Match List - I with List - II. Given V1 and V2 are initial and final volumes respectively. List - I (Isothermal process) Reversible expansion Free expansion Irreversible Compression Cyclic reversible List - II (Expression) q=0 q = nRT ln V2/V1 w = -pext (V1 - V2) qrev/T = 0 Choose the correct answer from the options given below :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q54 – JEE-MAIN 2026

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Home Chemistry ThermodynamicsQuestion #631

Match List - I with List - II. Given

V_{1}

and

V_{2}

are initial and final volumes respectively.

\begin{array}{lcl} List - I & List - II \\ A. (Isothermal process) Reversible expansion & I. q = 0 \\ B. Free expansion & II. q = n R T \ln \frac{V_{2}}{V_{1}} \\ C. Irreversible Compression & III. w = - p_{e x t} (V_{1} - V_{2}) \\ D. Cyclic reversible & IV. \frac{q_{r e v}}{T} = 0 \end{array}

Choose the correct answer from the options given below :

Clue 1 of 3: Understand Thermodynamic Processes

Recall the definitions and characteristics of isothermal, free expansion, irreversible compression, and cyclic reversible processes.

🧠 Full Solution Path

Step 1: Analyze Isothermal Reversible Expansion and Free Expansion✦ Active

For an isothermal reversible expansion of an ideal gas, the change in internal energy is zero, $Δ U = 0$ . According to the first law of thermodynamics, $Δ U = q + w$ . Thus, $q = - w$ . The work done in a reversible expansion is $w = - n R T \ln \frac{V_{2}}{V_{1}}$ . Therefore, $q = n R T \ln \frac{V_{2}}{V_{1}}$ . This matches A with II.

For free expansion, the external pressure is zero ( $P_{e x t} = 0$ ). Consequently, the work done is $w = - P_{e x t} Δ V = 0$ . If the free expansion occurs in a vacuum, it is also adiabatic, meaning $q = 0$ . This matches B with I.

💡 Teacher's Secret Hint

Remember that for an ideal gas, $Δ U = 0$ for any isothermal process, not just reversible ones.

Step 2: Analyze Irreversible Compression and Cyclic Reversible Process○ Expand

For irreversible compression, the work done is generally given by $w = - P_{e x t} Δ V = - P_{e x t} (V_{f i n a l} - V_{i n i t i a l})$ . Given $V_{1}$ is initial and $V_{2}$ is final, $w = - P_{e x t} (V_{2} - V_{1})$ . The expression in List-II, III is $w = - P_{e x t} (V_{1} - V_{2})$ . While there might be a sign convention difference (as this is equivalent to $w = P_{e x t} (V_{2} - V_{1})$ ), the form $P_{e x t} \times volume change$ is characteristic of irreversible work. By elimination, this is the intended match. This matches C with III.

For any cyclic process, the system returns to its initial state, so the net change in state functions is zero. For a reversible cyclic process, the change in entropy of the system is zero, $Δ S_{s y s} = \oint \frac{d q_{r e v}}{T} = 0$ . This matches D with IV.

💡 Teacher's Secret Hint

Be careful with sign conventions for work. The expression $w = - P_{e x t} (V_{1} - V_{2})$ is equivalent to $w = P_{e x t} (V_{2} - V_{1})$ . If $V_{1}$ is initial and $V_{2}$ is final, the standard work done by the system is $w = - P_{e x t} (V_{2} - V_{1})$ . The given expression has the opposite sign.

Step 3: Consolidate Matches and Select Option○ Expand

Based on the analysis:

\begin{array}{l} A \to II \\ B \to I \\ C \to III \\ D \to IV \end{array}

Comparing these matches with the given options, Option 3 (A-II, B-I, C-III, D-IV) is the correct answer.

💡 Teacher's Secret Hint

Always double-check all matches, especially when there might be subtle ambiguities in definitions or sign conventions.

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Chemistry Question 54 – JEE-MAIN 2026

Clue 1 of 3: Understand Thermodynamic Processes

🧠 Full Solution Path