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Heat and Thermodynamics question

2018 · Shift 2 · Q52
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Heat and Thermodynamics question

2018 · Shift 2 · Q52

JEE AdvancedPhysicsHeat and ThermodynamicsMCQ+3 / −1
One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PVPVPV-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-I with the corresponding statements in List-II.

JEE Advanced 2018 Paper 2 Offline Physics - Heat and Thermodynamics Question 55 English

LIST - I LIST - II
P. In process I 1. Work done by the gas is zero
Q. In process II 2. Temperature of the gas remains
unchanged
R. In process III 3. No heat is exchanged between
the gas and its surroundings
S. In process IV 4. Work done by the gas is 6P0V0
  1. A
    P→4;Q→3;R→1;S→2P \to 4;Q \to 3;R \to 1;S \to 2P→4;Q→3;R→1;S→2
  2. B
    P→1;Q→3;R→2;S→4P \to 1;Q \to 3;R \to 2;S \to 4P→1;Q→3;R→2;S→4
  3. C
    P→3;Q→4;R→1;S→2P \to 3;Q \to 4;R \to 1;S \to 2P→3;Q→4;R→1;S→2
  4. D
    P→3;Q→4;R→2;S→P \to 3;Q \to 4;R \to 2;S \toP→3;Q→4;R→2;S→
View written solutionFree

Correct answer: C

  1. Identify each process from the PVPVPV diagram qualitatively

    In a PVPVPV diagram:

    • Isochoric process ightarrowV= ightarrow V=ightarrowV= constant ightarrow ightarrowightarrow vertical line
    • Isobaric process ightarrowP= ightarrow P=ightarrowP= constant ightarrow ightarrowightarrow horizontal line
    • Isothermal process ightarrowPV= ightarrow PV=ightarrowPV= constant ightarrow ightarrowightarrow gentler falling curve
    • Adiabatic process ightarrowPVγ= ightarrow PV^\gamma=ightarrowPVγ= constant ightarrow ightarrowightarrow steeper falling curve than isothermal

  2. Match the statements in List-II with thermodynamic meanings

    • Statement 1: Work done by gas is zero W=∫P dV=0W=\int P\,dV=0W=∫PdV=0 This happens for an isochoric process.

    • Statement 2: Temperature remains unchanged This is an isothermal process.

    • Statement 3: No heat exchanged Q=0Q=0Q=0 This is an adiabatic process.

    • Statement 4: Work done by gas is 6P0V06P_0V_06P0​V0​ For an isobaric process, W=PΔVW=P\Delta VW=PΔV so this corresponds to the horizontal line whose area under curve is 6P0V06P_0V_06P0​V0​.

  3. Read the diagram-based process labels

    From the given schematic:

    • Process I is the adiabatic curve (steeper falling curve)
    • Process II is the isobaric process
    • Process III is the isochoric process
    • Process IV is the isothermal curve

  4. Now match them

    • PPP : Process I →\rightarrow→ adiabatic →3\rightarrow 3→3
    • QQQ : Process II →\rightarrow→ isobaric →4\rightarrow 4→4
    • RRR : Process III →\rightarrow→ isochoric →1\rightarrow 1→1
    • SSS : Process IV →\rightarrow→ isothermal →2\rightarrow 2→2

    Hence, P→3,Q→4,R→1,S→2P\to 3,\quad Q\to 4,\quad R\to 1,\quad S\to 2P→3,Q→4,R→1,S→2

  5. Correct option

    This corresponds to Option C.

  6. Comparison with stored answer

    Stored correct answer: C
    Derived answer: C
    So they agree.

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