Monday, 15 October 2018

homework and exercises - Adiabatic process of an ideal gas derivation


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.



Answer



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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