JEE MAIN - Chemistry (2022 - 29th June Evening Shift - No. 14)
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.)
Explanation
We know, 760 Torr = 1 atm
$$\therefore$$ 32 Torr = $${{32} \over {760}}$$ 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) = $${{0.9} \over {18}}$$
Pressure (P) = $${{32} \over {760}}$$ 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
$$ \Rightarrow {{32} \over {760}} \times v = {{0.9} \over {18}} \times 0.082 \times 300$$
$$ \Rightarrow v = 29$$ L
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