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Heat and Thermodynamics question

2016 · Shift 0 · Q57
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Heat and Thermodynamics question

2016 · Shift 0 · Q57

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity CCC remains constant. If during this process the relation of pressure PPP and volume VVV is given by PVn=P{V^n} =PVn= constant, then nnn is given by (Here Cp{C_p}Cp​ and Cv{C_v}Cv​ are molar specific heat at constant pressure and constant volume, respectively:
  1. A
    n=Cp−CC−Cvn = {{{C_p} - C} \over {C - {C_v}}}n=C−Cv​Cp​−C​
  2. B
    n=C−CvC−Cpn = {{C - {C_v}} \over {C - {C_p}}}n=C−Cp​C−Cv​​
  3. C
    n=CpCvn = {{{C_p}} \over {{C_v}}}n=Cv​Cp​​
  4. D
    n=C−CpC−Cvn = {{C - {C_p}} \over {C - {C_v}}}n=C−Cv​C−Cp​​
View written solutionFree

Correct answer: D

  1. Given a reversible quasi-static process for an ideal gas with constant molar heat capacity CCC.

    We use the definition

     dQ=C dT\, dQ = C\, dTdQ=CdT

  2. Use the first law of thermodynamics for 1 mole of an ideal gas:

    dQ=dU+P dV=CvdT+P dVdQ = dU + P\,dV = C_v dT + P\,dVdQ=dU+PdV=Cv​dT+PdV

    Since also dQ=CdTdQ = C dTdQ=CdT, we get

    CdT=CvdT+PdVC dT = C_v dT + P dVCdT=Cv​dT+PdV (C−Cv)dT=PdV(C - C_v)dT = P dV(C−Cv​)dT=PdV

  3. Relate dTdTdT to dPdPdP and dVdVdV using the ideal gas law for 1 mole:

    PV=RTPV = RTPV=RT

    Differentiating,

    PdV+VdP=RdTP dV + V dP = R dTPdV+VdP=RdT dT=PdV+VdPRdT = \frac{P dV + V dP}{R}dT=RPdV+VdP​

  4. Substitute into

    (C−Cv)dT=PdV(C - C_v)dT = P dV(C−Cv​)dT=PdV

    to get

    (C−Cv)PdV+VdPR=PdV(C - C_v)\frac{P dV + V dP}{R} = P dV(C−Cv​)RPdV+VdP​=PdV

    Since for an ideal gas,

    R=Cp−CvR = C_p - C_vR=Cp​−Cv​

    this becomes

    (C−Cv)(PdV+VdP)=(Cp−Cv)PdV(C - C_v)(P dV + V dP) = (C_p - C_v)P dV(C−Cv​)(PdV+VdP)=(Cp​−Cv​)PdV

  5. Expand and rearrange:

    (C−Cv)PdV+(C−Cv)VdP=(Cp−Cv)PdV(C - C_v)P dV + (C - C_v)V dP = (C_p - C_v)P dV(C−Cv​)PdV+(C−Cv​)VdP=(Cp​−Cv​)PdV (C−Cv)VdP=[(Cp−Cv)−(C−Cv)]PdV(C - C_v)V dP = [(C_p - C_v) - (C - C_v)]P dV(C−Cv​)VdP=[(Cp​−Cv​)−(C−Cv​)]PdV (C−Cv)VdP=(Cp−C)PdV(C - C_v)V dP = (C_p - C)P dV(C−Cv​)VdP=(Cp​−C)PdV

  6. Divide by PVPVPV:

    (C−Cv)dPP=(Cp−C)dVV(C - C_v)\frac{dP}{P} = (C_p - C)\frac{dV}{V}(C−Cv​)PdP​=(Cp​−C)VdV​ dPP=Cp−CC−CvdVV\frac{dP}{P} = \frac{C_p - C}{C - C_v}\frac{dV}{V}PdP​=C−Cv​Cp​−C​VdV​

  7. For a polytropic process

    PVn=constantPV^n = \text{constant}PVn=constant

    taking logarithmic differential,

    dPP+ndVV=0\frac{dP}{P} + n\frac{dV}{V} = 0PdP​+nVdV​=0 dPP=−ndVV\frac{dP}{P} = -n\frac{dV}{V}PdP​=−nVdV​

  8. Compare with the result from thermodynamics:

    −n=Cp−CC−Cv-n = \frac{C_p - C}{C - C_v}−n=C−Cv​Cp​−C​ n=−Cp−CC−Cv=C−CpC−Cvn = -\frac{C_p - C}{C - C_v} = \frac{C - C_p}{C - C_v}n=−C−Cv​Cp​−C​=C−Cv​C−Cp​​

  9. Therefore,

    n=C−CpC−Cv\boxed{n = \frac{C - C_p}{C - C_v}}n=C−Cv​C−Cp​​​

  10. Option check

  • A: Cp−CC−Cv\dfrac{C_p-C}{C-C_v}C−Cv​Cp​−C​ ❌ missing minus sign
  • B: C−CvC−Cp\dfrac{C-C_v}{C-C_p}C−Cp​C−Cv​​ ❌ reciprocal
  • C: CpCv\dfrac{C_p}{C_v}Cv​Cp​​ ❌ this is only γ\gammaγ, not general here
  • D: C−CpC−Cv\dfrac{C-C_p}{C-C_v}C−Cv​C−Cp​​ ✅

Hence the correct option is D.

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