# How do we maintain polytropic processes?

In polytropic processes for an ideal gas,

$$PV^{\alpha}=constant$$ where $$\alpha \neq 0,1,\gamma$$

And $$\gamma$$ is adiabatic exponent of gas

So, how these processes are maintained?

What things are done to initiate this process?

• You are asking about a reversible polytropic process that is neither isothermal, isochoric, isobaric nor adiabatic? Jan 3 at 19:13
• Yeah, exactly ! Jan 3 at 19:15
• You realize there are an infinite number of possible values of $\alpha$ other than these. Jan 3 at 19:47
• Yes, I am already aware. Jan 3 at 19:48
• Check out this: en.wikipedia.org/wiki/… Jan 3 at 19:53

You add or remove heat to change the temperature along the polytropic path in such a way that the exponent $$\alpha$$ remains constant. You have $$d\ln{P}+\alpha d\ln{V}=0$$and $$d\ln{P}+d\ln{V}=d\ln{T}$$So $$dln{T}=(1-\alpha)d\ln{V}$$or$$TV^{\alpha-1}=const$$
By definition a a polytropic process is one for which $$TdS=\mathcal K dT$$ and $$\mathcal K$$ is a constant. Using the $$dU=TdS-pdV$$ equation it follows that $$\frac{dp}{p}+\alpha \frac{dV}{V}=0 \tag{1}\label{1}$$ and upon integration you get $$pV^{\alpha}=K_0\tag{2}\label{2}$$ where $$\alpha=\frac{C_p-K}{C_V-K}$$. This means that if you change the volume by an amount of, say, $$\delta V$$ then you have to change the pressure by $$\delta p = -p \alpha \frac{dV}{V}$$. This can be achieved by absorbing $$\delta S=\frac{\mathcal K}{T}\delta T$$ entropy from a thermal reservoir at temperature $$T+\delta T$$ where $$pV=RT$$ and $$\delta T=({V \delta p+p \delta V})/R\alpha=\frac{1-\alpha}{R\alpha}p \delta V$$