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Question

In a thermodynamic process on an ideal monoatomic gas, the infinitesimal heat absorbed by the gas is given by TΔX, where T is temperature of the system and ΔX is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monoatomic ideal gas X=32R ln(TTA)+R ln(VVA). Here, R is gas constant, TA and VA are constants. The List - I below gives some quantities involved in a process and List - II gives some possible values of these qunatitites.
List - IList - II(i) Work done by the system in process 123(P)13RT0ln2(ii)Change in internal energy in process123(Q)13RT0(iii)Heat absorbed by the system in process123(R)RT0(iv) Heat absorbed by the system in process12(S)43RT0(T)ΔQ=RT0+RT03ln 2
If the process carried out on one mole of monoatomic ideal gas is as shown in the TV - diagram with P0V0=13RT0. The correct match is -

A
iP,iiT,iiiQ,ivT
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B
iS,iiT,iiiQ,ivU
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C
iP,iiR,iiiT,ivS
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D
iP,iiR,iiiT,ivP
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Solution

The correct option is D iP,iiR,iiiT,ivP
Process 12) (Isothermal)
Process 23 (Isochoric)
W.D. (123)
=nRT ln(VfVi)+W23
=RT03ln2+0
ΔU(123)
ΔU=f2nR(TfTi)
=32R(T0T03)
ΔU=RT0
For 123
ΔQ=ΔU+W
ΔQ=RT0+RT03ln 2
For 12
ΔQ=ΔU12+W12
=f2nR(TfTi)+nRT ln(VfVi)
=0+RT03ln 2

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