Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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)

AIPMTAIPMT 2012Thermodynamics

Solution:

$H _{2} O _{(1)} \rightarrow H _{2} O _{( g )}$

$\Delta H =\Delta E +\Delta nRT$

$40.66=\Delta E +1 \times \frac{8.314}{1000} \times 373$

$\Delta E =37.5\, kJ$