The correct option is B γ>α>1
In a polytropic process the molar heat capacity of the process doesnot vary.
Given PVα=constant
We need to find of values of α for which the molar heat capacity is negative.
By the definition of molar specific heat capacity, we get
ΔQ=nCΔT
C=1nΔQΔT=1ndQdT (For instantaneous value)
Using first law of thermodynamics, dQ=dU+dW in the above equation,
C=1ndU+dWdT
Substituting, dU=nCv dT and dW=PdV
C=1n(nCv dTdT+PdVdT)
=Cv+PdVndT....(i)
Using the ideal gas equation, PV=nRT
Differentiaing, PdV+VdPR=ndT
Substituting this in equation (i) we have,
C=Cv+PdVPdV+VdPR
C=Cv+R1+VdPPdV....(ii)
From the given equation,PVα=k⇒P=kvα
Differenciating, dPdV=−αkVα+1
Substituting this in equation (ii) we have,
C=Cv+R1+VdPPdV=Cv+11+Vk(−k)αVα+1Vα
In general Cv=Rγ−1
C=Rγ−1+R1−α
We want this value to be negative,
∴C=Rγ−1+R1−α<0
Simplyfing the above equation, (α−γ)(γ−1)(α−1)<0
γ−1 is positive.
∴(α−γ)(α−1)<0
Case 1: (α−γ)<0 and (α−1)>0
α<γ and α>1⇒γ>α>1
Case 2:(α−γ)>0 and (α−1)<0
α>γ and α<1
This possibility is not possible as γ cannot be less than 1
∴γ>α>1 is correct.