Question

The molar enthalpy change for $H_{2}O\left(l\right)\rightleftharpoons H_{2}O\left(g\right)$ at $373$K and $1$ atm is $41kJ/mol$. Assuming ideal behavior, the internal energy change for vaporization of $1$ mol of water at $373K$ and $1$ atm in kJ mol${}^{-1}$ is:

• A. $30.2$
• B. $41.0$
• C. $48.1$
• D. $37.9$

Answer 1

The correct option is D $37.9$

$\Delta H=\Delta U+\Delta \left(PV\right)$

$\Delta H=\Delta U+RT\Delta n$

$41=\Delta U+8.314\times 373\times 1\times {10}^{-3}$

$\Delta U=37.9kJ/mol$

Hence option $D$ is correct.