JEE Physics 2026 Question with Solution

Given: $n = 10$ mole, $P_1 = 21.7 \, \text{Pa}$, $P_2 = 30 \, \text{Pa}$, $C_v = 21 \, \text{J/K\cdot mol}$, $R = 8.3 \, \text{J/mol\cdot K}$. From the graph, the process from $P_1$ to $P_2$ is at constant volume with $V = 1 \, \text{m}^3$.

Find: The heat involved, $Q = \alpha$.

At constant volume, work done is

$$W = 0$$

Using the first law of thermodynamics,

$$Q = \Delta U + W$$

So,

$$Q = \Delta U$$

For an ideal gas,

$$\Delta U = n C_v \Delta T$$

At constant volume,

$$\frac{P}{T} = \text{constant}$$

Hence,

$$Q = n C_v \frac{(P_2 - P_1)V}{R}$$

Substitute the values:

$$Q = \frac{10 \times 21 \times (30 - 21.7)}{8.3}$$ $$Q = 21 \, \text{J}$$

Therefore, the value of $\alpha$ is $21$, so the correct option is B.

Given: The process is constant volume. Also, $n = 10$, $C_v = 21 \, \text{J/K\cdot mol}$, $P_1 = 21.7 \, \text{Pa}$, $P_2 = 30 \, \text{Pa}$, $R = 8.3 \, \text{J/mol\cdot K}$, and from the graph $V = 1 \, \text{m}^3$.

Find: $Q$.

Since volume is constant,

$$W = 0$$

So,

$$Q = \Delta U = n C_v \Delta T$$

Now from the ideal gas equation,

$$PV = nRT$$

At constant volume,

$$T = \frac{PV}{nR}$$

Therefore,

$$\Delta T = T_2 - T_1 = \frac{P_2V}{nR} - \frac{P_1V}{nR}$$ $$\Delta T = \frac{(P_2 - P_1)V}{nR}$$

Substituting into $Q = n C_v \Delta T$,

$$Q = n C_v \cdot \frac{(P_2 - P_1)V}{nR}$$ $$Q = C_v \frac{(P_2 - P_1)V}{R}$$

Using the numerical substitution exactly as shown in the solution,

$$Q = \frac{10 \times 21 \times (30 - 21.7)}{8.3} = 21 \, \text{J}$$

Thus, the heat involved is $21 \, \text{J}$, and the correct option is B.