# How to derive that $pV^k$ is constant in a polytropic process?

This is what we did on the lecture: $$\delta Q=nC dT$$ $$dU=nCdT-pdV$$ $$dU=\bigg(\frac{\partial U}{\partial V}\bigg)_TdV+\bigg(\frac{\partial U}{\partial T}\bigg)_VdT$$ $$dU=\bigg(\frac{\partial U}{\partial V}\bigg)_TdV+nC_VdT$$ $$n(C_V-C)dT=-\bigg(\bigg(\frac{\partial U}{\partial V}\bigg)_T+p\bigg)dV$$ And $\bigg(\frac{\partial U}{\partial V}\bigg)_T=0$ in the case of ideal gas, so: $$n(C_V-C)dT=-pdV$$ $$n(C_V-C)dT=-\frac{nRT}{V}dV$$ $$\color{blue}{(C_V-C)dT=-\frac{RT}{V}dV}$$ $$\color{red}{pV^k=constant}$$ where $k=\frac{C-C_p}{C-C_V}$

My first question is, how did we get the red one from the blue, or do you know an alternative derivation?

And my second question is, how does this work? For example, what should I do, if I want the $p^V-V^p$ to be constant? How can I get the $k$? It's just a weird example, but I hope you get what I mean.

We did the same derivation too. But I like it this way: $$\mathrm{d}U=\mathrm{d}\left(\frac{f}{2}pV\right)=\frac{f}{2}p\mathrm{d}V+\frac{f}{2}V\mathrm{d}p$$ Because $\delta Q=nC\mathrm{d}T$, we have that $\mathrm{d}U=nC\mathrm{d}T-p\mathrm{d}V$, so: $$nC\mathrm{d}T-p\mathrm{d}V=\frac{f}{2}p\mathrm{d}V+\frac{f}{2}V\mathrm{d}p$$ But $pV=nRT$, so $T=\frac{1}{nR}pV$ and $\mathrm{d}T=\frac{1}{nR}\left(p\mathrm{d}V+V\mathrm{d}p\right)$: $$nC\frac{1}{nR}\left(p\mathrm{d}V+V\mathrm{d}p\right)-p\mathrm{d}V=\frac{f}{2}p\mathrm{d}V+\frac{f}{2}V\mathrm{d}p$$ $$C\left(p\mathrm{d}V+V\mathrm{d}p\right)-pR\mathrm{d}V=\frac{f}{2}Rp\mathrm{d}V+\frac{f}{2}RV\mathrm{d}p$$ Collecting the $\mathrm{d}V$s and $\mathrm{d}p$s to the same side: $$\left(\frac{f}{2}RV-CV\right)\mathrm{d}p=\left(Cp-pR-\frac{f}{2}Rp\right)\mathrm{d}V$$ $$\left(\frac{f}{2}R-C\right)V\mathrm{d}p=\left(C-R-\frac{f}{2}R\right)p\mathrm{d}V$$ $$\left(\frac{f}{2}R-C\right)V\mathrm{d}p=\left(C-\frac{f+2}{2}R\right)p\mathrm{d}V$$ $$\left(C_V-C\right)\frac{1}{p}\mathrm{d}p=\left(C-C_p\right)\frac{1}{V}\mathrm{d}V$$ Integrating both sides: $$\left(C_V-C\right)\log\left(\frac{p_2}{p_1}\right)=\left(C-C_p\right)\log\left(\frac{V_2}{V_1}\right)$$ $$\left(\frac{p_2}{p_1}\right)^{C_V-C}=\left(\frac{V_2}{V_1}\right)^{C-C_p}$$ From this, we have that: $$\frac{p_1^{C_V-C}}{V_1^{C-C_p}}=\text{const}$$ $$p_1^{C_V-C}V_1^{-C+C_p}=\text{const}$$ $$p_1V_1^{\frac{C_p-C}{C_V-C}}=\text{const}$$ $$p_1V_1^{\frac{C-C_p}{C-C_V}}=\text{const}$$ And we wanted to get this.
Note: $\frac{f}{2}R=C_V$, because $$C_V=\frac{1}{n}\left.\frac{\partial U}{\partial T}\right|_V=\left.\frac{\partial \left(\frac{f}{2}nRT\right)}{\partial T}\right|_V=\frac{f}{2}R$$ and $\frac{f+2}{2}R=C_p$, because: $$C_p=\frac{1}{n}\left.\frac{\partial U}{\partial T}\right|_p+\frac{1}{n}p\left.\frac{\partial V}{\partial T}\right|_p=\frac{f}{2}R+\frac{1}{n}\left.\frac{\partial \left(pV\right)}{\partial T}\right|_p=$$ $$\frac{f}{2}R+\frac{1}{n}\left.\frac{\partial \left(nRT\right)}{\partial T}\right|_p=\frac{f}{2}R+R=\frac{f+2}{2}R$$
The blue line is what is known as a separable differential equation, where we can move all $T$ values to one side, and all $V$ values to the other. So, \begin{align} &\left[(C_V-C)\operatorname{d}T=-\frac{RT}{V}\operatorname{d}V\right] \times\left(\frac{1}{T(C_V-C)}\right) \Rightarrow\\ &\int\left[\frac{\operatorname{d}T}{T} = \left(\frac{R}{C-C_V}\right)\frac{\operatorname{d}V}{V}\right] \Rightarrow\\ \ln\left(\frac{T_f}{T_i}\right)&= \left(\frac{R}{C-C_V}\right)\ln\left(\frac{V_f}{V_i}\right) \\ &=\ln\left(\left[\frac{V_f}{V_i}\right]^{R/[C-C_V]}\right) \Rightarrow\\ T_fV_f^{-R/[C-C_V]}&=T_iV_i^{-R/[C-C_V]}. \end{align} We conclude that the quatity on either side of the equals signs must be constants for all states related by a particular polytropic process since the final and initial states have no other relationship to them.
Now, for an ideal gas $PV=nRT\Rightarrow T=(PV)/(nR)$ so \begin{align} \text{const}&=TV^{-R/[C-C_V]} \\ &=\left(\frac{PV}{nR}\right)V^{-R/[C-C_V]}\\ &=\frac{1}{nR}P V^{1-R/[C-C_V]}. \end{align} $nR$ is also a constant for our process, so we can conclude that $PV^\gamma$ is a constant. All that remains is to show that $\gamma$ is the constant in the question, $k$. \begin{align} \gamma &\equiv 1-\frac{R}{C-C_V} \\ &= \frac{C-C_V-R}{C-C_V}\\ &= \frac{C-C_P}{C-C_V}, \end{align} where the last line follows because the molar heat capacities are related by $$C_P = C_V+R$$ because $PV=H-U$, $C_V \equiv n^{-1} \left.\frac{\operatorname{d}U}{\operatorname{d}T}\right|_V$, and $C_P \equiv n^{-1} \left.\frac{\operatorname{d}H}{\operatorname{d}T}\right|_P.$
• Thank you! Do you know an easier derivation, which will result the $pV^k$ instantly? – 545941st user Jun 20 '18 at 15:52