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Q.

The normal boiling point of water is 373 K. 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

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a

290 K

b

250 K

c

291.4 K

d

230 K

answer is B.

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Detailed Solution

Given P1=19 mm Hg,P2=760 mm Hg ΔHyap. =40670 J/mol
Applying Clausius - Clapeyron's equation 
log⁡P2P1=ΔHvap 2.303×RT2−T1T1T2 log⁡76019=406702.303×8.3373−T1T1×373

T1 = 291.4 k

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