Question

# A compound having molar heat of vaporization 8.314 kJ/mol has a normal boiling point of 127∘C If the vapour pressure of same compound at 27∘C is denoted as P2. Find the value of loge(P2).

A
56
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B
+56
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C
23
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D
+23
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Solution

## The correct option is A −56Given, Temperature, T1=127∘C=400 K Temperature, T2=27∘C=300 K The normal boiling point of a liquid is the temperature at which its vapour pressure is equal to one atmosphere. thus, Vapour pressure at 400 K, P1=1 atm Molar heat of vaporisation, ΔH=8.314 kJ/mol Effect of temperature on the vapour​ pressure of a liquid is given by​ Clausius – Clapeyron equation, ln (P2P1)=ΔHR(1T1−1T2) we get, ln (P21)=(83148.314)(1400 − 1300) lnP2−ln1=−10001200lnP2=−56

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