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Thermo - PV work and mechanical work

  1. Mar 16, 2014 #1

    I`m 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?

    Thanks in advance!
  2. jcsd
  3. Mar 16, 2014 #2
    Let's rewrite the equation as follows:
    If A is the area of the piston, then Adx = dV

    Therefore, [tex]F = PA[/tex]
    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

  4. Mar 17, 2014 #3


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    Instead of dw=F dx you should rather write δw=F dx. Work alone is in general not a total differential.
  5. Mar 17, 2014 #4
    Thanks for the replies.
    Chestermiller, I did these steps but I couldn`t 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?
  6. Mar 17, 2014 #5
    They're both different ways of writing exactly the same thing. Unless you are dealing with a conservative force field, F is path dependent.

    Did you read my blog?

  7. Mar 17, 2014 #6
    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
  8. Mar 17, 2014 #7
    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.

  9. Mar 17, 2014 #8
    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.

    Attached Files:

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

  11. Mar 17, 2014 #10
    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?
  12. Mar 17, 2014 #11
    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?

  13. Mar 17, 2014 #12

    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.
  14. Mar 17, 2014 #13
    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.
    Same answer for this case.
  15. Mar 17, 2014 #14
    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
  16. Mar 17, 2014 #15
    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)

  17. Mar 17, 2014 #16
    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.

    Last edited: Mar 17, 2014
  18. Mar 17, 2014 #17
    Aside from the sign, I get the same answer. I've re-expressed it as:
    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
  19. Mar 17, 2014 #18
    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?
  20. Mar 17, 2014 #19
    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?

  21. Mar 17, 2014 #20
    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
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