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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250 K

291.4 K

230 K

290 K

answer is B.

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Detailed Solution

Given P1=19 mm Hg,P2=760 mm Hg ΔHyap. =40670 J/molApplying Clausius - Clapeyron's equation log⁡P2P1=ΔHvap 2.303×RT2−T1T1T2 log⁡76019=406702.303×8.3373−T1T1×373T1 = 291.4 k

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