JEE Main · 2022 · 27 Jul 2022 · Shift 2 · States of Matter

For a real gas at

2 5^{\circ} C

temperature and high pressure (99 bar) the value of compressibility factor is 2, so the value of Vander Waal's constant 'b' should be

____

\times 1 0^{- 2} L mol^{- 1}

(Nearest integer)
(Given

R = 0.083 L

bar

K^{- 1} mol^{- 1}

)

Answer: 25

Solution

For a real gas at high pressure, the compressibility factor is given by:

Z = 1 + \frac{P b}{R T}

Given Z = 2, we can solve for b:

2 = 1 + \frac{P b}{R T}

1 = \frac{P b}{R T}

b = \frac{R T}{P} = \frac{0.083 \times 298}{99}

b = \frac{24.734}{99} = 0.2498 \approx 0.25

L mol⁻¹

b = 25 \times 1 0^{- 2}

L mol⁻¹

Therefore, x = 25

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer

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