JEE Physics 2025 Question with Solution

Given: mass of CO$_2$ gas = $50 \, \text{g}$, temperature = $17^\circ \text{C}$.

Find: total translational kinetic energy of the molecules.

For an ideal gas, the total translational kinetic energy is

$$E = \frac{3}{2}nRT$$

where $n$ is the number of moles.

First, calculate the number of moles:

$$n = \frac{50}{44} = 1.136 \, \text{mol}$$

Convert temperature to Kelvin:

$$T = 17 + 273 = 290 \, \text{K}$$

Now substitute in the formula:

$$E = \frac{3}{2} \times 1.136 \times 8.314 \times 290$$ $$E = 4102.8 \, \text{J}$$

Therefore, the translational kinetic energy is $4102.8 \, \text{J}$. The correct option is A.

Given: mass of CO$_2$ gas = $50 \, \text{g}$, temperature = $17^\circ \text{C}$.

Find: total translational kinetic energy of all molecules.

The translational kinetic energy per molecule is

$$\frac{3}{2}kT$$

and total translational kinetic energy is

$$E = \left(\frac{3}{2}kT\right) \times N$$

The number of molecules is obtained from moles:

$$N = \frac{50}{44} \times 6.023 \times 10^{23}$$

Using this relation gives a value close to $4102.8 \, \text{J}$.

The first approach in the source shows $4108.644 \, \text{J}$, but the detailed substitution with $E = \frac{3}{2}nRT$ gives $4102.8 \, \text{J}$, which matches option A and is the defensible answer.