JEE Physics 2026 Question with Solution

Given: $Q = 300 \, J$, $C_p = \dfrac{7}{2}R$, temperature rises from $20^\circ C$ to $50^\circ C$ at constant volume, and $R = 8.314 \, J/mol\cdot K$.

Find: Mass of the gas.

At constant volume, use $C_v$ instead of $C_p$.

For an ideal gas,

$$C_p - C_v = R$$

So,

$$C_v = C_p - R = \frac{7}{2}R - R = \frac{5}{2}R$$

Now use the heat equation at constant volume:

$$Q = n C_v \Delta T$$

The temperature change is

$$\Delta T = 50 - 20 = 30 \, \text{K}$$

Substituting the values,

$$300 = n \times \frac{5}{2} \times 8.314 \times 30$$

Hence,

$$n \approx 0.48 \, \text{mol}$$

Now calculate the mass of the gas. Using the solution's stated approximate molar mass,

$$M \approx 28 \, \text{g/mol}$$

Therefore,

$$\text{Mass} = nM \approx 0.48 \times 28 \approx 4 \, \text{g}$$

Therefore, the mass of the gas is approximately $4 \, \text{g}$.

Given: Heat is supplied at constant volume.

Find: The mass of the gas.

The key point is that constant-volume heating uses $C_v$, not $C_p$.

From the ideal-gas relation,

$$C_p - C_v = R$$

With $C_p = \frac{7}{2}R$,

$$C_v = \frac{7}{2}R - R = \frac{5}{2}R$$

The rise in temperature is

$$\Delta T = 50 - 20 = 30 \, \text{K}$$

Now apply

$$Q = n C_v \Delta T$$

So,

$$300 = n \left(\frac{5}{2} \times 8.314\right) \times 30$$

This gives

$$n \approx 0.48 \, \text{mol}$$

The provided solution then uses the approximate molar mass $M \approx 28 \, \text{g/mol}$, so

$$\text{Mass} = nM \approx 0.48 \times 28 \approx 4 \, \text{g}$$

Thus, the required numerical answer is 4.