# Thermo - PV work and mechanical work

1. Mar 16, 2014

### LoopInt

Hello,

Im a bit confused about PV work. Here is the situation I am facing.

δw = P dV and dw = F dx

δw is an inexact differential, because it's path dependent. I read some articles that equal δw with dw.

saying that

P dV = F dx

For what path is that True? In other words, When δw=dw?

2. Mar 16, 2014

### Staff: Mentor

Let's rewrite the equation as follows:
$$PdV=Fdx=\frac{F}{A}(Adx)$$
If A is the area of the piston, then Adx = dV

So
$$PdV=Fdx=\frac{F}{A}(Adx)=\frac{F}{A}dV$$
Therefore, $$F = PA$$
For a more detailed discussion of how to properly calculate the work at the interface between the system and surroundings, see my Blog at https://www.physicsforums.com/member.php?u=345636

Chet

3. Mar 17, 2014

### DrDu

Instead of dw=F dx you should rather write δw=F dx. Work alone is in general not a total differential.

4. Mar 17, 2014

### LoopInt

Thanks for the replies.
Chestermiller, I did these steps but I couldnt figure out why Pdv=Fdx if they are path dependent.
And yes DrDu you are right. dw is only the case when the force is conservative, right?

dw=PdV is right when the process is adiabatic (the force will be conservative now)
so, if the process is adiabatic I can say dw=dw, and therefore PdV=Fdx
Am I right?

5. Mar 17, 2014

### Staff: Mentor

They're both different ways of writing exactly the same thing. Unless you are dealing with a conservative force field, F is path dependent.

Chet

6. Mar 17, 2014

### LoopInt

Yes, I did. The confusion i'm making is the following:

I'm considering work the area under the p-v diagram curve. If I have a piston and pressurize it, the force will be F=P.A.
Now for the work point of view. w=\integral P.dv wich may vary depending on the path. Therefore consider 2 different paths that done w1 and w2.
w1 = F1.dx
w2 = F2.dx
if w1 > w2 then F1 >F2. But for a given pressure on the piston, the force is F regardless of the path.

More precisely, I'm trying to figure out the force that a membrane exerts. For that I am using F=Pdv/dx (dv/dx is the rate of which the volume changes when I pull it dx). But I am afraid I am calculating the wrong F, because it may depend on which process I do that.

For a given pressure, the membrane should exert the same force regardless of how it reached that state (pressure and volume).

Last edited: Mar 17, 2014
7. Mar 17, 2014

### Staff: Mentor

As I said in my blog, for an irreversible path between the initial and final equilibrium states, the work is not the area under the equilibrium curve on the p-v diagram. The pressure within the system for an irreversible path is typically not even uniform. So what pressure do you use? The correct pressure to use is the pressure at the interface with the surroundings (which differs from the equilibrium pressure at the volume v). Only for a reversible path is the work equal to the area under the equilibrium curve on the p-v diagram. This is because, for a reversible path, the pressure within the system is uniform, and equal to the pressure at the interface with the surroundings.

To help solidify your understanding, why don't you dream up a specific problem that we can solve together.

Chet

8. Mar 17, 2014

### LoopInt

Sure, that's very nice of you.

Let's consider a balloon that is fixed in one end at the ceiling and the other end has a load F=mg. The balloon has volume V1 and initial pressure P1. The mass is at height 0 (reference point). We then blow air into the balloon and the pressure inside it increases. The mas then raises a distance x. What is that distance? What about if the pressure is kept constant and take the load off instead?

I draw something to make it clear.

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Last edited: Mar 17, 2014
9. Mar 17, 2014

### Staff: Mentor

I'm requesting that you think of a simpler problem than this one (one that is also capable of illustrating your issue). If we're dealing with a balloon, then we have to consider the stress/deformation behavior of the balloon material (which affects the force, pressure, and movement of the mass). We would also have to consider the non-linear kinematics of the balloon deformation, which is going to involve large changes in balloon curvature and large inhomogeneous deformations of the balloon membrane. I would like to avoid this complexity.

Might I suggest some sort of simple problem involving a gas, piston, and cylinder, with possibly surroundings pressure or force external to the piston.

Chet

10. Mar 17, 2014

### LoopInt

Sure, sorry!

Then imagine a piston at the floor pointing up (vertical position). It has a mass at the top. Pressure inside is P1 and volume V1. We pressurize it and the mass goes up by x. The new pressure is now P2 and V2. The piston is a cylinder. What is the distance x? And what happens if we take the load out keeping the pressure P1?

11. Mar 17, 2014

### Staff: Mentor

How exactly do we pressurize it? Increase the temperature? Add more gas to the cylinder?
What is the condition on the other side of the piston, air at atmospheric pressure?

How exactly do we take the load off while maintaining the pressure at P1, add a force in place of the weight?

Chet

12. Mar 17, 2014

### LoopInt

We put more gas in it. Outside it is atmospheric pressure. Temperature is kept constant.

While we take the load off we put gas in, so that pressure stays constant.

13. Mar 17, 2014

### Staff: Mentor

You are not going to be able to increase the pressure to P2 this way. The weight of the mass is mg, and the outside pressure is 1 atm., so, at final equilibrium, the pressure will be P2=P1 = mg/A + 1 atm and the weight will rise by ΔV/A.

14. Mar 17, 2014

### LoopInt

You are right.

Then initially its P1. Then we increase the mass until there is P2 in the cylinder. What's the distance x it dropped?

I found that

x= (L(A(P1-P0)-(m+M)g))/(P0A+(m+M)g)

where L is the distance between the bottom and the top of the piston, m is the mass at the beginning, M is the adicional mass.

The way I did it was:

- The sum of forces are 0, because it is in equilibrium;
- As the temperature is constant, P1V1=P2V2
- Isolate P2 for the above equation
- Consider V1=LA and V2=(L+x) I should had put (L-x) because P2 is higher, but that's not so relevant because I can choose x to increase downwards.
- Substitute V1 and V2 in the P2 equation
- Sum of forces when the mass is increased = 0
- Substitute P2 in the sum of forces equation
- Isolate x

If that is correct, consider two separate cases. We know that when we have P1 and V1 and press the piston by x, the force that the piston exerts is (m+M)g. If we press very fast and stop when P2 is reached, the force we apply to keep the system in P2 (equilibrium) is (m+M)g? In other words, I can do whatever I want to the system in any way, and at the state P2, V2 and the load will be the same always?

Last edited: Mar 17, 2014
15. Mar 17, 2014

### Staff: Mentor

I think you mean "decrease the mass", correct?
This analysis was not done correctly. Even if the temperature is constant, you decreased the number of moles. According to the ideal gas law, P1(V2-V1)=ΔnRT. So, here again the pressure doesn't change, but the volume does. For this case:
x = (V1-V2)/A=ΔnRT/(P1A)=ΔnRT/(mg+P0A)

Chet

16. Mar 17, 2014

### Staff: Mentor

OK. In the previous post, I misinterpreted what you were saying. I thought by mass, you were talking about the mass of gas in the cylinder. I will re-evaluate what you did, and get back with you.

Chet

Last edited: Mar 17, 2014
17. Mar 17, 2014

### Staff: Mentor

Aside from the sign, I get the same answer. I've re-expressed it as:
$$x=\frac{MgL}{((M+m)g+P_0A)}$$
Yes, as long as you do what you need to to make sure that the final temperature is the same.

Last edited: Mar 17, 2014
18. Mar 17, 2014

### LoopInt

Ok, I got it! Thanks!
Now, concerning that case of the ballon. The rubber would make it more complex, but is it the same case? I can do whatever I want, and still have same results since P, V and T are same?

19. Mar 17, 2014

### Staff: Mentor

Sure, as long as the balloon material is perfectly elastic.
Now I have a problem for you:

Suppose you attach a handle to the top of the mass M so you can lower the mass by hand. If you lower it gradually, how much work does piston do on the gas (assuming that the temperature is held constant, and you lower it slowly enough for the process to be reversible)? How much work does the piston do on your hand as you lower the weight? How much work is done on the piston in this process?

Chet

20. Mar 17, 2014

### LoopInt

The work the piston does on the gas is

W = A(nRT Ln(L/(L-d))-P0d)

For lowering the weight I do work of Mgd. Work done on the piston..... It is all the time in equilibrium, so There is no net force. Work should be 0 on the piston.

I do work of -Mgd, gravity does Mgd. The mass does A(nRT Ln(L/(L-d))-P0d) to the piston, P0 does P0Ad to the piston and gas does -AnRT Ln(L/(L-d)) to the piston.

Last edited: Mar 17, 2014
21. Mar 17, 2014

### Staff: Mentor

This is a pretty good start. You recognized that this is a reversible process, so the work on the gas is the area under the p-v curve. I got a slightly different answer:
$$W=nRT\ln{\left(\frac{L}{L-x}\right)}= nRT\ln{\left(\frac{V_1}{V_2}\right)}=nRT\ln{\left(\frac{P_2}{P_1}\right)}$$
Now let's continue addressing the rest of my questions. We're going to start out by doing a force balance on the combination of piston m and mass M. Since the system is always just slightly removed from equilibrium, we won't have a mass times acceleration term. Here is a list of the forces acting on the combination at some point during the lowering of the mass M:

F = upward force exerted by your hand on the top of M
P0A = downward force exerted by the atmospheric pressure on the top of M
PA = upward force exerted by the gas on the bottom of m
w = (M+m)g = downward force of gravity on m+M

Please write out this force balance. Also write out what this force balance reduces to time t = 0, when the mass M just begins to get lowered, m and M are just starting to kiss, and the volume of gas has not yet changed.

Please bear with me. I know that doing a force balance seems a little hinkey, but it will help us determine the amount of work being done.

Chet

22. Mar 18, 2014

### LoopInt

I didn't do that with only 1 FBD, but with 2 separate cases instead.
Doing the force balance on the piston gives me:

-Fatm+Fhand-Fmass+Fgas=0

Initially, my hand is lifting the whole mass so Fhand=mg, and therefore Fgas=Fatm.
Now, as I lower the weight

Fgas=Fatm+Fmass-Fhand

Where Fhand goes from mg to 0.
The work the piston does on the gas is the same as the work the gas does on the piston but with opposite signs, right?
So I can calculate the work of this expression
\integral F(x) dx = \integral (Fatm+Fmass-Fhand) dx
Now I need to express how the force varies with x
I will use PV=nRT to do that, because this equal the gas force.
P=nRT/V = nRT/(LA)
the force, then, is:
F=PA = nRT/L
But as I lower the weight by x, the height is L-x so,
F=nRT/(L-x)
Carrying the integration from x varying from 0 to d (maximum distance)
W = nRT Ln(L/(L-d))

23. Mar 18, 2014

### Staff: Mentor

Your force balance(s) do not look correct. Let C be the downward contact force exerted by the mass M on the piston. Then the force balance on the piston of mass m is:
$$PA-mg-C=0$$
And the force balance on the mass M is:
$$C-Mg+F-P_0A=0$$
At time zero, C=P0A and P=P1, so

$P_1A-P_0A=mg$ at time t = 0 and
$F=Mg$ at time t = 0
If we add the force balance on m to the force balance on M, we get the overall force balance on the "stack:"
$$PA-mg-Mg+F-P_0A=0$$
or
$$F=(m+M)g-PA+P_0A$$
Now, since downward movement is taken as positive, the differential work that the hand does when the stack moves down dx is -Fdx. Therefore, lets multiply the previous equation by -dx:
$$dW_F=-Fdx=-(m+M)gdx-PdV+P_0dV$$
Now, the next step is to integrate this equation from x = 0 to x = x (the final x) to get WF the work that the hand does on the stack, and -WF the work that the stack does on the hand.

After we finish this problem, we can do the irreversible problem of letting the mass M spontaneously fall from its initial position just kissing the top of the piston until it reaches its final equilibrium position. The results will not be the same. Now, won't that be fun?

Chet

24. Mar 18, 2014

### LoopInt

What I did was do the overall force balance directly. It equal to your force balance, but I considered the piston to be massless, wrongly.
The work done by the piston on the gas should be nRT Ln(L/(L-d)), right?
What you calculate is the work your hand is doing, isn't it?

25. Mar 18, 2014

### Staff: Mentor

Oh. OK. You had mentioned something about 2 force balances, so I got confused. So I think we now have the same force balance, correct?
Correct.

As I said, "WF the work that your hand does on the stack, and -WF the work that the stack does on the hand." So the answer to this question is Yes.

I think you've gotten what I wanted you to get out of this problem. I leave it up to you to complete the other integrations on the right hand side of the equation. But, just as a matter of summerizing, one good (practically fool proof) way of figuring out the amounts of work done (so that you get all the signs correct) is to start out with the force balance and then integrate the equation over the displacement.

If you are interested in continuing, we can do the irreversible problem in which, rather than lowering the weight gradually by hand, we just release the mass M and let it drop. We would be focusing on determining the work done by the gas on the piston, which is different for this case from what we got when we lowered the mass gradually. Any interest in continuing? If not I understand.

Chet