JEE Physics 2023 Question with Solution

Given: Heat supplied is $Q = 735 \, \text{J}$. The gas is diatomic, expands at constant pressure, rotates but does not oscillate.

Find: Increase in internal energy $\Delta U$.

For a diatomic gas with rotational degrees of freedom active and vibrational modes absent,

$$f = 5$$

So,

$$C_V = \frac{f}{2}R = \frac{5}{2}R, \qquad C_P = C_V + R = \frac{7}{2}R$$

At constant pressure,

$$Q = n C_P \Delta T$$

and the increase in internal energy is

$$\Delta U = n C_V \Delta T$$

Therefore,

$$\Delta U = \frac{C_V}{C_P} Q = \frac{\frac{5}{2}R}{\frac{7}{2}R} \times 735 = \frac{5}{7} \times 735 = 525 \, \text{J}$$

Therefore, the increase in internal energy is $525 \, \text{J}$. The correct option is A.

The solution contains a discrepancy because it labels the correct option as C, but the worked value is $525 \, \text{J}$, which matches option A.

Given: The supplied heat is $735 \, \text{J}$ at constant pressure.

For a diatomic gas,

$$Q = n C_P \Delta T$$

and

$$\Delta U = n C_V \Delta T$$

Hence,

$$\frac{\Delta U}{Q} = \frac{C_V}{C_P}$$

For a diatomic gas without vibration,

$$C_V = \frac{5}{2}R, \qquad C_P = \frac{7}{2}R$$

So,

$$\frac{\Delta U}{Q} = \frac{5}{7}$$

Thus,

$$\Delta U = \frac{5}{7} \times 735 = 525 \, \text{J}$$

Therefore, the internal energy increases by $525 \, \text{J}$.