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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?

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  • $\begingroup$ You are asking about a reversible polytropic process that is neither isothermal, isochoric, isobaric nor adiabatic? $\endgroup$
    – Bob D
    Commented Jan 3, 2023 at 19:13
  • $\begingroup$ Yeah, exactly ! $\endgroup$
    – Leibniz-Z
    Commented Jan 3, 2023 at 19:15
  • $\begingroup$ You realize there are an infinite number of possible values of $\alpha$ other than these. $\endgroup$
    – Bob D
    Commented Jan 3, 2023 at 19:47
  • $\begingroup$ Yes, I am already aware. $\endgroup$
    – Leibniz-Z
    Commented Jan 3, 2023 at 19:48
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    $\begingroup$ Check out this: en.wikipedia.org/wiki/… $\endgroup$
    – Bob D
    Commented Jan 3, 2023 at 19:53

2 Answers 2

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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$$TV^{\alpha-1}=const$$

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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 $\delta T=({V \delta p+p \delta V})/R\alpha=\frac{1-\alpha}{R\alpha}p \delta V$

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