Question

Standard enthalpy of vapourisation $\Delta_{\text {vap }} H ^{\Theta}$ for water at $100^{\circ} C$ is $40.66\, kJ\, mol ^{-1}$. The internal energy of vaporisation of water at $100^{\circ} C$ (in $kJmol ^{-1}$) is : (Assume water vapour to behave like an ideal gas)

• A. 43.76
• B. 40.66
• C. 37.56
• D. -43.76

Answer 1

Answer: (C)

37.56

Answer 2

$H _{2} O _{(1)} \rightarrow H _{2} O _{( g )}$
$\Delta H =\Delta E +\Delta nRT$
$40.66=\Delta E +1 \times \frac{8.314}{1000} \times 373$
$\Delta E =37.5\, kJ$