0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume $10 \mathrm{~m}^{3}$ at 1000 K. Given R is the gas constant in $\mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{x}$ is

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · States of Matter

0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume

10 m^{3}

at 1000 K. Given R is the gas constant in

JK^{- 1} mol^{- 1}, x

(A) $\frac{2 R}{4 + R}$
(B) $\frac{2 R}{4 - R}$
(C) $\frac{4 + R}{2 R}$
(D) $\frac{4 - R}{2 R}$ ✓ Correct

Answer: (D)

Solution

Total moles follow from the ideal gas law,

n = \frac{P V}{R T} = \frac{200 \times 10}{R \times 1000} = \frac{2}{R}

. This total is

0.5 + x

, so

x = \frac{2}{R} - 0.5 = \frac{4 - R}{2 R}

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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