The question is:
A monoatomic gas undergoes a process $PV^3=k$ where k is a constant. Find the heat capacity of the gas.
Problem:
I know how to calculate the heat capacity at constant pressure ($C_p$) and ($C_v$) of a gas but this problem does not ask any of them.
I came to know about this formula:
For a process $PV^n=k$, $C = (\frac{1}{\gamma - 1}-\frac{1}{n-1})R$
as the solution but found no proof or anything. How can I start to prove it myself?
If $PV^n=k$, then $$P=\frac{P_0V_0^n}{V^n}$$and $$PdV=\frac{P_0V_0^n}{V^n}dV=-\frac{P_0V_0^n}{(n-1)}d\left(\frac{1}{V^{n-1}}\right)$$So, $$\int_{V_0}^V{PdV}=\frac{P_0V_0^n}{(n-1)}\left[\frac{1}{V_0^{n-1}}-\frac{1}{V^{n-1}}\right]=\frac{P_0V_0^n}{(n-1)}\left[\frac{V_0}{V_0^{n}}-\frac{V}{V^{n}}\right]=\frac{P_0V_0-PV}{(n-1)}$$From this, it follows that, for this particular process path $$PdV=-\frac{d(PV)}{(n-1)}=-\frac{RdT}{(n-1)}$$So, from the first law: $$dU=C_vdT=dQ+\frac{RdT}{(n-1)}$$So, $$dQ=\left[C_v-\frac{R}{(n-1)}\right]dT$$ The authors refer to the term in brackets as the "heat capacity," but it should not really be considered heat capacity, since, in thermodynamics, we regard heat capacity is a fundamental physical property of the gas (independent of process path).
