I found this formula for a polytropic process, defined by $PV^n = {\rm constant}$, in a book:
$$C = \frac R{\gamma-1} + \frac R{1-n} $$ where $C$ is molar specific heat and $\gamma$ is adiabatic exponent. I do not know how it was derived, can someone guide me?
That $C$ is the specific heat for the given cycle, i.e. $$dQ=nCdT$$ This is for $n$ moles of gas.(not the $n$ you stated in question)
I will assume $$PV^z=\text{constant}$$
$$nCdT=dU+PdV$$ $$\int nCdT=\int nC_vdT+\int PdV$$
$$nC\Delta T=nC_v \Delta T+\int \frac{PV^z}{V^z}dV$$
As numerator is a constant, take it out!
Also note that $$P_iV_i^z=P_fV_f^z$$
$i = \text{initial}$
$f=\text{final}$
Focusing on integral only,
$$PV^z\int V^{-z}dV$$
$$PV^z\left[\frac{V^{-z+1}}{-z+1}\right]^{V_f}_{V_i}$$
Note that the $PV^z$ is same for initial and final step. So, we multiply it inside and do this ingenious work:
$$-\frac{P_iV_i^zV_i^{-z+1}}{-z+1}+\frac{P_fV_f^zV_f^{-z+1}}{-z+1}$$
$$-\frac{P_iV_i}{-z+1}+\frac{P_fV_f}{-z+1}$$
Note that $PV=nRT$
$$\frac{nR\Delta T}{-z+1}$$
where $\Delta T=T_f-T_i$
Final equation :
$$nC\Delta T=nC_v \Delta T+\frac{nR\Delta T}{-z+1}$$
$$C=C_v+\frac{R}{1-z}$$
This will bring you the original equation, you can find $C_v$ by
$$C_p/C_v=\gamma$$
$$C_p-C_v=R$$
Using $C_p=\gamma C_v$,
$$C_v\left(\gamma-1\right)=R$$
$$C_v=\frac{R}{\gamma-1}$$
Substituting in original equation,
$$C=\frac{R}{\gamma-1}+\frac{R}{1-z}$$
We can also derive the result without integration:
$ PV^n= \text{constant}$ can be written as $ TV^{n-1}=\text{constant} $ $$ nCdT=dU+PdV $$ Dividing this equation throughout by $ dT $, differentiating $ TV^{n-1} = \text{constant} $ with respect to temperature, and plugging $ \frac{dV}{dT} $ into the equation will give the desired result.
$$ nCdT=dU+PdV $$ $$ nC = \frac{dU}{dT}+ P\frac{dV}{dT} $$ $$ TV^{n-1} = \text{constant} $$ $$ V^{n-1} + (n-1)TV^{n-2} \frac{dV}{dT} = 0$$ $$\frac{dV}{dT} = -\frac{V^{n-1}}{(n-1)TV^{n-2}} = \frac{V}{T(1-n)}$$ $$ nC = \frac{dU}{dT}+ P\left(\frac{V}{T(n-1)}\right) $$
Using $PV = nRT$ and $dU = nC_VdT$, $$ nC = \frac{dU}{dT}+ \frac{PV}{T(n-1)}=nC_V + \frac{nR}{1-n}$$
Now $C_V = \frac{R}{\gamma - 1}$ $$ nC =n\frac{R}{\gamma - 1} + \frac{nR}{1-n}$$ $$ C =\frac{R}{\gamma - 1} + \frac{R}{1-n}$$
Though there are already answers, this was the high school way I was taught:
$$\Delta U=q+w \\ NC_v \Delta T=NC_m \Delta T + \frac{NR\Delta T }{n-1} \\ NC_v \Delta T + \frac{NR\Delta T }{1-n}=NC_m \Delta T \\ C_v + \frac{R }{1-n}=C_m.$$
Note that $N$ is the number of moles while $n$ is the $n$ in $pv^n=\text{constant}$.
