What is the mathematical significance of $\int dPdV$?

Question

Today in our chemistry class we derived the pressure-volume work done on an ideal gas. Our foremost assumption was that $P_{ext}=P_{int}+dP$ so that all the time the system remains (approximately) in equilibrium with the surrounding and the process occurs very slowly (its a reversible process).

Now $$W_{ext}=\int P_{ext}dV$$ $$\Rightarrow W_{ext}=\int (P_{int}+dP) dV$$ $$W_{ext}=\int P_{int}dV$$ (He reasons this as since $ dPdV$ is very small $\int dPdV =0$(isn't dV itself infinitesimal?)).

• What is the mathematical significance of $\int dPdV$? Does it have any meaning attached to it (say in multiple variable calculus, if so what?)?

[Note: I ain't yet formally introduced to the concept of integration in mathematics, whatsoever knowledge that I have is from what I was taught in during the introduction of physics]

Answer 1

These calculations are not much rigorous. It is actually more like an intuitive description of the behavior of the system than a mathematical one. To me, this integral $\int dP dV$ is not well-defined. But there is another way to atack your problem which is more "down to earth" and enlightening. Consider a closed system with some gas on it and a piston, which can move freely. For simplicity, let us suppose this is a large cylinder with a piston. Suppose this piston has area $A$. Now, remember that the pressure due to a force $F$ on some surface with area $A$ is given by $P = \frac{F}{A}$, so $F = PA$. Now, if for some reason the gas expands and pushes the piston upward (say this is the positive direction), this gas has done some work. Let us suppose that the gas pushes the piston from position $x$ to position $x+dx$, where $dx$ is "infinitesimal". Thus, the work done by the gas during this procedure is also infinitesimal and given by $dW = Fdx$. But we know that $F = PA$ so $dW = PAdx$. But, because $A$ is the surface area of the piston, $Adx$ is just the infinitesimal volume of a cylinder with basis area $A$ and height $dx$. Thus, we may write $Adx = dV$, where $dV$ stands for this infinitesimal volume, so that $dW = PdV$. Thus $$W = \int_{V_{0}}^{V} PdV$$ Note that this is just a motivation for the formula, once this does not represent the most general system one could have. This, however, is an intuitve approach, with physical meaning and terms like $\int dP dV$ play no role.

Answer 2

Here's a more rigorous form of that same argument:

First, take the case that the pressures are different by a non-infinitesmal amount, that is, $$P_{\text{ext}} = P_{\text{int}} + \Delta P$$ Note that here $\Delta P$ is not infinitesimal.

Then you can calculate the work done by the external gas by $$W_{\text{ext}} = \int P_{\text{ext}}\,\mathrm dV = \int P_{\text{int}}\,\mathrm dV+\int \Delta P\,\mathrm dV = W_{\text{int}}+\int\Delta P\,\mathrm dV$$ where $W_{\text{int}}$ is the work done on the internal gas. The last term is then the energy “lost” e.g. by friction, that is, it is not reversible.

To get the reversible process, you now take the limit $\Delta P\to 0$ (well, strictly speaking I'm not completely rigorous here, because in general $\Delta P$ will be a function, too, but let's for simplicity assume $\Delta P$ is a constant throughout the process). Clearly, in this equation neither $W_{\text{ext}}$ nor $W_{\text{int}}$ depends on $\Delta P$. And $\lim_{\Delta P\to 0}\int \Delta P\,\mathrm dV=0$.

Therefore, in the reversible limit, $W_{\text{ext}} = W_{\text{int}}$.