JEE Challenger
More from Thermodynamics
PhyMedAdv 2011 • Paper 2

Match Thermodynamic Processes for Monatomic Ideal Gas Cycle

One mole of a monatomic ideal gas is taken through a cycle ABCDA as shown in the P-V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.

Diagram 0

Column I

(A) Process A→BA \rightarrow B

(B) Process B→CB \rightarrow C

(C) Process C→DC \rightarrow D

(D) Process D→AD \rightarrow A

Column II

(p) Internal energy decreases.

(q) Internal energy increases.

(r) Heat is lost.

(s) Heat is gained.

(t) Work is done on the gas.

Options

A

(A)→(P,R),  (B)→(P,R),  (C)→(Q,S),  (D)→(R,T)(\text{A}) \rightarrow (\text{P}, \text{R}),\; (\text{B}) \rightarrow (\text{P}, \text{R}),\; (\text{C}) \rightarrow (\text{Q}, \text{S}),\; (\text{D}) \rightarrow (\text{R}, \text{T})

B

(A)→(P,R,T),  (B)→(P,R),  (C)→(Q,S),  (D)→(R,T)(\text{A}) \rightarrow (\text{P}, \text{R}, \text{T}),\; (\text{B}) \rightarrow (\text{P}, \text{R}),\; (\text{C}) \rightarrow (\text{Q}, \text{S}),\; (\text{D}) \rightarrow (\text{R}, \text{T})

Correct
C

(A)→(P,R,T),  (B)→(R),  (C)→(S),  (D)→(R,T)(\text{A}) \rightarrow (\text{P}, \text{R}, \text{T}),\; (\text{B}) \rightarrow (\text{R}),\; (\text{C}) \rightarrow (\text{S}),\; (\text{D}) \rightarrow (\text{R}, \text{T})

D

(A)→(R,T),  (B)→(P,R),  (C)→(Q),  (D)→(R)(\text{A}) \rightarrow (\text{R}, \text{T}),\; (\text{B}) \rightarrow (\text{P}, \text{R}),\; (\text{C}) \rightarrow (\text{Q}),\; (\text{D}) \rightarrow (\text{R})

Topics & Concepts

ThermodynamicsFirst Law of ThermodynamicsKinetic Theory of Gases

Step-by-Step Solution

To determine the correct match between Column I and Column II, we analyze each thermodynamic process for one mole (n=1n=1) of a monatomic ideal gas (CV=32RC_V = \frac{3}{2}R, CP=52RC_P = \frac{5}{2}R).

The internal energy of a monatomic ideal gas at any state (P,V)(P, V) is given by:
U=nCVT=32PVU = n C_V T = \frac{3}{2} P V

From the given PP-VV diagram, the state variables (P,V)(P, V) and internal energies at the four vertices are:

  • State A: PA=3PP_A = 3P, VA=3V  ⟹  UA=32(3P)(3V)=272PVV_A = 3V \implies U_A = \frac{3}{2}(3P)(3V) = \frac{27}{2} PV
  • State B: PB=3PP_B = 3P, VB=1V  ⟹  UB=32(3P)(1V)=92PVV_B = 1V \implies U_B = \frac{3}{2}(3P)(1V) = \frac{9}{2} PV
  • State C: PC=1PP_C = 1P, VC=1V  ⟹  UC=32(1P)(1V)=32PVV_C = 1V \implies U_C = \frac{3}{2}(1P)(1V) = \frac{3}{2} PV
  • State D: PD=1PP_D = 1P, VD=9V  ⟹  UD=32(1P)(9V)=272PVV_D = 9V \implies U_D = \frac{3}{2}(1P)(9V) = \frac{27}{2} PV

Analysis of Processes:

(A) Process A→BA \rightarrow B (Isobaric Compression):

  • Internal Energy Change:
    ΔUAB=UB−UA=92PV−272PV=−9PV<0\Delta U_{AB} = U_B - U_A = \frac{9}{2} PV - \frac{27}{2} PV = -9 PV < 0
    Since ΔU<0\Delta U < 0, internal energy decreases (p).

  • Work Done:
    WAB=PA(VB−VA)=3P(1V−3V)=−6PV<0W_{AB} = P_A (V_B - V_A) = 3P (1V - 3V) = -6 PV < 0
    Since W<0W < 0, work is done on the gas (t).

  • Heat Exchange:
    QAB=ΔUAB+WAB=−9PV−6PV=−15PV<0Q_{AB} = \Delta U_{AB} + W_{AB} = -9 PV - 6 PV = -15 PV < 0
    Since Q<0Q < 0, heat is lost (r).

Thus, (A) matches with (p), (r), (t).

(B) Process B→CB \rightarrow C (Isochoric Cooling):

  • Internal Energy Change:
    ΔUBC=UC−UB=32PV−92PV=−3PV<0\Delta U_{BC} = U_C - U_B = \frac{3}{2} PV - \frac{9}{2} PV = -3 PV < 0
    Since ΔU<0\Delta U < 0, internal energy decreases (p).

  • Work Done:
    WBC=0(since volume is constant)W_{BC} = 0 \quad (\text{since volume is constant})

  • Heat Exchange:
    QBC=ΔUBC+WBC=−3PV+0=−3PV<0Q_{BC} = \Delta U_{BC} + W_{BC} = -3 PV + 0 = -3 PV < 0
    Since Q<0Q < 0, heat is lost (r).

Thus, (B) matches with (p), (r).

(C) Process C→DC \rightarrow D (Isobaric Expansion):

  • Internal Energy Change:
    ΔUCD=UD−UC=272PV−32PV=12PV>0\Delta U_{CD} = U_D - U_C = \frac{27}{2} PV - \frac{3}{2} PV = 12 PV > 0
    Since ΔU>0\Delta U > 0, internal energy increases (q).

  • Work Done:
    WCD=PC(VD−VC)=1P(9V−1V)=8PV>0W_{CD} = P_C (V_D - V_C) = 1P (9V - 1V) = 8 PV > 0
    Work is done by the gas.

  • Heat Exchange:
    QCD=ΔUCD+WCD=12PV+8PV=20PV>0Q_{CD} = \Delta U_{CD} + W_{CD} = 12 PV + 8 PV = 20 PV > 0
    Since Q>0Q > 0, heat is gained (s).

Thus, (C) matches with (q), (s).

(D) Process D→AD \rightarrow A (Isothermal Compression along PV=constantPV = \text{constant}):

  • Internal Energy Change:
    Since PDVD=PAVA=9PVP_D V_D = P_A V_A = 9 PV, the temperatures at DD and AA are equal (TD=TAT_D = T_A).
    ΔUDA=UA−UD=0\Delta U_{DA} = U_A - U_D = 0

  • Work Done:
    WDA=nRTln⁡(VAVD)=(9PV)ln⁡(3V9V)=−9PVln⁡(3)<0W_{DA} = nRT \ln\left(\frac{V_A}{V_D}\right) = (9 PV) \ln\left(\frac{3V}{9V}\right) = -9 PV \ln(3) < 0
    Since W<0W < 0, work is done on the gas (t).

  • Heat Exchange:
    QDA=ΔUDA+WDA=0−9PVln⁡(3)<0Q_{DA} = \Delta U_{DA} + W_{DA} = 0 - 9 PV \ln(3) < 0
    Since Q<0Q < 0, heat is lost (r).

Thus, (D) matches with (r), (t).

Final Answer:

  • (A) →\rightarrow p, r, t
  • (B) →\rightarrow p, r
  • (C) →\rightarrow q, s
  • (D) →\rightarrow r, t
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