SolutionHard
Question
The molar latent heat of vaporization of a certain liquid is 4.9 kcal/mol. The normal boiling point of the liquid is 77°C. The change in boiling point per atm increase in pressure at the normal boiling point of the liquid is
Options
A.50 deg/atm
B.0.02 deg/atm
C.5.0 deg/atm
D.0.05 deg/atm
Solution
Clausius–Clapeyron equation:
$\frac{dT}{dP} = \frac{RT^{2}}{\Delta H.P} = \frac{2 \times (350)^{2}}{4.9 \times 10^{3}} = 50\text{ K/atm}$
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