JEE Physics 2024 Question with Solution

Given: An ideal gas undergoes a cyclic process A → B → C → A on a $P$-$V$ diagram.

Find: The total work done in the cycle.

The solution identifies the coordinates as:

• A: $P_A = 10 \, \text{kPa}, \; V_A = 10 \, \text{dm}^3$
• B: $P_B = 10 \, \text{kPa}, \; V_B = 30 \, \text{dm}^3$
• C: $P_C = 30 \, \text{kPa}, \; V_C = 10 \, \text{dm}^3$

Use

$$W = \int P \, dV$$

and add the work for each segment:

$$W_{\text{total}} = W_{A \to B} + W_{B \to C} + W_{C \to A}$$

For A → B, pressure is constant:

$$W_{A \to B} = P_A (V_B - V_A)$$ $$W_{A \to B} = (10 \, \text{kPa})(30 - 10 \, \text{dm}^3) = 200 \, \text{J}$$

For C → A, volume is constant, so

$$W_{C \to A} = 0$$

For B → C, the path is a straight line and the work equals the trapezoid area under the line. Since volume decreases, the work done by the gas is negative:

$$|W_{B \to C}| = \frac{1}{2}(P_B + P_C)(V_B - V_C)$$ $$|W_{B \to C}| = \frac{1}{2}(10 + 30)(30 - 10) = 400 \, \text{J}$$

Therefore,

$$W_{B \to C} = -400 \, \text{J}$$

Now sum all contributions:

$$W_{\text{total}} = 200 + (-400) + 0 = -200 \, \text{J}$$

The negative sign indicates that the net work is done on the gas. Therefore, the total work done by the gas in the cycle is $-200 \, \text{J}$.

The answer key shows $200 \, \text{J}$, but the detailed working in the solution concludes $-200 \, \text{J}$. By the given extraction rule, the solution is treated.

Given: The process is cyclic on a $P$-$V$ diagram.

Find: Net work done in one complete cycle.

For a cyclic process, the net work done equals the signed area enclosed by the loop on the $P$-$V$ graph. The enclosed figure is a right triangle with base

$$30 - 10 = 20 \, \text{dm}^3$$

and height

$$30 - 10 = 20 \, \text{kPa}$$

So the magnitude of enclosed area is

$$W = \frac{1}{2} \times 20 \times 20 = 200 \, \text{kPa} \cdot \text{dm}^3 = 200 \, \text{J}$$

However, the traversal A → B → C → A is anticlockwise, so the work done by the gas is negative.

$$W_{\text{total}} = -200 \, \text{J}$$

Therefore, the total work done in the process is $-200 \, \text{J}$.