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A liquid is in equilibrium with its vapo...

A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(1) What is the initial effect of the change on vapour pressure?
(2) How do rates of evaporation & condensation change initially?
(3) What happens when equilibrium is restored finally and what will be the final vapour pressure?

Text Solution

Verified by Experts

Equilibrium: `H_(2)O (l) hArr H_(2)O(g)`
(1) Increasing the volume of the container at contant temperature reduces the vapour pressure.
(1) Increasing the volume of the container at constant temperature reduces the vapour pressure.
(2) According to Le chatelier's principle, the increase in volume of the system at constant temperature increases the rate of evaporation compared with that of the rate of condensation.
(3) When equilibrium is restored, the rates of evaporation and condensation because equal. at new equilibrium, vapour pressure will be the same as that of original equilibrium as vapour pressure of a liquid is an intensive property which is independent of the size of the system.
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Knowledge Check

  • Statement I : A cylinder fitted with a movable piston contains a certain amount of liquid in equilibrium with its vapour. The temperature of the system is kept constant with the help of a thermostat. When the volume of the vapour is decreased by moving the piston inwards, the vapour pressure does not increase. Statement II : Vapour in equilibrium with its liquid, at a constant temperature, does not obey Boyle's law.

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    Statement I is true, statement II is true , statement II is a correct explanation for statement I.
    B
    Statement I is true, statement II is true , statement II is not a correct explanation for statement I.
    C
    Statement I is true, statement II is false.
    D
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  • Final pressure is higher than initial pressure of a container filled with an ideal gas at constant temperature. What will be the value of equilibrium constant ?

    A
    K = 1.0
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    `K lt 1`
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    `K gt 1.0`
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