Question: Normal boiling point of water is 373 K (at 760 mm). Vapor pressure of water at 298 K is 23 mm. If the enthalpy of evaporation is 40.656 kJ/mol, the boiling point of water at 23 mm pressure will be:
Options:
A) 250 K
B) 294 K
C) 51.6 K
D) 12.5 K
Answer:
Correct Answer: B
Solution:
[b] Applying ClausiusClapeyron equation, we get $ \log \frac{P_2}{P_1}=\frac{\Delta H_{v}}{2.303R}[ \frac{T_2-T_1}{T_1\times T_2} ] $ $ \log \frac{760}{23}=\frac{40656}{2.303\times 8.314}[ \frac{373-T_1}{373T_2} ] $ This gives $ T_1 $ = 294.4 K.
