JEE Main 2026 — Kinetic Theory of Gases Question with Solution

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

$n=\frac{PV}{RT}$

Substituting the given values:
$P=100\text{ kPa}=10^5\text{ Pa}$
$V=8310{\text{ cm}}^3=8.31\times 10^{-3}{\text{ m}}^3$
$T=300\text{ K}$
$R=8.31\text{ J/(mol}\cdot \text{K)}$

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

Let $n_1$ and $n_2$ be the number of moles of $C{O}_2$ and $O_2$ respectively.

$n_1+n_2=\frac{1}{3}$

The total mass of the mixture is $13.2\text{ g}$. Using the molar masses of $C{O}_2$ ($44\text{ g/mol}$) and $O_2$ ($32\text{ g/mol}$):

$44n_1+32n_2=13.2$

Substituting $n_2=\frac{1}{3}-n_1$ into the mass equation:

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

$12n_1+\frac{32}{3}=13.2$

$12n_1=\frac{39.6-32}{3}=\frac{7.6}{3}$

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

The number of moles of $O_2$ is:

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

The number of moles of $C{O}_2$ and $O_2$ are $0.21$ and $0.12$ respectively.

Answer: $0.21$ and $0.12$