Why do we use $P_{\rm ext}$ in the formula of $\int p \, dv $ work?

Question

To what I can understand-in a piston cylinder arrangement, piston moves out due to net force experienced by piston due to difference between internal pressure of system and external pressure. So, the formula of work done by system on piston, should be

Work = $\int (P_{\rm int}-P_{\rm ext})\cdot{\rm area \,}\cdot dx$

then only we should be able to account for the net force acting on piston but then why do we define work as $\int P_{\rm ext} \cdot {\rm area} \cdot dx$?

To be clear, I understand the concept of reversible processes (where $P_{\rm ext}=P_{\rm int}$ as the system is undergoing a quasistatic process, so in such a case work $= \int P_{\rm int} \cdot dV $= $\int P_{\rm ext} \cdot dV)$ as well as irreversible processes (where work $=\int P_{\rm ext} \cdot dv$).

What I don't understand is why $P_{\rm ext}$ is used in the formula of work done instead of the pressure difference (i.e., $ P_{\rm int} - P_{\rm ext}$)?

Thanks.

Answer 1

These are two different forces acting on the piston: one due to external factors (e.g., a person pushing the piston) and the other due to the gas contained inside the cylinder. Now, as in mechanics, we can calculate work done by each of these forces or work done by the net force. The latter is of little interest to us, whereas the former determines how much the energy is added to the gas due to the work done by the external force. (The work done by the net force tells us how much the energy of the piston itself has changed - this is of little interest, and in quasistatic processes is zero.)

Answer 2

First law of thermodynamics states that

Work done on system = Increase in internal energy + Heat released by the system.

It is just a way of writing conservation of energy in macroscopic scale.

Here, $P_{ext}$ leads to a external force on the system which makes it move and does work on the system.

On the other hand, $P_{int}V$ difference between start and end state of the system is accounted in variation in internal energy of the system. Moreover, $P_{int}$ is not well defined when system is not in equilibrium.

But work done by internal forces are taken into account while writing the change in internal energy as potential energy change is negative of the work done by internal forces.

So, in a way, we do get $P_{int}$ in the final equations. At the end of the day, what you need to satisfy is conservation of energy and the equations will turn out to be correct.

Answer 3

The net work on the piston includes all external forces; your equation is correct if there is no gravity and no friction. The net work on the piston is the change in kinetic energy of the piston.

If you want the work done by the system on the surroundings and you include the piston as part of the system, that work is integral Pext dV.

For a quasi-static, quasi-equilibrium process, the piston moves very slowly and the work done by the system on the surroundings is integral Pext dV = integral Pint DV.

See my answer to Work done by a gas on this exchange.

Answer 4

By Newton's 2nd law of motion, for the case of a massless, frictionless piston, the outside pressure $P_{ext}$ must be equal to the average compressive normal stress of the gas on the inside face of the piston $\bar{\sigma}=\frac{1}{A}\int{\sigma dA}$ where $\sigma$ is the local compressive normal stress at the piston face. From Newtonian fluid mechanics, we learn that the local normal stress is the linear sum of a compressive isotropic ideal gas pressure $P=\frac{RT}{v}$ and a viscous normal stress component $\tau$, where T, v, and P vary with position dA at the piston, and the viscous stress component varies not with the amount of gas deformation (local specific volume change), bu the rate of gas deformation (determined by local velocity gradients in the gas).

In reversible deformations, the viscous contribution is negligible, but in rapid irreversible deformations, the viscous contribution $\tau$is significant. In particular, for irreversible expansion, the average value of P at the piston face is higher than the external pressure, but the average value of $\tau$ is negative, such that $$\frac{1}{A}\int{(P+\tau)dA}=P_{ext}$$For irreversible compression, P at the piston is lower than the external pressure, but the average value of $\tau$ is positive, so again $$\frac{1}{A}\int{(P+\tau)dA}=P_{ext}$$

So my points are:

1. The external pressure is actually equal to the mean compressive normal stress of the gas on the piston (from a massless, frictionless piston)

2. For an irreversible process, the compressive normal stress is not equal to the pressure determined by the ideal gas law (or other equilibrium equation of state). You can't determine the contribution of viscous stresses to the compressive normal stress without a detailed fluid mechanics analysis on the gas during the process. So you are stuck having to know or specify the external pressure $P_{ext}$.