External Pressure vs. Internal Pressure when considering work

[MaestroBach]

Summary:: Basically the title: It seems that one of the formulas I use considers the pressure in PdV to be internal, when as far as I know, it is external.

So to my understanding, in w = PdV, the pressure is the external pressure.

However, I get tripped up because in my textbook, for a reversible adiabatic process, it states that:

U = -w (which makes sense)

and then goes to say that:

w = -PdV, but P = (nRT/V) from the ideal gas law and so they replace P with that expression in the integral and integrate from V1 to V2.

But wouldn't that really be the equivalent of looking at the change in volume of the external volume, given that the P comes from an external pressure? Assuming pretty common situations, shouldn't the external pressure always just be constant?

(Sorry about my lack of LaTeX, I haven't used the forums in a while and have completely forgot how to turn it on...)

 

[ergospherical]

The pressure used to calculate work is always the external pressure, yes: $\delta w = -p_{\mathrm{ext}} dV$. However, in a reversible process such as the reversible expansion of a gas inside a piston, it's assumed that the piston is always in equilibrium (i.e. the pressure on one side of the piston is only infinitesimally greater than on the other side). In such processes the external pressure is equal to the internal pressure (given by some equation of state, e.g. the ideal gas law), so you can use this instead.

 

[Chestermiller]

See my responses in the following thread: https://www.physicsforums.com/threa...lways-greater-than-irreversible-work.1007230/

 

[MaestroBach]

The pressure used to calculate work is always the external pressure, yes: $\delta w = -p_{\mathrm{ext}} dV$. However, in a reversible process such as the reversible expansion of a gas inside a piston, it's assumed that the piston is always in equilibrium (i.e. the pressure on one side of the piston is only infinitesimally greater than on the other side). In such processes the external pressure is equal to the internal pressure (given by some equation of state, e.g. the ideal gas law), so you can use this instead.

This makes a lot of sense, thanks a ton!

 

[DrDu]

This also depends on which of the two pressures internal/ external is larger and obviously, far from equilibrium, work may even for gasses not be a function of pressure even if there is only volume work. So dw=-pdV is unproblematic only in quasistatic processes.

 

[Chestermiller]

Here are a couple of true-false questions to stimulate the discussion. True or false: In a rapid irreversible expansion or compression of an ideal gas,

1. the force per unit area exerted by the gas on the inside face of the piston is described by the ideal gas law.

2. The external pressure exerted by the inside face of the piston on the gas is equal to the internal force per unit area exerted by the gas on the inside face of the piston.

 

[MaestroBach]

Here are a couple of true-false questions to stimulate the discussion. True or false: In a rapid irreversible expansion or compression of an ideal gas,

1. the force per unit area exerted by the gas on the inside face of the piston is described by the ideal gas law.

2. The external pressure exerted by the inside face of the piston on the gas is equal to the internal force per unit area exerted by the gas on the inside face of the piston.

1. True
2. I'm not quite sure what "the external pressure exerted by the inside face" means- does that mean the pressure that the internal gas exerts on the exterior? If so, I would say True but only in a reversible process.

 

[Chestermiller]

1. True

This answer is incorrect. The ideal gas law (or any real gas equation of state) only applies to a gas at thermodynamic equilibrium (or to a gas experiencing a reversible expansion or compression, which consists of a continuous sequence of thermodynamic equilibrium states). In a gas experiencing a rapid irreversible deformation, the properties of density (inertia) and viscosity come into play; both of these cause the force per unit area exerted by the gas on the inside face of the piston to deviate significantly from the pressure calculated from the ideal gas law. The inertia of the gas allows compression and expansion waves to travel through the gas so that the compressive stress in the gas is not even spatially uniform. The viscosity of the gas allows anisotropic viscous tensile or compressive stresses to develop within the gas that are proportional to the derivatives of the gas velocity with respect to spatial position and add to the equilibrium thermodynamic pressure. You will learn about viscous stresses when you study fluid mechanics. The bottom line is that the ideal gas law cannot be used to calculate the force per unit area on the piston face in a rapid irreversible deformation.

2. I'm not quite sure what "the external pressure exerted by the inside face" means- does that mean the pressure that the internal gas exerts on the exterior? If so, I would say True but only in a reversible process.

Incorrect again. By Newton's 3rd law, the force per unit area exerted by the inside face of the piston (which we usually call the "external pressure") on the gas is exactly equal to the force per unit area exerted by the gas on the inside face of the piston. But, as we've already noted, the force per unit area exerted by the gas on the inside face of the piston is not equal to that predicted by the ideal gas law. So to solve problems involving rapid irreversible expansions or compression of ideal gases, we have a dilemma. How are we supposed calculate the work done by the gas in applying the first law of thermodynamics? Thoughts?

I hope that this exercise so far has improved your understanding of internal and external "pressure" in a rapid irreversible deformation.