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I am working through the derivation of an adiabatic process of an ideal gas $pV^{\gamma}$ and I can't see how to go from one step to the next. Here is my derivation so far which I understand:

$$dE=dQ+dW$$ $$dW=-pdV$$ $$dQ=0$$ $$dE=C_VdT$$

therefore

$$C_VdT=-pdV$$

differentiate the ideal gas equation $pV=Nk_BT$

$$pdV+Vdp=Nk_BdT$$

rearrange for $dT$ and substitute into the 1st law:

$$\frac{C_V}{Nk_B}(pdV+Vdp)=-pdV$$.

The next part is what I am stuck with I can't see how the next line works specifically how to go from $\frac{C_V}{C_p-C_V}=\frac{1}{\gamma -1}$

using the fact that $C_p-C_V=Nk_B$ and $\gamma = \frac{C_p}{C_V}$ it can be written

$$\frac{C_v}{Nk_B}=\frac{C_V}{C_p-C_V}=\frac{1}{\gamma -1}$$.

If this could be explained to me, I suspect it is some form of algebraic rearrangement that I am not comfortable with that is hindering me.

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I think the last line does not follow from the previous steps. It is used to show how $\gamma$ comes in place, so I extrapolated a bit and show the next few steps:

Since $$ \frac{C_V}{Nk_B} = \frac{C_V}{C_p-C_V} = \frac{\frac{C_V}{C_V}}{\frac{C_p}{C_V}-\frac{C_V}{C_V}}=\frac{1}{\gamma-1} $$ Therefore, $$ \frac{C_V}{Nk_B} (pdV+Vdp)= \frac{1}{\gamma-1} (pdV+Vdp) = -pdV $$ Dividing both sides with $pdV$: $$ \frac{1}{\gamma-1}(1+\frac{V}{p}\frac{dp}{dV})=-1 $$ Continue to simplify the expressions and you will reach your result of $pV^\gamma$is constant.

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Hint: Divide both numerator and denominator by $C_V$: $$\frac{C_V}{C_p-C_V} = \frac{1}{C_p/C_V -1}$$

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$$dU =\frac f2 NKBdT$$ $$dW= pdv$$ $$d(pv) =pdv +vdp$$ $$pdv +vdp =NKBT = - \frac 2f pdv$$ $$1+\frac 2f pdv +vdp =0$$ $$ˠ= 1 +\frac 2f$$ So, $$ˠpdv +vdp = \text{constant}$$ $$ʸIn (dv2/v1) – In (d p2/p1)$$ $$ˠIn (d v2/v1) = In (d p2/p1)$$ $$(v2/v1)ˠ = (p2/p1)$$ Now take the exponent on both sides: $$(p1v1ˠ) = (p2v2ˠ) = \text{constant}$$ Therefore, $pvˠ$ is a constant.

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protected by ACuriousMind Oct 24 '17 at 10:53

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