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Question

A compound having molar heat of vaporization 8.314 kJ/mol has a normal boiling point of 127C
If the vapour pressure of same compound at 27C 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 56
Given,
Temperature, T1=127C=400 K
Temperature, T2=27C=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(1T11T2)
we get,
ln (P21)=(83148.314)(1400 1300)

lnP2ln1=10001200lnP2=56

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