External pressure when internal pressure is constant

[Termodinamico]

Homework Statement

The homework was about a piston-cylinder system in which i know that the internal pressure of the substance is known and constant. I haven't got any info on the reversibility of the process nor if there are mechanical equilibriums at the beggining or ending, even though I could make some assumptions and solve for expansion work, I'd like to derive a general statement for when this happens.

Relevant Equations

Internal pressure: Pint
External pressure: Pext
Expansion work: Wexp

dPint = 0
dWexp = Pext dV
dU = dQ - dWexp
T dS = dU + Pint dV
If process is reversible then Pext = Pint (+-) dPint

So, when I try to solve for expansion work, I don't know nothing about Pext. I've seen multiple times that they just replace Pext with Pint but I don't really get why. I thought about a few situations and I'd like for you to correct me if I'm wrong.

If the system in vaccum: the simplest one, Pext = 0 and there isn't any expansion work being done.

If the system is inside an atmosphere: at the beggining of the process there is mechanical equilibrium and Pext = Pint, Pint is constant and so is Pext.

If the Pext is variable: I can only think about having a variable Pext and constant Pint if I make changes to the system through heat or other kind of works at the same rate that Pext changes, that kinda sounds like a reversible? Not sure really but let say it does, then Pext = Pint (+-) dPint and the latter being 0, Pext = Pint.

I thought of another possible experiments but they don't seem to admit constant Pint or they result similar to the previous ones.

So, the question is if I have constant internal pressure, can I say that the external pressure will be equal to it?
Even though it is for different reasons.

 

[etotheipi]

So, when I try to solve for expansion work, I don't know nothing about Pext. I've seen multiple times that they just replace Pext with Pint but I don't really get why.

If the process is quasistatic, i.e. happens so slowly that it is always (effectively) in equilibrium, then $P_{ext} = P_{int}$. You can get this just by applying $\sum F_x = 0$ to the piston.

As you say, expansion work is always $W = -\int P_{ext} dV$. You can derive this from applying the work energy theorem to the piston between two rest points. If the process is quasi-static (i.e. a reversible expansion), then you have $P_{int} = P_{ext}$ so the formula $W = -\int P_{int} dV$ also now applies. And this is useful, since you can relate $P_{int}$ to quantities like $V$ (i.e. $P_{int}V = nRT$) and derive lots of useful formulae. In introductory thermodynamics, many of the examples are reversible so using the internal pressure is valid and the point you raise is not often discussed!

However, for a non-quasistatic expansion, only $W = -\int P_{ext} dV$ applies. The reason is that the internal pressure is not well defined for such an expansion, and the actual force per unit area on the inside piston face is not equal to $\frac{nRT}{V}$.

I will also tag @Chestermiller since I'm not entirely sure why the internal pressure is undefined/not equal to the force per unit area at the inside piston face for a non-quasistatic process; I believe it has to do with viscous stresses in the gas, but hopefully he will be able to add more insight!

 

[Chestermiller]

If the process is quasistatic, i.e. happens so slowly that it is always (effectively) in equilibrium, then $P_{ext} = P_{int}$. You can get this just by applying $\sum F_x = 0$ to the piston.

As you say, expansion work is always $W = -\int P_{ext} dV$. You can derive this from applying the work energy theorem to the piston between two rest points. If the process is quasi-static (i.e. a reversible expansion), then you have $P_{int} = P_{ext}$ so the formula $W = -\int P_{int} dV$ also now applies. And this is useful, since you can relate $P_{int}$ to quantities like $V$ (i.e. $P_{int}V = nRT$) and derive lots of useful formulae. In introductory thermodynamics, many of the examples are reversible so using the internal pressure is valid and the point you raise is not often discussed!

However, for a non-quasistatic expansion, only $W = -\int P_{ext} dV$ applies. The reason is that the internal pressure is not well defined for such an expansion, and the actual force per unit area on the inside piston face is not equal to $\frac{nRT}{V}$.

I will also tag @Chestermiller since I'm not entirely sure why the internal pressure is undefined/not equal to the force per unit area at the inside piston face for a non-quasistatic process; I believe it has to do with viscous stresses in the gas, but hopefully he will be able to add more insight!

Everything you said here is correct. It is important for the OP to recognize that, in a rapid (non-quasistatic) expansion or compression, the gas under consideration does not obey its equation of state (e.g., the ideal gas law). The force that the gas exerts on the piston depends not only on the volume change, but also on the rate of volume change. This is, as you have indicated, because of viscous stresses exerted by the gas. To get a better understanding of how this all works, one needs to study a little fluid mechanics. For that purpose, I highly recommend the textbook Transport Phenomena by Bird, Stewart, and Lightfoot. This is a masterful piece of work. Most of the information relevant to what we are discussing is laid out in Chapter 1.

@Thermodinamico Please avoid using differentials in the state functions when you are looking at large irreversible deformations. This will only lead to trouble and confusion. Please reserve the use of differentials of the state functions only to reversible deformations. Also, please provide us with an exact word-for-word statement of the problem you are trying to solve. Thank you.