Ideal gas derivations

For an ideal gas, how can someone obtain this:


$$ \bar{V}dP = RdT- P \bar{dV}$$


from which it can de deducted, that for a constant temperature


$$ \frac{dP}{\bar{dV}} = -RT $$




This looks like the differential of the ideal gas equation. I'm going to use $V\_m$ for the molar volume.


$$PV\_m=RT$$


Now we convert each side to the differential. We keep $P$ and $V\_m$ together since they are both variable. We can move $R$ out of the differential because it is a constant.


$$d\left(PV\_m\right)=RdT$$


We can evaluate $d\left(PV\_m\right)$ using the [product rule](http://en.wikipedia.org/wiki/Product_rule):


$$d\left(PV\_m\right) = PdV\_m + V\_mdP$$


Now:


$$PdV\_m + V\_mdP = RdT$$


Which is equivalent to your first equation.


\*\*However, at constant temperature $dT=0$, so at constant temperature, the equation simplifies to:


$$PdV\_m + V\_mdP=0$$
$$V\_mdP = - PdV\_m$$
$$\frac{dP}{dV\_m}=-\frac{P}{V\_m}$$


Which is not equal to your second equation. The closest I can get is the following:
$$\frac{dP}{dV\_m}=-\frac{P}{V\_m}\cdot\frac{V\_m}{V\_m}=-\frac{PV\_m}{V^2\_m}=-\frac{RT}{V^2\_m}$$


Which is what we would get by evaulating:


$$\left(\frac{\partial}{\partial V\_m}\right)\_T \left(P=\frac{RT}{V\_m}\right)$$


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