Change in entropy in a polytropic process

@The myth buster • 19 Oct, 2012 well i believe that instead of (k-n) term you should be having k^2-1 term in the expression..rest of it is fine..!!

@Kyle Schieber • 21 May, 2016 Hello, for those who are interested in the answer:
It all begins with these four expressions:

dQ=dU-dW (1) ->1st law (the minus sign in dW will be useful later);
dQ=TdS (2) ->from definition of entropy for reversible processes;
dW=PdV (3) -> Compresion-expansion work;
dU=CvdT (4) -> for ideal gases U=U(T);

substituting (2) and (3) in (1):
dS=(dU/T)-(PdV/T),
inserting (4) in this expresion, and using ideal gases (PV=NRT) yields:
dS=(CvdT/T)-(NRdV/V)
integrating:
S2-S1=Cvln(T2/T1)-NRln(V2/V1)

now, using T2/T1=(V2/V1)^(1-n) (from polytropic process definition coupled with ideal gas eq.), and knowing that Cp-Cv=NR (for ideal gases):
S2-S1=ln(V2/V1)*(Cv(1-n)-NR)=ln(V2/V1)*(Cp-nCv)
the definition of adiabatic factor is k=Cp/Cv, so:
S2-S1=Cv*ln(V2/V1)*(k-n)
if Cv' is the constant-volume specific heat per unit mass:

S2-S1=mCv'(k-n)*ln(V2/V1)

regards!