JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Heat & Thermodynamics

A cylindrical container of volume

4.0 \times 1 0^{- 3} m^{3}

contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K . The pressure of the mixture of gases is: [Take gas constant as

8.3 J mol^{- 1} K^{- 1}

]

(A)24.9 Pa
(B) $24.9 \times 1 0^{3} Pa$
(C) $24.9 \times 1 0^{5} Pa$ ✓ Correct
(D) $249 \times 1 0^{1} Pa$

Answer: (C)

Solution

The partial pressures of the two gases add, so the ideal gas equation applies to the total number of moles,

n = 1 + 2 = 3

P = \frac{n R T}{V} = \frac{3 \times 8.3 \times 400}{4.0 \times 1 0 ^{- 3}} = \frac{9960}{4.0 \times 1 0 ^{- 3}} = 24.9 \times 1 0^{5} Pa

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option

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