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Thermodynamics work done question

  1. Oct 18, 2006 #1
    hey guys got this question that i have been stuck on for a while.

    air is expanded from 1M Pa at 327 degrees celcius to 200kPa in a closed piston cylinder device. for the process PV^1.2 = constant. calculate the work done in kJ/kg during this process

    i hav managed to work out the temperature after expansion using T2/T1 = (P1/P2)^(n-1/n) but i dont know how to work out the work done without knowing the mass or any of the volumes

    if work = (p1v1 - p2v2)/ 1-n

    then surely i need the volumes to work out the work done

    any help will be welcomed

    cheers
     
    Last edited: Oct 18, 2006
  2. jcsd
  3. Oct 18, 2006 #2

    quasar987

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    If the gaz can be considered ideal, then by conservation of the number of moles of gaz,

    [tex]\nu_i = \nu_f[/itex]

    you must have

    [tex]V_f=\frac{p_fT_i}{p_iT_f}V_i[/tex]

    So

    [tex]W=\int_{V_i}^{\frac{p_fT_i}{p_iT_f}V_i}pdV = \int_{V_i}^{\frac{p_fT_i}{p_iT_f}V_i} \frac{\alpha}{V^{1.2}}dV[/tex]

    And substitude back [itex]\alpha = p_iV_i^{1.2}[/itex] at the end.
     
  4. Oct 18, 2006 #3
    but i dont know any of the volumes so this method would not work
     
  5. Oct 18, 2006 #4

    Andrew Mason

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    If [itex]PV^\alpha = K[/itex] where [itex]\alpha = 1.2[/itex] (note: this is not the [itex]\gamma[/itex] for air which is 1.4), then substituting V = nRT/P gives:

    [tex]P^{1-\alpha}T^\alpha = K/n^\alpha R^\alpha = K'[/tex]

    So:

    [tex]P_1^{1-\alpha}T_1^\alpha = P_2^{1-\alpha}T_2^\alpha = K'[/tex]

    From that, work out PdV in terms of K' and T and integrate from T1 to T2

    AM
     
    Last edited: Oct 18, 2006
  6. Oct 18, 2006 #5

    quasar987

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    If you succed, would you please post the answer junglep?
     
  7. Oct 19, 2006 #6

    Andrew Mason

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    Find T2 from the relationship:

    [tex]P_1^{(1-\alpha)}T_1^\alpha = P_2^{(1-\alpha)}T_2^\alpha[/tex]

    so:

    [tex]T_2 = \left(P_1^{(1-\alpha)}T_1^\alpha/P_2^{(1-\alpha)}\right)^{1/\alpha}[/tex]

    Use PV=nRT to find V:

    [tex]V_1 = nRT_1/P_1[/tex]

    [tex]V_2 = nRT_2/P_2[/tex]

    Integrating PdV from V1 to V2 using [itex]P = K/V^\alpha[/itex]:

    [tex]W = \int_{V_1}^{V_2} PdV = \int_{V_1}^{V_2} KdV/V^\alpha[/tex]

    You just have to work that out.

    AM
     
    Last edited: Oct 19, 2006
  8. Oct 19, 2006 #7
    what is n in the equation

    pv = nRT?

    i thought the perfect gas eqn was pv = mass * R * T

    also i am not given a value for the gas constant (R). if it is any help the answer that is given in the book is in kJ/kg not in J.
     
  9. Oct 19, 2006 #8

    Andrew Mason

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    n is the number of moles of the gas. R is in units of J/mole K.

    This problem does not give you n or V, so assume n = 1 in which case: PV = MRT where M is the mass of one mole of air (29 g/mole). Essentially, you are working out and using the volume for one mole of air.

    AM
     
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