JEE Challenger
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PhyMedAdv 2017 • Paper 1

Thermodynamic Process Representation for First Law Formula

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}

Which of the following choices gives the only valid representation of a process governed by ΔU=ΔQ−PΔV\Delta U = \Delta Q - P \Delta V?

Question Diagram 1

Options

A

(II) (iv) (R)

B

(III) (iii) (P)

C

(II) (iii) (S)

D

(II) (iii) (P)

Correct

Topics & Concepts

ThermodynamicsFirst Law of ThermodynamicsHeat Transfer

Step-by-Step Solution

To determine the correct representation of the thermodynamic process governed by the equation ΔU=ΔQ−PΔV\Delta U = \Delta Q - P \Delta V, we analyze the first law of thermodynamics and match the corresponding entries across the three columns.

  1. First Law of Thermodynamics: The general statement of the First Law of Thermodynamics for a system is given by: ΔU=ΔQ+W1→2\Delta U = \Delta Q + W_{1 \rightarrow 2} where ΔQ\Delta Q is the heat supplied to the system, ΔU\Delta U is the change in internal energy, and W1→2W_{1 \rightarrow 2} is the work done on the system.

  2. Matching Column 1: Comparing the given equation ΔU=ΔQ−PΔV\Delta U = \Delta Q - P \Delta V with the general First Law equation, we identify the work done on the system as: W1→2=−PΔV=−P(V2−V1)=−PV2+PV1W_{1 \rightarrow 2} = -P \Delta V = -P (V_2 - V_1) = -P V_2 + P V_1 This expression directly corresponds to (II) in Column 1.

  3. Matching Column 2: The work expression W1→2=−P(V2−V1)W_{1 \rightarrow 2} = -P (V_2 - V_1) assumes a constant pressure PP during the transition from state 1 to state 2. A process occurring at constant pressure is defined as an Isobaric process. This corresponds to (iii) in Column 2.

  4. Matching Column 3: On a P−VP-V diagram, an isobaric process moving from state 1 to state 2 is represented by a horizontal line (since pressure PP remains constant while volume VV changes).

    • In diagram (P), path 1→21 \rightarrow 2 is a horizontal line indicating constant pressure, which represents an isobaric process.
    • Diagram (Q) represents a curve (isothermal/adiabatic expansion).
    • Diagram (R) represents a curve (isothermal/adiabatic expansion).
    • Diagram (S) represents a vertical line (isochoric process at constant volume).

Therefore, diagram (P) correctly depicts the isobaric process.

Combining these matching entries yields: (II)→(iii)→(P)\mathbf{(II) \rightarrow (iii) \rightarrow (P)}

Thus, the correct option is D.

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