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MCQEasyJEE 2026Internal Energy & Enthalpy

JEE Chemistry 2026 Question with Solution

Match the LIST-I with LIST-II for an isothermal process of an ideal gas system.

List-IList-IIWork done (Vf>Vi)A.Reversible expansionI.w=0B.Free expansionII.w=−nRTln⁡ ⁣(VfVi)C.Irreversible expansionIII.w=−Pex(Vf−Vi)D.Irreversible compressionIV.w=−Pex(Vi−Vf)\begin{array}{|c|l||c|l|} \hline \text{List-I} & & \text{List-II} & \text{Work done } (V_f > V_i) \\ \hline \text{A.} & \text{Reversible expansion} & \text{I.} & w = 0 \\ \text{B.} & \text{Free expansion} & \text{II.} & w = -nRT\ln\!\left(\dfrac{V_f}{V_i}\right) \\ \text{C.} & \text{Irreversible expansion} & \text{III.} & w = -P_{ex}(V_f - V_i) \\ \text{D.} & \text{Irreversible compression} & \text{IV.} & w = -P_{ex}(V_i - V_f) \\ \hline \end{array}

Choose the correct answer from the options given below:

  • A

    A-II, B-I, C-III, D-IV

  • B

    A-IV, B-I, C-III, D-II

  • C

    A-I, B-III, C-II, D-IV

  • D

    A-IV, B-II, C-III, D-I

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: An isothermal process of an ideal gas system is to be matched with the corresponding work done expressions.

Find: The correct matching between LIST-I and LIST-II, and hence the correct option.

Step 1: Reversible isothermal expansion For a reversible isothermal expansion of an ideal gas, the work done is

w=−nRTln⁡ ⁣(VfVi)w = -nRT\ln\!\left(\frac{V_f}{V_i}\right)

So, A →\rightarrow II.

Step 2: Free expansion In free expansion, the external pressure is zero, so no work is done.

w=0w = 0

So, B →\rightarrow I.

Step 3: Irreversible expansion For irreversible expansion against constant external pressure,

w=−Pex(Vf−Vi)w = -P_{\text{ex}}(V_f - V_i)

So, C →\rightarrow III.

Step 4: Irreversible compression For irreversible compression,

w=−Pex(Vi−Vf)w = -P_{\text{ex}}(V_i - V_f)

So, D →\rightarrow IV.

Therefore, the correct matching is A-II, B-I, C-III, D-IV. The correct option is A.

Match by process type

Given: Four isothermal processes of an ideal gas and four work expressions.

Find: Which process corresponds to which work expression.

For an ideal gas undergoing an isothermal change:

  • Reversible expansion gives the logarithmic expression for work.
  • Free expansion occurs against zero external pressure, so work is zero.
  • Irreversible expansion against a constant external pressure uses the constant-pressure form.
  • Irreversible compression has the same constant external pressure idea, but with the volume change written as (Vi−Vf)(V_i - V_f).

Thus,

  • A →\rightarrow II
  • B →\rightarrow I
  • C →\rightarrow III
  • D →\rightarrow IV

Hence, the correct option is A.

Common mistakes

  • Confusing free expansion with irreversible expansion against constant external pressure. In free expansion, Pex=0P_{\text{ex}} = 0, so w=0w = 0. Use w=−PexΔVw = -P_{\text{ex}}\Delta V only when the gas expands or compresses against a finite constant external pressure.

  • Using the logarithmic expression for every isothermal process. The formula w=−nRTln⁡ ⁣(VfVi)w = -nRT\ln\!\left(\frac{V_f}{V_i}\right) is valid only for a reversible isothermal process, not for irreversible expansion or compression.

  • Missing the sign convention in compression. For compression, the volume decreases, so writing the expression carelessly can give the wrong sign. Follow the given form carefully and distinguish between (Vf−Vi)(V_f - V_i) and (Vi−Vf)(V_i - V_f).

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