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.
 A: From the first law of thermodynamics, you have: 
$$\Delta U = q + w$$
As $q$ is zero,
$$\mathrm{d}U = -P \, \mathrm{d}V$$
Now put $\mathrm{d}U = C_v \, \mathrm{d}T$ (molar heat capacity at constant volume). 
Therefore, 
$$C_v \, \mathrm{d}T = -P \, \mathrm{d}V$$
Using the ideal gas equation, substitute $P$ by $\frac{RT}{V}$.
$$-\frac{RT}{V} \mathrm{d}V = C_v \, \mathrm{d}T$$
Integrating the equation, you get
$$
\begin{align}
\int_{V_1}^{V_2} -\frac{RT}{V} \mathrm{d}V &  =  \int_{T_1}^{T_2} C_v \, \mathrm{d}T \\[10px]
\frac{RT}{V} \ln \left(\frac{V_1}{V_2}\right) &  =  C_v \left(T_2 - T_1 \right)
\end{align}
$$
A: The mathematical condition you're missing is work = -change in internal energy (which 
 you mentioned as dQ = 0).
dU is ideally written as n*Cv(dT).
take n=1 (for simplicity),
 equate PdV to -dU, bring in R = Cv (gamma - 1) (same thing as you mentioned), and 
 integrate it. It will be of a logarithmic form with a ton of constants. 
Hope this helps. Good Luck!
A: Before the hint, some conceptual issues:

since there is no heat exchange, the process is reversible

Not necessarily. An adiabatic process is irreversible if either (1) it is not carried out extremely slowly (quasi statically) or (2) mechanical friction is present. To be reversible it must be carried out quasi statically and without friction.

reversible means: $\mathrm{d}W = -P \, \mathrm{d}V$, ($W$ is for work)

Not necessarily. Only if the system is always in mechanical equilibrium with the surroundings so that the $P$ is both the gas and external pressure. That requires the process be carried out slowly. Otherwise $P$ is the external pressure only and the work is irreversible.

heat capacity is defined as $C_\text{V} = \left(
> \frac{\mathrm{d}Q}{\mathrm{d}T} \right)_V$, respectively $C_\text{P} =
> \left( \frac{\mathrm{d}Q}{\mathrm{d}T} \right)_\text{P}$

These are not the definitions of the heat capacities. The specific heat capacities are defined in terms of the specific internal energy and enthalpy, as follows:
$$c_{v}=\biggl (\frac{\partial u}{\partial T}\biggr)_v$$
$$c_{p}=\biggl (\frac{\partial h}{\partial T}\biggr)_P$$
Now, the hint. Your equation for a reversible adiabatic process is for an ideal gas. For an ideal gas, and only an ideal gas $du=c_{v}dT$ regardless of the process. Use this together with the ideal gas equation, the reversible work equation, and the first law, and you can derive the equation.
Hope this helps.
A: As an alternative to the popular proposal of using $dU = c_V dT$, I find it more intuitive to use the equation for the internal energy of an ideal gas, and its differential:
$$
U = \frac{3}{2} N k T = \frac{3}{2} p V\quad \rightarrow \quad dU =  \frac{3}{2} p \, dV + \frac{3}{2} V \, dp
$$
where I've assumed (in using "3") that the gas is monatomic. Then the differential form of the First Law becomes:
\begin{align}
dU &= q + w_\text{on sys}\\
dU &= 0 - p dV \\
\frac{3}{2} \left[ p \, dV + V \, dp\right] &= - p \, dV
\end{align}
Consolidating like variables and integrating leads to the answer.
