Verified Solution • 2025–26 Session moderate

QJEEM22Chb5a8: JEE Main 2022 Chemistry - Class 11 JEEM22Chb5a8 Solution: A box contains 0.90 g of liquid water in equilibrium with water …

A box contains 0.90 g of liquid water in equilibrium with water vapour at 27 ∘^\circ C. The equilibrium vapour pressure of water at 27 ∘^\circ C is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be __________ litre. [nearest integer]

(Given : R = 0.082 L atm K −- 1 mol −- 1)

(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)

Step-by-Step Verified Solution

We know, 760 Torr = 1 atm

∴\therefore 32 Torr = 32760{{32} \over {760}} atm

As all the liquid water evaporates so entire water is in gaseous state.

∴\therefore Weight of water vapour = 0.9 g

∴\therefore Moles of water vapour (n) = 0.918{{0.9} \over {18}}

Pressure (P) = 32760{{32} \over {760}} atm

Temperature (T) = (27 + 273) K = 300 K

R = 0.082 L atm K −- 1 mol −- 1

Given water vapour act as an ideal gas, so we can apply ideal gas equation.

From ideal gas equation,

PV = nRT

⇒32760×v=0.918×0.082×300\Rightarrow {{32} \over {760}} \times v = {{0.9} \over {18}} \times 0.082 \times 300

⇒v=29\Rightarrow v = 29 L

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