The normal boiling point of water is 373 K . Vapour pressure of water at temperature $T$ is 19 mm Hg

Question

The normal boiling point of water is 373 K . Vapour pressure of water at temperature $T$ is 19 mm Hg . If enthalpy of vaporisation is $40.67 \mathrm{~kJ} / \mathrm{mol}$, then temperature $T$ would be (Use : $\log 2=0.3, R: 8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ ):

Professor PC Thomas Classes · Answer

(b) Given $P_1=19 \mathrm{~mm} \mathrm{Hg}, P_2=760 \mathrm{~mm} \mathrm{Hg}$; 
$ \Delta H_{	ext {vap. }}=40670 \mathrm{~J} / \mathrm{mol} $
Applying Clausius-Clapeyron's equation 
$ \log \frac{P_2}{P_1}=\frac{\Delta H_{\mathrm{vap}}}{2.303 	imes R}\left(\frac{T_2-T_1}{T_1 T_2}ight) $
 or $\log \frac{760}{19}=\frac{40670}{2.303 	imes 8.3}\left(\frac{373-T_1}{T_1 	imes 373}ight)$ on solving, we get $T_1=291.4 \mathrm{~K}$