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Question

In a thermodynamic process on an ideal monatomic 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 monatomic ideal gas X=32Rln(TTA)+Rln(VVA). Here, R is gas constant, V is volume of gas, 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 quantities.

List-I List-II
(i)

Work done by the system in process 123

(P)

13RT0ln2

(ii)

Change in intermal energy in process 123

(Q)

13RT0

(iii)

Heat absorbed by the system in process 123

(R)

RT0

(iv)

Heat absotbed by the system in process 12

(S)

43RT0

(T)

13RT0(3+ln2)

(U) 56RT0


If the process carricd out on one mole of monoatomic ideal gas is as shown in figure in the PV-diagram with P0V0=13RT0, the correct match is,


A

IP,IIT,IIIQ,IVT

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B

IP,IIR,IIIT,IVP

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C

IS,IIT,IIIQ,IVU

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D

IP,IIR,IIIT,IVS

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Solution

The correct option is B

IP,IIR,IIIT,IVP


(1) W123=RT03ln(V2V1)+0=RT03ln2

(2) ΔU=32R(T0T03)=RT0

(3) ΔQ=RT03m2+RT0
(4) ΔQ=RT03ln(2)

IP,IIR, III T,IVP


Process 12 is isothermal (temprature constant)
Process 23 is isochoric (volume constant)

(1) Work done in 123
W=W12+W23

=nRTlnVfVi+W23

=RTo3ln2VoVo+0

W=RTo3ln2 option P
(ii) ΔU in 123
ΔU=f2nR(TfTi)
=32R(ToTo3)

=32R(2To3) ΔU=RTo option R

(iii) For any system, first law of thermodynamics for 123
ΔQ=ΔU+ΔW
ΔQ=RTo+RTo3ln2
ΔQ=RTo3(3+ln2) option T

(iv) For process 12 (isothermal)
ΔQ=ΔU+W
=f2nR(TfTi)+nRTln(Vf/Vi)
=0+R(To3)ln(2VoVo)
ΔQ=RTo3ln2 option P

Hence the correct option is IP,IIR,IIIT,IVP

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