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Q.

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

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a

291.4K

b

250 K

c

230K

d

290K

answer is B.

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Detailed Solution

Given P1=19mmHg,P2=760mmHg;
ΔHvap. =40670J/mol
Applying Clausius-Clapeyron's equation
log⁡P2P1=ΔHvap2.303×RT2−T1T1T2
or log⁡76019=406702.303×8.3373−T1T1×373
on solving, we get T1=291.4K

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