Work in a polytropic transformation $PV^k=\text{const}$

Question 1. The polytropic relation can be written

$$PV^k=C, \qquad P(V)=CV^{-k}, \qquad C=P_AV_A^k=P_BV_B^k.$$

Using the banker's convention, the work received by the gas is

$$W_{A\to B} =-\int_{V_A}^{V_B}P(V)\,\mathrm{d}V =-C\int_{V_A}^{V_B}V^{-k}\,\mathrm{d}V.$$

For $k\neq1$, an antiderivative of $V^{-k}$ is $V^{1-k}/(1-k)$. Thus,

$$W_{A\to B} =-\frac{C}{1-k} \left(V_B^{1-k}-V_A^{1-k}\right) =\frac{CV_A^{1-k}-CV_B^{1-k}}{1-k}.$$

Now

$$CV_A^{1-k}=P_AV_A, \qquad CV_B^{1-k}=P_BV_B.$$

Finally,

$$\boxed{ W_{A\to B} =\frac{P_AV_A-P_BV_B}{1-k} =\frac{P_BV_B-P_AV_A}{k-1} }.$$

Question 2. For $k=0$, the relation $PV^0=C$ becomes $P=C$: the transformation is isobaric. Since $P_A=P_B=P$,

$$W_{A\to B} =P_AV_A-P_BV_B =P(V_A-V_B) =-P(V_B-V_A) =-P\Delta V.$$

This recovers the direct expression for isobaric work.

Question 3. For an isochoric transformation, $V_B=V_A$ and $\mathrm{d}V=0$ at all times. Consequently,

$$W_{A\to B}=-\int_A^B P\,\mathrm{d}V=0.$$

A nontrivial isochoric transformation does not correspond to any finite value of $k$: if $V$ is constant, the relation $PV^k=\text{const}$ would also require $P$ to be constant. The isochoric transformation can only be regarded as the formal limit $k\to\infty$.

Question 4. For $k=1$, the integrand is proportional to $1/V$: the antiderivative is no longer a power but a logarithm. The integration must therefore be performed separately. The polytropic relation becomes

$$PV=C.$$

For a closed ideal gas, $PV=nRT$; the temperature is therefore constant and $C=nRT$. The work received is then

$$W_{A\to B} =-\int_{V_A}^{V_B}\frac{nRT}{V}\,\mathrm{d}V =-nRT\ln\!\left(\frac{V_B}{V_A}\right).$$

For a compression, $V_B/V_A<1$, so the logarithm is negative and $W_{A\to B}>0$, as expected.