JEE Challenger
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PhyMedAdv 2017 • Paper 1

Matching Work Done Process and Curve Combination

Comprehension Passage

Answer questions 16, 17, and 18 by correctly matching the information given across the three columns in the following table.

An ideal gas undergoes a thermodynamic cyclic process along different paths as depicted in the corresponding P−VP-V diagrams in Column 3. Consider only the path from state 1 to state 2. Here, WW denotes the work done on the system. Standard notations are used for equations and plots in thermodynamics, where γ\gamma represents the ratio of specific heat capacities (Cp/CvC_p / C_v) and nn is the amount of gas in moles.

Column 1Column 2Column 3(I) W1→2=1γ−1(P2V2−P1V1)(i) Isothermal(P)(II) W1→2=−PV2+PV1(ii) Isochoric(Q)(III) W1→2=0(iii) Isobaric(R)(IV) W1→2=−nRTln⁡(V2V1)(iv) Adiabatic(S)\begin{array}{|l|l|l|} \hline \text{Column 1} & \text{Column 2} & \text{Column 3} \\ \hline \text{(I) } W_{1 \rightarrow 2} = \frac{1}{\gamma - 1} (P_2 V_2 - P_1 V_1) & \text{(i) Isothermal} & \text{(P)} \\ \hline \text{(II) } W_{1 \rightarrow 2} = -P V_2 + P V_1 & \text{(ii) Isochoric} & \text{(Q)} \\ \hline \text{(III) } W_{1 \rightarrow 2} = 0 & \text{(iii) Isobaric} & \text{(R)} \\ \hline \text{(IV) } W_{1 \rightarrow 2} = -n R T \ln \left(\frac{V_2}{V_1}\right) & \text{(iv) Adiabatic} & \text{(S)} \\ \hline \end{array}

Identify the option that represents a correct combination among the three columns:

Question Diagram 1

Options

A

(IV) (ii) (S)

B

(III) (ii) (S)

Correct
C

(II) (iv) (P)

D

(II) (iv) (R)

Topics & Concepts

ThermodynamicsFirst Law of ThermodynamicsHeat Transfer

Step-by-Step Solution

To determine the correct combination among the three columns, we analyze the definition of work done on the system (WW) for different thermodynamic processes along the path 1→21 \rightarrow 2.

By definition, the work done on the system during a transition from state 1 to state 2 is given by: W1→2=−∫V1V2P dVW_{1 \rightarrow 2} = -\int_{V_1}^{V_2} P \, dV

1. Isochoric Process (Column 2: (ii))

  • Definition: An isochoric process occurs at a constant volume (V=constantV = \text{constant}), meaning dV=0dV = 0.

  • Work Done: W1→2=−∫V1V2P dV=0W_{1 \rightarrow 2} = -\int_{V_1}^{V_2} P \, dV = 0 This matches Column 1: (III).

  • P−VP-V Diagram: In a P−VP-V diagram, a process at constant volume is depicted as a vertical straight line parallel to the pressure axis. In Column 3: (S), the path 1→21 \rightarrow 2 is a vertical line indicating constant volume.

2. Verification of the Combination (III) (ii) (S):

  • Column 1 (III): W1→2=0W_{1 \rightarrow 2} = 0
  • Column 2 (ii): Isochoric process
  • Column 3 (S): Diagram representing a vertical path from state 1 to state 2 (V=constantV = \text{constant})

All three entries consistently describe an isochoric process.

Thus, the correct combination is (III) (ii) (S).

Correct Option: B

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