Question

If water vapour is assumed to be an ideal gas and the molar enthalpy change for vaporization of one mole of water at $1$ bar and $100^{\circ} C$ is $41 kJ / mol$. The internal energy change when $1$ mol of water is vaporised

• A. $37.904 \,kJ / mol$
• B. $35\,kJ / mol$
• C. $40.5\,kJ / mol$
• D. $39\,kJ / mol$

Answer 1

Answer: (A)

$H _{2} O _{(l)} \rightarrow H _{2} O _{(g)} ; \Delta H=41\, kJ / mol$

$\Delta H=\Delta E+\Delta n_{g} R T ; \Delta E$

$=\Delta H-\Delta n_{g} R T$

$\Delta n_{g}=1-0=1, R=8.314, T=373 K$

$\Delta E=41-\left(1 \times 8.314 \times 373 \times 10^{-3}\right)$

$=41-3.1=37.9\, kJ / mol$