JEE Challenger
More from Thermodynamics
PhyHardAdv 2019 • Paper 2

Matching Thermodynamic Quantities for Ideal Gas in TV Diagram

Comprehension Passage

Consider a thermodynamic process performed on an ideal monatomic gas where infinitesimal heat absorbed by the gas is represented by TΔXT \Delta X, in which TT denotes the system's temperature and ΔX\Delta X represents an infinitesimal variation in a thermodynamic quantity XX. For one mole of a monatomic ideal gas, X=32Rln⁡(TTA)+Rln⁡(VVA)X = \frac{3}{2} R \ln \left(\frac{T}{T_A}\right) + R \ln \left(\frac{V}{V_A}\right), where RR is the universal gas constant, VV is the volume, and TA,VAT_A, V_A are constant parameters.

List-I details certain quantities in a process, whereas List-II provides potential values for these quantities.

List-I: (I) Work done by the system during process 1→2→31 \rightarrow 2 \rightarrow 3 (II) Internal energy change during process 1→2→31 \rightarrow 2 \rightarrow 3 (III) Heat absorbed by the system during process 1→2→31 \rightarrow 2 \rightarrow 3 (IV) Heat absorbed by the system during process 1→21 \rightarrow 2

List-II: (P) 13RT0ln⁡2\frac{1}{3} R T_0 \ln 2 (Q) 13RT0\frac{1}{3} R T_0 (R) RT0R T_0 (S) 43RT0\frac{4}{3} R T_0 (T) 13RT0(3+ln⁡2)\frac{1}{3} R T_0 (3 + \ln 2) (U) 56RT0\frac{5}{6} R T_0

If one mole of a monatomic ideal gas undergoes the process depicted in the given T−VT-V diagram with P0V0=13RT0P_0 V_0 = \frac{1}{3} R T_0, select the correct matching combination.

Question Diagram 1

Options

A

I→P,II→R,III→T,IV→S\text{I} \rightarrow \text{P}, \text{II} \rightarrow \text{R}, \text{III} \rightarrow \text{T}, \text{IV} \rightarrow \text{S}

B

I→P,II→T,III→Q,IV→T\text{I} \rightarrow \text{P}, \text{II} \rightarrow \text{T}, \text{III} \rightarrow \text{Q}, \text{IV} \rightarrow \text{T}

C

I→P,II→R,III→T,IV→P\text{I} \rightarrow \text{P}, \text{II} \rightarrow \text{R}, \text{III} \rightarrow \text{T}, \text{IV} \rightarrow \text{P}

Correct
D

I→S,II→T,III→Q,IV→U\text{I} \rightarrow \text{S}, \text{II} \rightarrow \text{T}, \text{III} \rightarrow \text{Q}, \text{IV} \rightarrow \text{U}

Topics & Concepts

ThermodynamicsFirst Law of ThermodynamicsKinetic Theory of Gases

Step-by-Step Solution

To determine the correct matching combination for the given thermodynamic process, we analyze each step in the T−VT-V diagram for 1 mole1\text{ mole} of a monatomic ideal gas (n=1n = 1 and Cv=32RC_v = \frac{3}{2}R).

Analysis of the Process:

From the given T−VT-V diagram:

  • State 1: T1=T03T_1 = \frac{T_0}{3}, V1=V0V_1 = V_0
  • State 2: T2=T03T_2 = \frac{T_0}{3}, V2=2V0V_2 = 2V_0
  • State 3: T3=T0T_3 = T_0, V3=2V0V_3 = 2V_0

Step-by-Step Calculations:

(I) Work done by the system during process 1→2→31 \rightarrow 2 \rightarrow 3 (W123W_{123}):

  • Process 1→21 \rightarrow 2 (Isothermal Expansion at T=T03T = \frac{T_0}{3}): W12=nRTln⁡(V2V1)=1⋅R⋅(T03)ln⁡(2V0V0)=13RT0ln⁡2W_{12} = n R T \ln \left(\frac{V_2}{V_1}\right) = 1 \cdot R \cdot \left(\frac{T_0}{3}\right) \ln\left(\frac{2V_0}{V_0}\right) = \frac{1}{3} R T_0 \ln 2

  • Process 2→32 \rightarrow 3 (Isochoric Heating at V=2V0V = 2V_0): Since volume remains constant, no work is done: W23=0W_{23} = 0

  • Total Work Done (W123W_{123}): W123=W12+W23=13RT0ln⁡2W_{123} = W_{12} + W_{23} = \frac{1}{3} R T_0 \ln 2

This matches with (P). Thus, I→P\text{I} \rightarrow \text{P}.

(II) Internal energy change during process 1→2→31 \rightarrow 2 \rightarrow 3 (ΔU123\Delta U_{123}):

Internal energy depends solely on temperature: ΔU123=nCv(T3−T1)=1⋅(32R)(T0−T03)=32R(2T03)=RT0\Delta U_{123} = n C_v (T_3 - T_1) = 1 \cdot \left(\frac{3}{2} R\right) \left(T_0 - \frac{T_0}{3}\right) = \frac{3}{2} R \left(\frac{2 T_0}{3}\right) = R T_0

This matches with (R). Thus, II→R\text{II} \rightarrow \text{R}.

(III) Heat absorbed by the system during process 1→2→31 \rightarrow 2 \rightarrow 3 (Q123Q_{123}):

By the First Law of Thermodynamics: Q123=ΔU123+W123Q_{123} = \Delta U_{123} + W_{123} Q123=RT0+13RT0ln⁡2=13RT0(3+ln⁡2)Q_{123} = R T_0 + \frac{1}{3} R T_0 \ln 2 = \frac{1}{3} R T_0 (3 + \ln 2)

This matches with (T). Thus, III→T\text{III} \rightarrow \text{T}.

(IV) Heat absorbed by the system during process 1→21 \rightarrow 2 (Q12Q_{12}):

For an isothermal process, ΔU12=0\Delta U_{12} = 0. By the First Law of Thermodynamics: Q12=W12=13RT0ln⁡2Q_{12} = W_{12} = \frac{1}{3} R T_0 \ln 2

This matches with (P). Thus, IV→P\text{IV} \rightarrow \text{P}.

Conclusion:

The correct matches are:

  • I→P\text{I} \rightarrow \text{P}
  • II→R\text{II} \rightarrow \text{R}
  • III→T\text{III} \rightarrow \text{T}
  • IV→P\text{IV} \rightarrow \text{P}

This corresponds to Option C.

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