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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 $P_{1} = 19 mmHg, P_{2} = 760 mmHg$ ;
${ΔH}_{vap.} = 40670 J / mol$
Applying Clausius-Clapeyron's equation
$\log \frac{P_{2}}{P_{1}} = \frac{{ΔH}_{vap}}{2 .303 \times R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
or $\log \frac{760}{19} = \frac{40670}{2.303 \times 8.3} (\frac{373 - T_{1}}{T_{1} \times 373})$
on solving, we get $T_{1} = 291 .4 K$

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