Physics Q33 – JEE-MAIN 2025

Home Physics ThermodynamicsQuestion #529

An ideal gas exists in a state with pressure

P_{0}

, volume

V_{0}

. It is isothermally expanded to 4 times of its initial volume(

V_{0}

), then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is

Clue 1 of 3: Cyclic Process

For a cyclic process, the net change in internal energy is zero.

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Ninja Strategy•Component Analysis and Sign Check

Calculate the work done for each process and sum them. The net work must be positive, and the specific terms for isothermal work ( $2 P_{0} V_{0} \ln 2$ ) and isobaric work ( $- 0.75 P_{0} V_{0}$ ) can be directly matched to the correct option.

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Step 1: Identify States and Processes✦ Active

Let the initial state be State 1: $(P_{0}, V_{0}, T_{0})$ . The process consists of three steps:

1. **Isothermal Expansion (1 $\to$ 2):** The gas expands from $V_{0}$ to $V_{2} = 4 V_{0}$ at constant temperature $T_{0}$ . Using $P_{0} V_{0} = P_{2} V_{2}$ , we find $P_{2} = P_{0} / 4$ . So, State 2 is $(P_{0} / 4, 4 V_{0}, T_{0})$ .

2. **Isobaric Compression (2 $\to$ 3):** The gas is compressed from $4 V_{0}$ to $V_{3} = V_{0}$ at constant pressure $P_{2} = P_{0} / 4$ . Using $\frac{V_{2}}{T_{2}} = \frac{V_{3}}{T_{3}}$ , we find $\frac{4 V_{0}}{T_{0}} = \frac{V_{0}}{T_{3}}$ , so $T_{3} = T_{0} / 4$ . Thus, State 3 is $(P_{0} / 4, V_{0}, T_{0} / 4)$ .

3. **Isochoric Heating (3 $\to$ 1):** The gas is heated at constant volume $V_{0}$ to return to its initial state $(P_{0}, V_{0}, T_{0})$ . This is consistent as $\frac{P_{3}}{T_{3}} = \frac{P_{0} / 4}{T_{0} / 4} = \frac{P_{0}}{T_{0}}$ .

Step 2: Calculate Work Done for Each Process○ Expand

The total heat exchanged in a cyclic process is equal to the net work done by the system ( $Q_{c y c l e} = W_{c y c l e}$ ). We calculate the work done for each step:

1. **Work done in isothermal expansion (1 $\to$ 2):**

W_{12} = n R T_{0} \ln (\frac{V_{2}}{V_{0}}) = P_{0} V_{0} \ln (\frac{4 V_{0}}{V_{0}}) = P_{0} V_{0} \ln 4 = 2 P_{0} V_{0} \ln 2

2. **Work done in isobaric compression (2 $\to$ 3):**

W_{23} = P_{2} (V_{3} - V_{2}) = (P_{0} / 4) (V_{0} - 4 V_{0}) = (P_{0} / 4) (- 3 V_{0}) = - 3 P_{0} V_{0} / 4 = - 0.75 P_{0} V_{0}

3. **Work done in isochoric heating (3 $\to$ 1):**

W_{31} = 0 (since volume is constant)

💡 Teacher's Secret Hint

Remember that for an ideal gas, $n R T = P V$ can be used to substitute terms.

Step 3: Calculate Total Heat Exchanged○ Expand

The total work done in the cyclic process is the sum of work done in each step:

W_{c y c l e} = W_{12} + W_{23} + W_{31}

W_{c y c l e} = 2 P_{0} V_{0} \ln 2 - 0.75 P_{0} V_{0} + 0

W_{c y c l e} = P_{0} V_{0} (2 \ln 2 - 0.75)

Since $Q_{c y c l e} = W_{c y c l e}$ for a cyclic process, the total heat exchanged is:

Q_{c y c l e} = P_{0} V_{0} (2 \ln 2 - 0.75)

💡 Teacher's Secret Hint

Ensure to correctly identify the sign of work done (positive for expansion, negative for compression).

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Physics Question 33 – JEE-MAIN 2025

Clue 1 of 3: Cyclic Process

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