# 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:

$PV^\gamma = constant$

with $\gamma = \frac{C_p}{C_V}$

where P is pressure, V is Volume, and C is 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: $dQ = 0$, (Q for heat)
• since there is no heat exchange, the process is reversible
• reversible means: $dW = -PdV$, (W is for work)
• heat capacity is defined as $C_V = \left( \frac{dQ}{dT} \right)_V$, respectively $C_P = \left( \frac{dQ}{dT} \right)_P$

If I draw a PV diagram for this situation, it looks like this:

Now I want to show that $PV^\gamma=const$ by going from state 1 to state 2 in the PV diagram.

I've started like this:

$W = -\int_{V_1}^{V_2}PdV = -\int_{V_1}^{V_2} \frac{RT}{V}dV = RT ln (V_2/V2)$

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.

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You have to use $dU=c_V dT$. The answer can be found in any book. – jinawee Feb 10 '14 at 16:07
NB: "since there is no heat exchange, the process is reversible". This is not true as far as I'm aware. The mixing of different substances is irreversible, but does not correspond to heat exchange. (On the flip side, there do exist plenty of reversible processes that involve heat exchange). When we talk about adiabatic expansions/compressions, we typically mean an expansion which is both a) reversible and b) involves no heat exchange. The key point is that the latter on its own does not imply the former. – gj255 Feb 10 '14 at 17:30