Question

what is molar heat capacity in different process

Answer 1

Hey friend here is your answer

That CC is the specific heat for the given cycle, i.e.

dQ=nCdTdQ=nCdT

This is for nn moles of gas.(not the nnyou stated in question)

I will assume

PVz=constantPVz=constant

nCdT=dU+PdVnCdT=dU+PdV

∫nCdT=∫nCvdT+∫PdV∫nCdT=∫nCvdT+∫PdV

nCΔT=nCvΔT+∫PVzVzdVnCΔT=nCvΔT+∫PVzVzdV

As numerator is a constant, take it out!

Also note that

PiVzi=PfVzfPiViz=PfVfz

i=initiali=initial

f=finalf=final

Focusing on integral only,

PVz∫V−zdVPVz∫V−zdV

PVz[V−z+1−z+1]VfViPVz[V−z+1−z+1]ViVf

Note that the PVzPVz is same for initial and final step. So, we write multiply it inside and do this ingenious work :

−PiVziV−z+1i−z+1+PfVzfV−z+1f−z+1−PiVizVi−z+1−z+1+PfVfzVf−z+1−z+1

−PiVi−z+1+PfVf−z+1−PiVi−z+1+PfVf−z+1

Note that PV=nRTPV=nRT

nRΔT−z+1nRΔT−z+1

where ΔT=Tf−TiΔT=Tf−Ti

Final equation :

nCΔT=nCvΔT+nRΔT−z+1nCΔT=nCvΔT+nRΔT−z+1

C=Cv+R1−zC=Cv+R1−z

This will bring you the original equation, you can find CvCv by

Cp/Cv=γCp/Cv=γ

Cp−Cv=RCp−Cv=R

Using Cp=γCvCp=γCv,

Cv(γ−1)=RCv(γ−1)=R

Cv=Rγ−1Cv=Rγ−1

Substituting in original equation,

C=Rγ−1+R1−z


Hope it helps you.