Basic Thermodynamics: Quasistatic Adiabatic Process

I’m going through the exercises in a Thermodynamics book, just to revise and build my intuition. Right now, I’m working on:

Show that for a quasistatic adiabatic process in a perfect gas, with
constant specific heats:

$$PV^gamma = left[text{constant}right]$$

with $gamma = frac{C_P}{C_V}$

where $P$ is pressure, $V$ is volume, and $C_V$ is the constant-volume heat capacity.

I’m not looking for the answer, just for a hint (I’m stuck and want to find the solution myself).

So those are my thoughts:

• perfect gas means: $PV = RT$, ($R$ is the universal gas constant)
• adiabatic means: $mathrm{d}Q = 0$, ($Q$ for heat)
• since there is no heat exchange, the process is reversible
• reversible means: $mathrm{d}W = -P , mathrm{d}V$, ($W$ is for work)
• heat capacity is defined as $C_text{V} = left( frac{mathrm{d}Q}{mathrm{d}T} right)_V$, respectively $C_text{P} = left( frac{mathrm{d}Q}{mathrm{d}T} right)_text{P}$

If I draw a $PV$ diagram for this situation, it looks like this:
$hspace{175px}$.

Now I want to show that $PV^gamma=left[text{const}right]$ by going from $text{State 1}$ to $text{State 2}$ in the $PV$ diagram.

I’ve started like this:
$$
W ~=~ -int_{V_1}^{V_2}P , mathrm{d}V ~=~ -int_{V_1}^{V_2} frac{RT}{V} mathrm{d}V ~=~ RT ln{left(frac{V_2}{V_2}right)}
$$

This leads me into the wrong direction though. I thought about using $R = C_P – C_V$ here, but it doesn’t seem to work. Any suggestions?

Please just give me a hint, not the solution.