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Work done on a spring by an ideal gas

  1. Feb 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the work done by an ideal gas on a hooke's spring when heat is added until volume and pressure triple

    2. Relevant equations
    Pv=nrT
    Q=ncΔt
    W=(1/2)kx^2
    F=kx=P*A

    3. The attempt at a solution
    I took the work equation for the spring and subbed in P*A to get w=(1/2)(PA)Δx and then i believe i can state that the work done on the spring is W=.5P2V2-.5P1V1 would you agree with this? I skipped some of the process so let me know if you need clarification.
     
    Last edited: Feb 22, 2013
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  3. Feb 22, 2013 #2

    Andrew Mason

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    I am not sure what the question is. Do we assume the spring is providing 0 pressure initially and during expansion the spring is compressed (or stretched). How does the spring length change with the volume? It depends on how the spring is connected to the gas container.

    AM
     
  4. Feb 23, 2013 #3
    Its pretty much a piston problem, where the spring is above the container of the gas connected to the lid and is initially at rest, then heat is added to the gas until it expands, working against the spring. Its mostly conjecture, I'm trying to anticipate one of the problems that is likely to be on my exam.
     
  5. Feb 23, 2013 #4

    Andrew Mason

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    The change in potential energy of the spring is equal to the work done on the spring by the gas. But in order to determine how much work is done by the gas on the spring, one has to know how much work is done on other bodies: eg the atmosphere; the piston.

    Can you give us the exact question?

    AM
     
    Last edited: Feb 23, 2013
  6. Feb 23, 2013 #5
    Yeah for sure I've attached one, thanks AM for being patient and responding to my incomplete posts. I think I can solve it I just need help with relating the mechanical work to the thermodynamic work.
     
    Last edited: Feb 23, 2013
  7. Feb 23, 2013 #6

    ehild

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    Originally the gas has Po pressure and Vo Volume. Assume slow heating, so the pressure of the gas is balanced by the atmospheric pressure and the pressure from the compressed spring. What is the pressure and the volume when the piston moves to the right by Δx, assuming k spring constant and A cross-section area of the cylinder?

    How is work calculated? How much work does the gas do while its volume increases three times the original volume?

    When the volume is 3 Vo the pressure is 3Po. What does it mean for the spring constant?



    ehild
     

    Attached Files:

  8. Feb 24, 2013 #7
    Well I think I'm starting to get there. I can state the following

    Wgas=PΔV=P*A*Δx=Wspring=-(1/2)k(Δx)^2

    or PAΔx=(1/2)k(Δx)^2

    I don't know where to go with this exactly, I know Δx is needed right? Work is calculated as pressure multiplied by the change in volume, but how do I use that when both pressure and volume are tripled?
     
  9. Feb 24, 2013 #8

    ehild

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    The gas does work against the surrounding atmosphere, too. The work ∫PdV from Vo to 3Vo. You need to express P in terms of V.

    What is the volume of the gas if the piston moves to the right by Δx?

    What is the pressure balanced by the spring and the atmospheric pressure if the spring is compressed by Δx?

    ehild
     
  10. Feb 24, 2013 #9
    Hmm. I used the ideal gas equation to get the integral into one variable and get the work done on the surroundings as W=nRTln(V2/V1), so that must be the total work done by the gas in expanding.

    The Volume of the gas moved by the piston is V+(A*Δx) or V2/3V1

    I don't understand where any of this is getting me I'm sorry. I don't mean to be high maintenance.


    is it just

    nrTln(V2/V1)=(1/2)kΔx^2

    or what. That's the work done in the expansion of the gas, but it doesn't take into effect the extra work the gas had to do to push the spring.
     
  11. Feb 24, 2013 #10

    ehild

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    The formula you use for work is for an isothermal process. Why should it be isothermal expansion when both volume and pressure increases three times? Think of the Ideal Gas Law.

    You need the work done on the spring but for that you need both k and Δx.

    You know that V= Vo+AΔx.

    How is the pressure related to Δx? What force is exerted on the piston from the spring?

    ehild
     
  12. Feb 24, 2013 #11
    I do not know what kind of process it is, that is the problem, its not given. Its not isobaric because the pressure triples. Its not isochoric because the volume triples. Its not adiabatic because heat is added. I know this stuff, but nowhere in any of my thermo chapters is there a problem like this. So how am I supposed to calculate work?? Unfortunately this cryptic stuff is just leading me down nonsense paths...


    anyways, why would I need to deal with k if its constant and given in the problem? and yeah I need Δx. Not sure how to get it though. I keep trying to set things equal different ways but not really seeing any progress.

    Pressure is related to work which is related to Δx, I think I established that, although I dont know if I did so correctly. and the force exerted by the spring is going to be F=kΔx
     
    Last edited: Feb 24, 2013
  13. Feb 24, 2013 #12

    ehild

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    You wrote already that the work done on the spring is 0.5 kΔx2.
    But neither k nor Δx are given.

    You can assume a quasi-static process when the total force of the gas exerted on the piston is equal to the external force. Yes, the spring exerts kΔx force on the piston, nut also the outside atmosphere exerts some force.


    Initially the pressure of the gas was equal to the atmospheric pressure.


    ehild
     
  14. Feb 24, 2013 #13
    oh wow you're right It is not provided. Goodness I think its time I went to sleep, its 1:30 a.m. here. Thank you for your help this evening, I'll be back to trying to figure this out tomorrow.
     
  15. Feb 24, 2013 #14
    Ok, let me go ahead and try to do a force summation of the internal and external And see if I can get down to two equations two unknowns. If anybody is willing to tell me what the bigger picture is it would really help me out. I haven't had a clue what I've been working towards this whole time.

    Let me just show you the process I tried to use before, this guessing game is getting me lost I think. I suck at being led to water I guess.

    so

    W=(1/2)kx^2 and kx=F
    W=(1/2)(F)*x=(1/2)(P*A)x and since Δx is the change in height of the chamber I can state the following, letting x represent the height of the container
    W=(1/2)P*A*(x+Δx)-(1/2)P*A*(x)
    W=(1/2)P2V2-(1/2)P1V1

    And If it has to do work against the outside pressure as well I'm not sure how to account for it. this could very well be wrong. Heres the force summation I was talking about, I found it in this thread https://www.physicsforums.com/showthread.php?t=256682


    Ftot=k*v+Fatm which forms a linear equation going through both my known points. Can I then solve for k using the slope formula? I dont understand why the spring force would be k*V.
     
    Last edited: Feb 24, 2013
  16. Feb 24, 2013 #15

    ehild

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    The spring force is not equal to PA.
    Initially, the gas was in equilibrium and the spring was unstretched. The initial pressure was equal to the atmospheric pressure, Po.
    The piston moves to the right by Δx, the spring gets shorter by Δx during the process.
    The piston exerts kΔx force on the spring, the spring exerts also kΔx force on the piston. The atmospheric pressure means APo force on the piston. So the external force on the piston is PoA +kΔx. The gas also exerts force on the piston: It is PA. The piston is stationary at the end: The inward and outward forces balance.

    PoA +kΔx=P(final)A and P(final)=3Po.

    You know that the final volume is Vf=Vi+AΔx, and Vf=3Vi.

    From these two equations you get expressions for k and Δx in terms of A and the pressure Po and volume Vi. Use them to determine the spring energy.

    ehild
     
    Last edited: Feb 24, 2013
  17. Feb 24, 2013 #16
    I've got it!! science be praised. Here's what I did

    Forces:

    PA=Patm+kx

    Divide both sides by A to get a linear function for pressure in terms of volume

    P=Patm+(k/A)x

    Then I can take the equation for volume and solve for x to eliminate it in the above equation

    V=V0+Ax
    x=(V-V0)/A

    So I have

    P=Patm+(k/A2)(V-V0)


    Now this is a linear function and the slope is defined by the term (k/A2)
    SO I can use the given initial and final pressures and volumes to calculate the slope m=ΔP/ΔV (getting a numerical value) thereby eliminating the not given values for k and A. Then the above equation can be integrated with respect to volume to give the work.

    ∫PdV=∫(Patm+(m)(V-V0))dV
    W=Patm(V-V0)+(1/2)m(V-V0)2

    man it seems so simple now that I see it all worked out like this. Oh well, as my instructor says, you gotta embrace the struggle. Thanks for your help ehild.
     
  18. Feb 25, 2013 #17

    ehild

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    It is even simpler than that :biggrin: You do not need to integrate to get the work on the spring,. It is 1/2 kx2.

    As you wrote correctly, P=Patm+(k/A)x, and x=(V-Vo)/A .

    P-Patm=kx/A = k(V-Vo)/A2, that is k=(P-Patm) A2/(V-Vo)

    Substitute for x and k in W=1/2 kx2 . A cancels.

    ehild
     
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