JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Heat & Thermodynamics

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The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T . Consider R as universal gas constant. The pressure of the mixture of gases is :

(A) $\frac{3 R T}{V}$
(B) $\frac{5}{2} \frac{RT}{V}$ ✓ Correct
(C) $\frac{88 R T}{V}$
(D) $\frac{4 R T}{V}$

Answer: (B)

Solution

The gases share the volume and the temperature, and their partial pressures add, so the mixture obeys

P V = n R T

with

n

the total number of moles.

n = \frac{16}{32} + \frac{28}{28} + \frac{44}{44} = 0.5 + 1 + 1 = 2.5

P = \frac{n R T}{V} = \frac{5}{2} \frac{R T}{V}

The pressure of the mixture is

\frac{5}{2} \frac{R T}{V}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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