What is constant in an isobaric process, internal pressure or external pressure?

Question

I was taking a course in thermodynamics, where the instructor started to derive the formula for work in case of isobaric process for an ideal gas It stated W = -∫ P_ext dV And then equated it with -∫PdV where P was the internal pressure. I am sure that it is not a misunderstanding because later this PdV was written as d(PV) (because P is stated to be constant) and further written as d(nRT) This brings me to my question.

1. Is it the external pressure or the internal pressure that is constant in a isobaric process? The above equations indicate that it is the internal pressure which us constant but for isobaric process, it is the external pressure vs volume graph that has a horizontal line

2. If it is internal pressure that is constant which seems to be the case to me, how is it, that we can equate internal pressure and external pressure?

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Answer 1

In a process in which the gas is considered the "system," the force per unit area exerted by the system on its surroundings (sometimes interpreted as the "internal pressure") is, by Newton's 3rd law of action-reaction, exactly equal to the "external pressure" of the surroundings on the gas. The problem is that, for a rapid irreversible expansion or compression, the "internal pressure" is not described by the ideal gas law. The ideal gas law applies only to equilibrium (and quasi-static) situations. Otherwise, the "internal pressure" includes a contribution from viscous stresses (which depend not on the overall deformation but on the rate of deformation), as described by the 3D tensorial version of Newton's law of viscosity. So, in lieu of detailed information on the local rate of gas deformation during the process, we can sometimes use the external pressure to calculate the work (if the external pressure is known as a function of the volume). In any case, the work done by the gas is always equal to the integral of $P_{ext}dV$, and, for what we call an isobaric process, both the "internal pressure" and the "external pressure" are constant. It's just that, if the process is irreversible, the "internal pressure" is not given by the ideal gas law.

Answer 2

In a laboratory experiment, you can’t control the inner pressure of a material. An “isobaric” quantity is measured by keeping the external pressure constant and allowing the material to respond.

If you want to measure an isobaric property at one atmosphere, and the pressure dependence is (suspected to be) weak, then you can just do your experiment on the tabletop without any vessel. That is the data which the models of thermodynamics are designed to explain.

Answer 3

1. Is it the external pressure or the internal pressure that is constant in a isobaric process?

It is always the external pressure for any process. Only if it is a reversible isobaric process, it is both the external and gas pressure. That's because for a reversible isobaric process the pressure of the gas is always in equilibrium with the external pressure. For the process to be reversible, it has to be carried out very slowly (quasi-statically).

So if you are looking at a PV diagram the pressure shown is always the external pressure. Only if your are told the process is reversible is when the pressure on the graph is also the internal gas pressure.

2. If it is internal pressure that is constant which seems to be the case to me, how is it, that we can equate internal pressure and external pressure?

The external and internal pressure are always the same at the boundary between the system and surroundings, per Newton's third law. For reversible process the pressure of the gas throughout is the same as the external pressure. For an irreversible process, pressure gradients exist in the gas.

Hope this helps.

Answer 4

In the case of a general process, the internal and external pressures do not need to be the same.

However, if the work $-\int P_{ext} dV$ was equated with $-\int P dV$, presumably, the context was that of a reversible quasistatic transformation. Indeed, a quasistatic transformation is required to be able to speak about a well-defined (and unique) internal pressure $P$.

A counterexample would be the case of a fast-moving piston, generating inhomogeneous pressure variations (pressure waves). Moreover, no friction is present if the two works are equated.