Solution

Since $PV^n$ is constant and so is $PV/T$ for ideal gas, it follows that $TV^{n-1}=const$. Differentiating both sides using the product rule $d(uv) = v\,du + u\,dv$, we obtain:

$$V^{n-1}\,dT+T\cdot(n-1)V^{n-2}\,dV=0\qquad\Rightarrow\qquad dV=\frac{V}{T(1-n)}\,dT.$$

Consequently, the work done by one mole of gas in an expansion over a small temperature change is

$$P\,dV=\frac{PV}{T(1-n)}\,dT=\frac{R}{1-n}\,dT.$$

At the same time, the change in internal energy of monatomic gas is $dU=(3R/2)\,dT$. Thus, the gas must absorb heat

$$dQ=P\,dV+dU=\frac{(5-3n)R}{2(1-n)}\,dT,$$

so the molar heat capacity is $(5-3n)R/(2(1-n))$. It is zero when $n=5/3$ and infinite when $n=1$.