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.
 A: Hint: Divide both numerator and denominator by $C_V$:
$$\frac{C_V}{C_p-C_V} = \frac{1}{C_p/C_V -1}$$
A: 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. 
A: $$dU =\frac f2 NKBdT$$
$$dW= pdv$$
$$d(pv) =pdv +vdp$$
$$pdv +vdp =NKBT = - \frac 2f pdv$$
$$1+\frac 2f pdv +vdp  =0$$
$$ˠ= 1 +\frac 2f$$
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.
