JEE Main 2026 — Physics Thermodynamics

2026mcqmedium

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 $]$

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$0.15$ and $0.18$
B
$0.25$ and $0.08$
C
$0.21$ and $0.12$
D
$0.13$ and $0.20$

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Solution

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 $0.12$

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Solution

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 2026 — Physics Thermodynamics

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Solution

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JEE Main 2026 — Physics Thermodynamics

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics