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The normal bo...

Question

The normal boiling point of water is $373 K$ . Vapour pressure of water at temperature 'T' is $19 m m H g$ . If enthalpy of vaporisation is $40.67 k J / m o l$ , then temperature 'T' would be:
(Use : $l o g 2 = 0.3, R = 8.3 J K^{- 1} m o l^{- 1})$

220.5 K

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291.4 K

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230.8 K

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200.1 K

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Solution

The correct option is B $291.4 K$
Given $P_{1} = 19 m m H g, P_{2} = 760 m m H g;$
$△ H_{v a p .} = 40670 J / m o l$

Applying Clausius-Clapeyron's equation :

$ln \frac{P_{2}}{P_{1}} = \frac{△ H_{v a p}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$

or $ln \frac{760}{19} = \frac{40670}{8.3} (\frac{373 - T_{1}}{T_{1} \times 373})$

$\frac{ln (40) \times 8.3}{40760} + \frac{1}{373} = \frac{1}{T_{1}}$

on solving, we get $T$ or $T_{1} = 291.4 K$

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Ideal Solutions

Standard XII Chemistry

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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: (Use : log 2=0.3, R=8.3 J K−1mol−1)

The correct option is B 291.4 KGiven P1=19 mm Hg,P2=760 mm Hg; △Hvap.=40670 J/mol Applying Clausius-Clapeyron's equation : lnP2P1=△HvapR(T2−T1T1T2) or ln76019=406708.3(373−T1T1×373) ln(40)×8.340760+1373=1T1 on solving, we get T or T1=291.4 K

The normal boiling point of water is $373 K$ . Vapour pressure of water at temperature 'T' is $19 m m H g$ . If enthalpy of vaporisation is $40.67 k J / m o l$ , then temperature 'T' would be:
(Use : $l o g 2 = 0.3, R = 8.3 J K^{- 1} m o l^{- 1})$