Question

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

• A. 250 K
• B. 298 K
• C. 51.6 K
• D. 12.5 K

Answer 1

The Correct Option is B

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 $T_{1}=294.4 \,K$.