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How do you calculate the work done in an adiabatic process?

  1. Jun 16, 2006 #1
    How do you calculate the work done in an adiabatic process?

    I know change in thermal energy = work done.

    which amounts to (f/2)Nk(change in T). But how do you get change in T?

    Or is there another way to compute the work directly. The trouble is both the volume and temperture vary during the process.
     
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  3. Jun 16, 2006 #2

    Andrew Mason

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    (1) [tex]W = \int PdV = nR\int TdV/V[/tex] (substituting P = nRT/V)

    But the adiabatic condition [itex]PV^\gamma = nRTV^{\gamma-1} = K[/itex] applies, so (1) becomes:

    [tex]W = K\int dV/V^{\gamma}[/tex]

    Determine that integral for a particular volume change and that is the work.

    AM
     
  4. Jun 16, 2006 #3
    But do you think it is easier to find the change in internal energy hence only change in temperture between the two states? This will give the work as Q=0.
     
  5. Jun 16, 2006 #4

    Andrew Mason

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    You could take that approach. You apply the adiabatic condition in terms of temperature and integrate [itex]\int nC_vdT[/itex] from initial to final temperature. But you have to apply the adiabatic condition to find the initial and final temperatures.

    That will give you the change in internal energy. You then equate that to the work done (since, as you point out, Q = 0, [itex]\Delta U = \Delta W[/itex].

    But because you have to use the adiabatic condition [itex]T = K/nRV^{\gamma-1}[/itex] it is essentially the same calculation that I have set out above.

    The integral:

    [tex]K \int_{V_i}^{V_f} V^{-\gamma}dV[/tex]

    is not that difficult to work out using antiderivative of [itex]V^{-\gamma}[/itex] = [itex]V^{-\gamma +1}/(-\gamma + 1)[/itex]

    [tex]W = K(V_f^{1-\gamma} - V_i^{1-\gamma})/1-\gamma[/tex]

    AM
     
  6. Jun 17, 2006 #5
    I don't take the integral. I just use the adiabat relation to find the initial and end tempertures in terms of one temperture that I know the value to such as the initial temperture.
     
  7. Jun 17, 2006 #6

    Andrew Mason

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    The integral evaluates to [tex]nC_v(T_f - T_i)[/tex]. Is that not what you use? How do you find the beginning and end temperatures?

    AM
     
    Last edited: Jun 17, 2006
  8. Jun 19, 2006 #7

    Andrew Mason

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    Just to do the math to make my point clearer:

    [tex]T= K/nRV^{\gamma-1}[/tex]

    So:

    (1) [tex]nC_v(T_f - T_i) = nC_v\frac{K}{nR}(V_f^{1-\gamma} - V_i^{1-\gamma})[/tex]

    But [tex]C_v = C_p-R = C_v\gamma - R[/tex] so

    [tex]C_v = \frac{R}{\gamma-1}[/tex]

    Substituting in (1):

    [tex]nC_v(T_f - T_i) = \frac{K}{(\gamma-1)}(V_f^{1-\gamma} - V_i^{1-\gamma})[/tex]

    Thus:

    [tex]\Delta U = -\Delta W[/tex]

    AM
     
    Last edited: Jun 19, 2006
  9. Jul 6, 2009 #8
    I have the same problem but is K a constant?
     
  10. Jul 6, 2009 #9
    Oh I wasn't sure! Thanks!
     
  11. Jul 6, 2009 #10
    is that little symbol 7/5 for diatomic? and 5/3 for monatomic?

    i got a really small number......
     
  12. Jul 6, 2009 #11
    whoops. silly me. k is just a constant, from adiabatic condition [itex]pV^\gamma[/itex]=constant
     
  13. Jul 6, 2009 #12
    let me try that and see how that works.
     
  14. Jul 6, 2009 #13
    hold on i have 2 differnt pressures and volumes and i have K=pV^alpha....which do i use or what do I do.
     
  15. Jul 6, 2009 #14
  16. Jul 6, 2009 #15
    there is my problem....
     
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