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Standard XII

Chemistry

Elevation in Boiling Point

The normal bo...

Question

The normal boiling of water is 373 K (at 760 mm). Vapour pressure of water at 298 K is 23 mm. If enthalpy of vaporization is $40.656 {k J m o l}^{- 1}$ , the boiling point of water at 23 mm atmospheric pressure will be:

250 K

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294 K

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51.6 K

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12.5 K

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Solution

The correct option is B $294 K$
By Clausius-Clayperon equation:-
$ln \frac{P_{2}}{P_{1}} = \frac{△ H}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
$∴ ln \frac{760}{23} = \frac{40656}{8.314} [\frac{1}{T_{b}} - \frac{1}{373}]$
$\Rightarrow T_{b} \approx 294 K$

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Similar questions

Normal boiling point of water is 373 K (at 760mm). Vapour pressure of water at 298 K is 23 mm. If the enthalpy of evaporation is 40.656 kJ/mole, the boiling point of water at 23 mm pressure will be

Q. 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})

Q. The normal boiling point of water is

373 K

. Vapour pressure of water at temperature

T

19 mm Hg

. If enthalpy of vaporisation is

40671.40 J/mol

, then temperature

T

would be:
(Use :

log 2 = 0.3, R : 8.3 {JK}^{- 1} {mo1}^{- 1}

)

Q. The normal boiling of water is

373 K

. Vapour pressure of water at temperature

T

is 19 mm

H g

. If enthalpy of evaporation 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}

)

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The normal boiling of water is 373 K (at 760 mm). Vapour pressure of water at 298 K is 23 mm. If enthalpy of vaporization is 40.656kJ mol−1 , the boiling point of water at 23 mm atmospheric pressure will be:

The correct option is B 294KBy Clausius-Clayperon equation:-lnP2P1=△HR[1T1−1T2]∴ln76023=406568.314[1Tb−1373]⇒Tb≈294K

The normal boiling of water is 373 K (at 760 mm). Vapour pressure of water at 298 K is 23 mm. If enthalpy of vaporization is $40.656 {k J m o l}^{- 1}$ , the boiling point of water at 23 mm atmospheric pressure will be:

The correct option is B $294 K$
By Clausius-Clayperon equation:-
$ln \frac{P_{2}}{P_{1}} = \frac{△ H}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
$∴ ln \frac{760}{23} = \frac{40656}{8.314} [\frac{1}{T_{b}} - \frac{1}{373}]$
$\Rightarrow T_{b} \approx 294 K$