Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The normal boiling point of water is 373 K(at 760 mm). Vapour pressure of water at 298 K is 23 mm. If enthalpy of vaporisation is 40.656 kJ the boiling point of water at 23 mm atmospheric pressure will be

AIPMTAIPMT 1995Solutions

Solution:

Applying clausius clapeytron equation
$\log \cdot \frac{P_{2}}{P_{1}}=\frac{\Delta H_{V}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} \times T_{2}}\right]$
$\log \cdot \frac{760}{23}=\frac{40656}{2.303 \times 8.314}\left[\frac{373-T_{1}}{373 T_{1}}\right]$
This gives $T_{1}=294.4 \,K$.