JEE Main · 2022 · 29 Jun 2022 · Shift 2 · States of Matter

A box contains 0.90 g of liquid water in equilibrium with water vapour at

2 7^{\circ} C

. The equilibrium vapour pressure of water at

2 7^{\circ} C

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.)

Answer: 29

Solution

Using the ideal gas equation

P V = n R T

:

Given:

Mass of water = 0.90 g
Molar mass of water = 18 g/mol
Number of moles: $n = \frac{0.90}{18} = 0.05$ mol
Temperature: $T = 27 + 273 = 300$ K
Pressure: $P = 32$ Torr $= \frac{32}{760}$ atm
$R = 0.082$ L atm K $^{- 1}$ mol $^{- 1}$

Substituting into the ideal gas equation:

V = \frac{n R T}{P} = \frac{0.05 \times 0.082 \times 300 \times 760}{32}

V = \frac{0.05 \times 0.082 \times 300 \times 760}{32} = 29.21 L

Rounding to the nearest integer: 29 litres

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 29 Jun 2022 · Shift 2 · States of Matter

Solution

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