# Expression for pressure Volume work

1. Jan 13, 2013

### amk_dbz

Work done by a gas (consider ideal for simplicity) is given by =
= -pressure(external) . Change in volume
Why is external pressure considered here?
Take for example a closed vessel with weight on apiston having gas enclosed with 200 atm pressure, if the weight is suddenly removed the gas expands against 1 atm external pressure
So why isn't the work done= integral (P(int).dV) which would be larger compared to P(ext).dV?
Any help would be appreciated. :)

2. Jan 13, 2013

### Andrew Mason

In a dynamic process, such as this, the gas is not in equilibrium. The gas does work on itself to create this dynamic energy, but this work ends up as internal (thermal) energy when the gas returns to an equilibrium state. So, although the gas does more than ∫PextdV work, the excess is not useful work. The W in the first law (Q = ΔU + W) refers to useful work.

AM

3. Jan 13, 2013

### amk_dbz

Thanks for the help but its not yet clear to me....
Also if i am compressing the gas by applying SUDDEN force to the the piston , which pressure should i consider for finding work on gas? P applied or Pressure of the gas?

4. Jan 13, 2013

### Staff: Mentor

For an irreversible compression like this, the situation is a little more complicated than you may think. The pressure within the gas inside the cylinder will not be constant from one end of the cylinder to the other, but rather will vary with axial position (and time). (So which pressure would one use in such a case?) To determine the detailed axial pressure variations within the cylinder, one would need to solve the differential gas dynamics equations. The solution to the gas dynamics equations tells us that the gas pressure within the cylinder will vary from the lowest values at the end of the cylinder away from the piston to the highest value at the face of the piston. If the piston has very little mass, than the gas pressure within the cylinder at the face of the piston will be essentially equal to the imposed external pressure on the piston, and will remain at that value throughout the compression. So the work done by the surroundings on the gas will be equal to both the external pressure times the change in volume (initial volume minus final volume) and to the gas pressure at the piston face times the change in volume. That is why we use the external pressure to calculate the work.

5. Jan 13, 2013

### Andrew Mason

The first law, Q = ΔU + W says that if Q = 0 (adiabatic) the work done on the surroundings by the system, or by the surroundings on the system, has to be equal in magnitude to the change in internal energy of the system.

In an expansion, work is done by the gas on the surroundings. The measure of that work is the force on the surroundings x distance through which the force acts, (or pressure x change in volume). So for an expansion you have to use the external pressure x change in volume.

In a compression, work is done by something outside the system on the system. The measure of the work done is the internal pressure of the gas x the change in volume. So for a compression to find the work done you have to use the internal pressure of the gas x change in volume.

Example: If there no external pressure at all (a free adiabatic expansion) the expanding gas does no external work at all (Q = W = 0). So there can be no change in internal energy.

Use internal pressure of the gas to determine work done during a compression.

I will assume the piston has non-negligible mass and you are referring to a sudden force that creates an external pressure exceeding the internal pressure of the gas for a period of time, Δt. The excess pressure will cause the piston to accelerate and gain kinetic energy (assuming no friction). This will cause the gas to continue being compressed after the external force is stopped (ie. after Δt) until the piston stops. The work done on the gas use the internal pressure x change in volume to determine the work done on the system in a compression.

AM

6. Jan 13, 2013

### Studiot

You always have difficulty applying a sudden deforming force to anything.

How do you propose to do this?

7. Jan 13, 2013

### Staff: Mentor

Are you asking us all, or just the OP?

8. Jan 13, 2013

### amk_dbz

Thank you guys for the help.....:)
Its not crystal clear yet but well better than it was before..

So do we take in consideration the 'resistive force' during the process (Pext for expansion and Pint for compression?). Then why isn't work done in pushing a block of mass on ground taken as, say Frictional force*displacement (Here frictional force is resistive right?)

And by sudden force, I meant like releasing a large weight of the piston in short time..I know that it will take some time to act but still the gas wont have time enough to gain equilibrium (irreversible process)..

Thanks again.

9. Jan 13, 2013

### Andrew Mason

Are you trying to find the work done by the surroundings on the gas or by the gas on the surroundings. They differ in sign.

If you are trying to find the work done by the pusher on the ground, you would use the applied force and the displacement which is in the direction of that applied force. If you are trying to find the work done by the ground on the pusher, you would use the force of the ground on the pusher (which is opposite in direction to the force of the pusher on the ground). The displacement is the same. So the work done by ground on pusher = - work done by pusher on ground.

AM

10. Jan 14, 2013

### Studiot

You can (very nearly) apply a sudden force to move an object as a whole, but if you apply a deforming force consider the following.

Consider the deformation / applied force curve.

This must be zero deformation at zero force or the body would be deforming on its own without the force.

If you suddenly release a support for your heavy piston so it moves down under gravity and compresses a gas in a chamber the weight of the piston will not instantaneously be applied to the gas in the chamber.
The piston will apply an increasing weight and the gas will 'push back' until its increased pressure produduces a force that balances the weight of the piston.
At this point the piston will be travelling at some velocity and will therefore continue until a further increase in gas pressure slows it to zero.
The force provided by the gas pressure then exceeeds the weight and thus the piston is accelerated back upwards.

In fact the piston oscillates and some of the energy goes into this oscillation and in the real world is dissipated.

Edit:
Once you have understood this, it leads to the answer to your question as to why we use the external pressure.

Last edited: Jan 14, 2013
11. Jan 14, 2013

### Staff: Mentor

Dear amk_dbz,

You should know that Andrew Mason and I have a fundamental disagreement on this point. My assertion is that you use Pext to calculate the work both for expansion and for compression. Here is my line of reasoning:

We can simplify things tremendously if we consider only the gas within the cylinder as our "system." Our system extends up to the inside face of the piston; and the surroundings end at the inside face of the piston.

Newton's third law (action-reaction) applies not only to forces between rigid bodies, but to forces between gases and solids as well. So, the force exerted by the inside face of the piston on the gas is PextA. The force exerted by the gas on the inside face of the piston must also be PextA. (Of course, it is in the opposite direction). The work done by the piston on the gas is -PextdV, and the work done by the gas on the surroundings must be +PextdV.

So, how come we don't use the pressure of the gas pgas to calculate the work. Well, during an irreversible process, the gas pressure within the cylinder varies with axial position. To calculate the work using the gas pressure, we would have to measure the gas pressure immediately adjacent to the piston face. This could be done by using a flush mounted pressure transducer or by having pressure transducers flush to the cylinder walls along the axis, and extrapolating to the instantaneous location of the inside piston face. Clearly, this is difficult to do, and it would be much easier just to use the external force per unit area pext. By doing this, we do not need to make any measurements inside the cylinder.

Overall, the point I want to emphasize is that, at the inside piston face, pgas=pext (always).

12. Jan 14, 2013

### Andrew Mason

Perhaps I am not understanding what you mean by pgas and pext, because it is only if the external pressure (ie the force provided by the piston /area of the piston) exceeds the internal pressure of the gas that the gas can compress. The irreversible process that we are talking about is one in which the piston suddenly accelerates, so there is a net (ie. unbalanced) force on the piston.

AM

13. Jan 14, 2013

### Staff: Mentor

The point that I've been trying to emphasize is that, during an irreversible compression or expansion, at any one time, the gas within the cylinder is not all at the same pressure. The gas pressure varies along the length of the cylinder (i.e, pgas = pgas(x,t) ). By Newton's third law of motion, the gas pressure at the inner face of the piston must exactly match the force per unit area exerted by the piston face on the gas. During an irreversible compression, the gas pressure at the piston face will be the highest at any location within the cylinder, and during an irreversible expansion, the gas pressure at the piston face will be the lowest at any location within the cylinder. Why is the gas pressure not uniform? Because of the inertia of the gas, which, during an irreversible process, is being rapidly accelerated.

Here is a rough description of what happens to the gas during an irreversible compression. Just prior to the onset of compression, the gas pressure within the cylinder is uniform, and the the force exerted by the inner piston face on the gas is equal to this initial gas pressure. However, when the piston starts to rapidly move toward the dead end of the cylinder, there is a narrow compression zone that forms within the gas immediately adjacent to the inner piston face. The pressure within this zone is significantly higher than the initial pressure within the cylinder, and is also equal to the force per unit area exerted by the piston face. Beyond this compression zone, the gas within the cylinder is still unaware that any compression has occurred. Its pressure is still equal to the initial gas pressure. There is essentially a sharp discontinuity between the compression zone and the, as yet, uncompressed gas closer to the dead end of the cylinder. As time progresses, the size of the compression zone increases, and the size of the uncompressed gas zone decreases until, finally, all the gas within the cylinder has been compressed. I should also mention that, within the compression zone, the gas temperature has risen above the initial temperature of gas in the cylinder, and within the uncompressed zone, the temperature is still at the initial temperature. So not only is the pressure varying discontinuously with position along the cylinder, but so also is the gas temperature. The rate of growth of the compression zone is very rapid, and the discontinuity advances at a speed comparable to the speed of sound in the gas.

Now, if you are really interested in all the details of this, I can present a derivation of the gas dynamics equations for irreversible compression (or expansion) within the cylinder, and can also present a solution to these gas dynamics equations (which includes all the physical features that I discussed in the previous paragraph). But, I'm hoping that this won't be necessary.

With regard to your statement that "The irreversible process that we are talking about is one in which the piston suddenly accelerates, so there is a net (ie. unbalanced) force on the piston," since I am not including the piston as part of the system (but only the gas within the cylinder), I only need to address what is happening at the inner face of the piston, where the force per unit area exerted by the inner piston face is always equal to the pressure of the gas immediately adjacent to the piston.

Is there anyone else out there whom what I've been saying makes sense to?

14. Jan 14, 2013

### Andrew Mason

It depends on how fast the compression is. A compression wave moves pretty quickly though a gas.

That is true, of course. But how do you measure that pressure exerted by the piston? ie. Pext? It depends on the pressure of the gas and the mass and speeds of the gas molecules. If there are no gas molecules at all, there is no force exerted by the piston so Pext=0

It makes perfect sense. You are making a valiant effort to analyse the physics of a dynamic process. But I am not sure it is useful. While the internal and external pressures are necessarily the same at the interface between the piston and the gas in the cylinder, this does not tell us what either pressure is. It is really complicated. Really complicated.

Thermodynamics, despite its name, is about equilibrium states. We only really care what the states are before and after the process has occurred.

So, for example, if I have an insulated cylinder filled with an ideal monatomic gas and drop a piston of mass m (m >> mass of gas) into the cylinder (friction is negligible), the work done on the gas is mgΔh where Δh is the difference between the initial and final resting height of the piston. This must equal ΔU (= 3RΔT/2) of the gas. Whether that is some complicated integral of some notional external or internal pressure x volume oscillating back and forth for a few milliseconds really doesn't matter, does it?

AM

15. Jan 15, 2013

### amk_dbz

Thank you for helping me sort this problem (it was bugging me continuously :) )
Its seems clear to me right now, I might trouble you guys more later in case I find something incomprehensible.
Again thank you.

16. Jan 15, 2013

### Studiot

@amkdbz

Are you still interested in this subject, there is much more to say?

@Andrew Mason

Do you not think that the fluid in the piston example forms an good example of a spring?

In which case there is a problem between the loss of potential energy of a massive piston dropped onto the gas (edit due to) the elastic constants of a volume of gas.

for instance

Last edited: Jan 15, 2013
17. Jan 15, 2013

### Staff: Mentor

"It depends on how fast the compression is. A compression wave moves pretty quickly though a gas. "

For a compression or expansion within a cylinder, this is what it takes for the pressure, temperature, and density to vary with position, and, in particular, for the pressure at the inside face of the piston to differ from the pressures at other locations within the gas. More precisely, the characteristic time for the compression or expansion multiplied by the speed of a sound in the gas must be on the same order or smaller than the length of the cylinder. Probably, this is why, in practice, if the piston is frictionless and the time scale for expansion or compression is relatively large (by this criterion), the compression or expansion can be regarded as reversible, the gas pressure within the cylinder will be nearly uniform; also, the gas pressure on the inside face of the piston pext will match the gas pressure everywhere else in the cylinder.

"That is true, of course. But how do you measure that pressure exerted by the piston? ie. Pext? It depends on the pressure of the gas and the mass and speeds of the gas molecules."

Lets suppose that, by brute force, you can control the motion of the outside face of the piston as a function of time (i.e., you control its velocity) and you can also measure the force required to impose this motion, (This way you don't have to make measurements of pressure within the cylinder). If the mass of the piston is negligible, then this force divided by the area of the piston must be equal to pressure of the gas on the inside face of the piston (by Newton's second law). But, what if the mass of the piston is not negligible. Then, if you do a force balance on the piston, you find that:
$$p_{ext} A= p_{gipf}A=F+m\frac{dv}{dt}=p_{opf}A+m\frac{dv}{dt}$$
where $p_{ext}$ is the force per unit area exerted by the inside piston face on the gas, $p_{gipf}$ is the pressure of the gas at the inside face of the piston, m is the mass of the piston, F is the imposed force on the outside piston face, $p_{opf}$ is the imposed force per unit area on the outside piston face (F/A) and v(t) is the imposed velocity variation of the piston.

If we take the above equation, multiply by the velocity v, and integrate between the initial and final equilibrium states, we obtain:
$$W = \int p_{gipf}dV=\int p_{opf}dV$$
According to this equation, it doesn't matter whether we calculate the work between the two equilibrium states by using the pressure history at the inside face of the piston or the pressure history at the outside face of the piston. You get the same result either way. This is because, the change in kinetic energy of the piston is zero between the two equilibrium states.

"It makes perfect sense. You are making a valiant effort to analyse the physics of a dynamic process. But I am not sure it is useful. While the internal and external pressures are necessarily the same at the interface between the piston and the gas in the cylinder, this does not tell us what either pressure is. It is really complicated. Really complicated."

The only reason I discussed the dynamics was to provide a rationale for how the pressure of the gas would vary within the cylinder during an irreversible compression or expansion. As far as the pressure at the interface between the piston and the gas in the cylinder are concerned, I showed in the previous paragraph how the pressure histories on both faces of the piston could be measured, and how they could be used to calculate the irreversible work. For obvious reasons, in calculating the irreversible work, my preference would be to employ the time history of the force on the outside face of the piston.

Last edited: Jan 15, 2013