# How to differentiate the ideal gas equation?

The equation states that $$p \cdot V=m \cdot R \cdot T$$ where R is a constant, m is mass, T is temperature, p is pressure, V is volume.

My textbook says "By differentiating the ideal gas equation we get:

$$p \cdot dV + V \cdot dp=m \cdot R \cdot dT$$

Later on it differentiates this: $$p1 \cdot V1=p2 \cdot V2=const$$

into this: $$\frac{dp}{dV}=-\frac{p}{V}$$

Now I don't understand how they are getting this. We learnt differentiation in math class but it looks nothing like this (?) so I'm hoping someone can explain this a little.

• Physicists have a bit of a reputation for playing fast and loose with the differential operators compared to how you see it presented in the math book. It is sound, it just isn't what you might be familiar with. – Doug M Jun 28 '17 at 22:46
• For the second result, use PV = constant, which means dT = 0 (no change in temperature). The first result can then be written $p dV + v dP = m R dT = 0$ and then you can get $p dV = - V dP$, namely $\frac{dP}{dV}=-\frac{V}{P}$. – jim Jun 28 '17 at 23:31

## 2 Answers

Denote $df/dt$ by $f^\prime$. Then, $d (pV)/dt = pV^\prime+p^\prime V$ by product rule, and $d (mRT)/dt = mRT^\prime$ since $R$ is constant and mass is assumed to be constant wrt time. Thus, multiplying by $dt$ on both sides gives $pdV+dpV = mRdT$.

The second part can be derived like so: $PV=k$ for some constant $k$. So $P=k/V$, and thus $\frac{dP}{dV}=-\frac{k}{V^2}=-\frac{k}{V}\cdot\frac{1}{V}=\frac{-P}{V}$ (since, again, $P=k/V$).

For the first part, you differentiate both sides with respect to time as AlexanderJ93 showed and "multiply out" $dt$. This comes from the fact that physicists like to view differentials like $dt$ as very very small, infinitesimal changes. Even though this is not extremely rigorous, it often works for practical purposes since the concept is so similar to that of a limit.