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Kinetic Theory

JEE Main level, single correct, medium

4 of 168

Question 4 of 168

A mixture of carbon dioxide and oxygen has volume $8310$ cm $^3$ , temperature $300$ K, pressure $100$ kPa and mass $13.2$ g. The number of moles of carbon dioxide and oxygen gases in the mixture respectively are _______.
(Assume both carbon dioxide and oxygen gases behave like ideal gases) $[R = 8.31$ J/mol.K $]$

Show answer and solution

Answer: (C) $0.21$ and $0.12$

Using the ideal gas equation, the total number of moles of the mixture is given by:

n = \dfrac{PV}{RT}

Substituting the given values:

P = 100 \text{ kPa} = 10^5 \text{ Pa}

V = 8310 \text{ cm}^3 = 8.31 \times 10^{-3} \text{ m}^3

T = 300 \text{ K}

R = 8.31 \text{ J/(mol}\cdot\text{K)}

n = \dfrac{10^5 \times 8.31 \times 10^{-3}}{8.31 \times 300} = \dfrac{100}{300} = \dfrac{1}{3} \text{ mol}

Let

n_1

and

n_2

be the number of moles of

CO_2

and

O_2

respectively.

n_1 + n_2 = \dfrac{1}{3}

The total mass of the mixture is

13.2 \text{ g}

. Using the molar masses of

CO_2

(

44 \text{ g/mol}

) and

O_2

(

32 \text{ g/mol}

44n_1 + 32n_2 = 13.2

Substituting

n_2 = \dfrac{1}{3} - n_1

into the mass equation:

44n_1 + 32\left(\dfrac{1}{3} - n_1\right) = 13.2

12n_1 + \dfrac{32}{3} = 13.2

12n_1 = \dfrac{39.6 - 32}{3} = \dfrac{7.6}{3}

n_1 = \dfrac{7.6}{36} = \dfrac{19}{90} \approx 0.21 \text{ mol}

The number of moles of

O_2

is:

n_2 = \dfrac{1}{3} - \dfrac{19}{90} = \dfrac{11}{90} \approx 0.12 \text{ mol}

The number of moles of

CO_2

and

O_2

are

0.21

and

0.12

respectively.

Answer:

0.21

and

0.12

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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