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
<br /> dP = \frac{-\gamma P_0 V_0^{\gamma}}{V^{\gamma + 1}}dV<br />