How to Derive the Pressure-Volume Relationship in Thermodynamics?

[dimensionless]

Homework Statement

Given that
$$P V^{\gamma} = P_{0} V_{0}^{\gamma}$$
Show that
$$dP \doteq - \frac{\gamma P_{0}}{V_{0}} dV$$

Homework Equations

$$c^{2} = \left( \frac{\partial P}{\partial \rho} \right)_{\rho_{0}}$$
$$P = P_{0} \frac{\rho}{\rho_{0}}$$
$$\gamma = \frac{c_{p}}{c_{v}}$$
$$P = \rho r T_{k}$$

The Attempt at a Solution

I'm not sure what the $$\doteq$$ means, but I can pressume that it is the same thing as an equal sign. Starting with the first equation I can get
$$P = \frac{P_{0} V_{0}^{\gamma}}{V^{\gamma} }$$
At this point I get stuck. I could try to relate the volume to the density, but I'm not sure that that would lead anywhere.

 

[dimensionless]

from my last equation I can get
$$dP = \frac{-\gamma P_{0} V_{0}^{\gamma}}{V^{\gamma-1} }dV$$
But this is not a huge help.

 

[Astronuc]

$$P V^{\gamma} = P_{0} V_{0}^{\gamma}$$

by the right hand side is a constant, so derivative or differential is zero.

So one starts with

$$d(P V^{\gamma})\,=\,0$$ and expand, and then put it in a form

dP = ?, but that seems where one go to.

Also I think its

$$dP = \frac{-\gamma P_{0} V_{0}^{\gamma}}{V^{\gamma+1} }dV$$

and then what happens if V ~ V_o?

 

[olgranpappy]

from my last equation I can get
$$dP = \frac{-\gamma P_{0} V_{0}^{\gamma}}{V^{\gamma-1} }dV$$
But this is not a huge help.

That's wrong. It should read
$$dP = \frac{-\gamma P_0 V_0^{\gamma}}{V^{\gamma + 1}}dV$$