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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290 K

250 K

291.4 K

230 K

answer is B.

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Detailed Solution

$Given P_{1} = 19 mm Hg, P_{2} = 760 mm Hg {ΔH}_{yap.} = 40670 J / mol$
Applying Clausius - Clapeyron's equation
$\log \frac{P_{2}}{P_{1}} = \frac{{ΔH}_{vap}}{2 .303 \times R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}}) \log \frac{760}{19} = \frac{40670}{2 .303 \times 8 .3} (\frac{373 - T_{1}}{T_{1} \times 373})$

T₁ = 291.4 k

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