# External Pressure vs. Internal Pressure when considering work

• MaestroBach
In summary, the force per unit area exerted by a gas on the inside face of a piston cannot be accurately described by the ideal gas law in a rapid irreversible expansion or compression. The properties of density and viscosity cause the force per unit area to deviate from the pressure calculated from the ideal gas law, as compression and expansion waves and anisotropic viscous stresses come into play. Therefore, the ideal gas law only applies to a gas at thermodynamic equilibrium or experiencing a reversible expansion or compression.
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...)

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.

MaestroBach
ergospherical said:
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!

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.

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.

Chestermiller said:
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.

MaestroBach said:
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.
MaestroBach said:
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.

Chestermiller said:
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.

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.
Hello. I’ve been trying to grasp this concept for some time. I thought that force exerted by the gas on the piston face was higher than the force exerted by the surroundings on the gas-Isn’t this why the expansion occurs in the first place? So I don’t get how Newton’s third law is relatable during irreversible expansion/compression. Intuitively I realize that higher the opposing pressure, greater the work done by the gas,but still want a clearer insight. I really want using Pext to make sense. Hopefully you will reply I would appreciate it.

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Neddy said:
Hello. I’ve been trying to grasp this concept for some time. I thought that force exerted by the gas on the piston face was higher than the force exerted by the surroundings on the gas-Isn’t this why the expansion occurs in the first place? So I don’t get how Newton’s third law is relatable during irreversible expansion/compression. Intuitively I realize that higher the opposing pressure, greater the work done by the gas,but still want a clearer insight. I really want using Pext to make sense. Hopefully you will reply I would appreciate it.
When a horizontal spring exerts a force on a mass A situated on a horizontal frictionless surface, is that force not equal in magnitude to the force the mass A exerts on the spring? And yet the mass accelerates horizontally even though the contact forces are equal. When we consider a force balance on A, we only consider the forces that other entities exert on A, not the reaction forces that A exerts on other entities.

Let's consider a massless piston. If we apply Newton's 2nd law to the motion of this piston, we obtain ##F_g-P_{ext}A=ma=0## or ##F_g=P_{ext}A##, where ##F_g## is the force the gas exerts on the inside face of the piston, right? So the question is, how could ##F_g## satisfy this equation if the gas also satisfies the ideal gas law?

Yes, now I do realize my mistake that surroundings exerting the same force on the gas can’t stop it from expanding(as they act on different objects).I also realize that imbalance of forces acting on the same object causes its acceleration, but I don’t think I fully comprehend what is causing this imbalance of forces on the gas. Correct me If I am wrong at any part. I didn’t quite understand your question. Thanks a lot!

Neddy said:
Yes, now I do realize my mistake that surroundings exerting the same force on the gas can’t stop it from expanding(as they act on different objects).I also realize that imbalance of forces acting on the same object causes its acceleration, but I don’t think I fully comprehend what is causing this imbalance of forces on the gas. Correct me If I am wrong at any part. I didn’t quite understand your question. Thanks a lot!
Are you saying that the ideal gas law describes the behavior of an ideal gas even in a very rapid deformation of the gas characteristic of an irreversible expansion against a constant external pressure ##P_{ext}##? How can it satisfy the ideal gas equation under these circumstances? If it doesn't satisfy the ideal gas equations under these circumstances, what equation does it satisfy?

Chestermiller said:
Are you saying that the ideal gas law describes the behavior of an ideal gas even in a very rapid deformation of the gas characteristic of an irreversible expansion against a constant external pressure ##P_{ext}##? How can it satisfy the ideal gas equation under these circumstances? If it doesn't satisfy the ideal gas equations under these circumstances, what equation does it satisfy?
Ideal gas equation describes the behavior of gas at equilibrium, so can’t be deployed during rapid expansion. I don’t know what else equation it satisfies,though.

The ideal gas law describes the state of isotropic compressive stress in an ideal gas only at thermodynamic equilibrium (or in a reversible process, which consists of a continuous sequence of thermodynamic equilibrium states):$$\sigma_1=\sigma_2=\sigma_3=\frac{RT}{v}$$where the ##\sigma##s are the three principal compressive stresses in the fluid, R is the ideal gas constant, and T and v are the local temperature and local specific volume, respectively (which at thermodynamic equilibrium are spatially uniform).

In a rapid irreversible deformation of a Newtonian (viscous) fluid like air or water, the compressive stresses are not isotropic (equal in all directions), and are given by $$\sigma_1=\frac{RT}{v}-\frac{4}{3}\mu\frac{\partial u_1}{\partial x_1}$$$$\sigma_2=\frac{RT}{v}-\frac{4}{3}\mu\frac{\partial u_2}{\partial x_2}$$$$\sigma_3=\frac{RT}{v}-\frac{4}{3}\mu\frac{\partial u_3}{\partial x_3}$$where the u's are the fluid velocity components in the three principal directions of strain rate, and the x's are the spatial distances in the three principal directions of strain rate (at a given location). So the compressive stresses depend not only on the specific volume and temperature (as in an ideal gas at equilibrium), but also on the rate at which the gas is deforming in each direction. You will learn about these equations for irreversible deformations of liquids and gases when you study fluid mechanics.

Right now, all you need to know is that you can't use the ideal gas law for rapid irreversible deformations of an ideal gas.

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berkeman
Chestermiller said:
The ideal gas law describes the state of isotropic compressive stress in an ideal gas only at thermodynamic equilibrium (or in a reversible process, which consists of a continuous sequence of thermodynamic equilibrium states):$$\sigma_1=\sigma_2=\sigma_3=\frac{RT}{v}$$where the ##\sigma##s are the three principal compressive stresses in the fluid, R is the ideal gas constant, and T and v are the local temperature and local specific volume, respectively (which at thermodynamic equilibrium are spatially uniform).

In a rapid irreversible deformation of a Newtonian (viscous) fluid like air or water, the compressive stresses are not isotropic (equal in all directions), and are given by $$\sigma_1=\frac{RT}{v}-\frac{4}{3}\mu\frac{\partial u_1}{\partial x_1}$$$$\sigma_2=\frac{RT}{v}-\frac{4}{3}\mu\frac{\partial u_2}{\partial x_2}$$$$\sigma_3=\frac{RT}{v}-\frac{4}{3}\mu\frac{\partial u_3}{\partial x_3}$$where the u's are the fluid velocity components in the three principal directions of strain rate, and the x's are the spatial distances in the three principal directions of strain rate (at a given location). So the compressive stresses depend not only on the specific volume and temperature (as in an ideal gas at equilibrium), but also on the rate at which the gas is deforming in each direction. You will learn about these equations for irreversible deformations of liquids and gases when you study fluid mechanics.

Right now, all you need to know is that you can't use the ideal gas law for rapid irreversible deformations of an ideal gas.
I highly doubt that I will study any of this at my university.Thanks again for your time and attention!

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Neddy said:
I highly doubt that I will study any of this at my university.
Looks like a pretty fun and useful subject, though.

Neddy
Neddy said:
I highly doubt that I will study any of this at my university.Thanks again for your time and attention!
Sorry if it seems complicated to you now. This is just basic Newtonian fluid mechanics.

berkeman

## 1. What is external pressure in the context of work?

External pressure in the context of work refers to the demands and expectations placed on an individual by external factors, such as their employer, colleagues, or clients. It can also include societal or cultural pressures, as well as financial or time constraints.

## 2. How does external pressure affect work performance?

External pressure can have both positive and negative effects on work performance. On one hand, it can motivate individuals to work harder and meet expectations. However, excessive external pressure can also lead to stress, burnout, and decreased productivity.

## 3. What is internal pressure in relation to work?

Internal pressure in relation to work refers to the self-imposed expectations and standards that individuals place on themselves. This can include personal goals, perfectionism, and a desire to succeed and excel in their work.

## 4. How does internal pressure impact work performance?

Internal pressure can have a significant impact on work performance. It can drive individuals to set high standards for themselves and strive for excellence. However, it can also lead to self-doubt, anxiety, and a fear of failure, which can hinder productivity and overall performance.

## 5. What is the difference between external and internal pressure?

The main difference between external and internal pressure is the source of the pressure. External pressure comes from outside factors, such as employers or societal expectations, while internal pressure is self-imposed. Additionally, external pressure is often tangible and measurable, while internal pressure is more subjective and based on personal beliefs and goals.

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