Why is $PV^\gamma$ constant in an adiabatic process? In non isolated systems where there is no adiabatic process, $PV$ is constant. But the graph gets steeper in adiabatic process because of the $\gamma$ over the $V$. Why is it there in adiabatic processes and why only over the $V$?
 A: For an ideal gas 
$$PV=RT$$
Since 
$$dU=dQ-dW$$
For adiabatic process
$$C_v dT = -{PdV}$$
Substituting $R dT = VdP+PdV$
$$VdP = -\frac{(R+C_v)}{C_v}PdV$$
Since $C_p -C_v =R$ and $\gamma= \frac{C_p}{C_v}$ the coefficient on RHS becomes $\gamma$. Integrating on both sides
$$ln(PC) =-\gamma ln(V)$$
Where $C$ is an integration constant. Rearranging
$$ PV^{\gamma} = const $$
A: Breaking the Question Down Into Parts to Understand It Better
To make it easier to arrive at an answer, it is often helpful to take a multi-part question and break it down into simpler parts.  So I address the three parts of the question individually.
The first part of the question is:
In non isolated systems where there is no adiabatic process, PV is constant.
I would write:
"The ideal gas law (Boyle's Law) holds that for a confined gas in equilibrium at contact Temperature $T$, that $PV$ is constant, where $P$ is the pressure and $V$ is the volume of the confinement."
The second part of your question was:
But the graph gets steeper in adiabatic process because of the γ over the V.
I would write:
"But the graph gets steeper in an adiabatic process because of the $γ>1$ over the $V$ in the polytropic process equation for an ideal gas."

In part three of your question, you finally asked:
Why is it there in adiabatic processes and why only over the V?
I would then ask:
"Why is the $\gamma$ there in adiabatic processes?"  and "Why is the $\gamma$ only over the $V$ in the formula $P V ^\gamma = Constant, \gamma > 1$ and not also over the $P$ (in example $P^\gamma V^\gamma = Constant$)?" 
The Answer
My answer is that from the graph below, that it shows constant temperature processes as isotherms and adiabatic process going from one initial isotherm to a different final isotherm.  In short, temperature is generally not constant (in the shown adiabatic transitions), so generally $P V \ne Constant$ for adiabatic transitions.

A: In short, because the heat capacity at constant volume and heat capacity at constant pressure are not equal, even for ideal gas. $\newcommand{\d}{\textrm{d}}$
$$C_p = \frac{V\,\d p}{n\,\d T}$$
$$C_v = \frac{P\,\d V}{n\,\d T}$$
$$\gamma = \frac{C_p}{C_v} = \frac{\displaystyle\frac{\d p}{p}}{\displaystyle\frac{\d V}{V}}$$
Integrating the last equation gives $pV^\gamma = \text{constant}$.
