2O(g)$
$\Delta n_{g}=1-0=1$
$\Delta H=\Delta U+\Delta n_{g}RT$
$\Delta U=\Delta H-\Delta n_{g}RT$
$=41-8.3\times10^{-3}\times373$
$=37.9\,kJ\,mol^{-1}$
Hence, (3) is correct.