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
250 K
b
291.4 K
c
230 K
d
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 logP2P1=ΔHvap 2.303×RT2−T1T1T2 log76019=406702.303×8.3373−T1T1×373T1 = 291.4 k
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