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Adiabatic expansion of a gas

  1. Mar 27, 2009 #1
    I'm deriving this formula for the adiabatic expansion of an ideal gas.

    [tex] PV^{\gamma} = Constant_2 [/tex]

    there are 3 ways to expand to end up at same internal energy dU.

    1. this is the direct adiabatic expansion from T1 to T3
    [tex] dU_{systA} = -dW_{systA} = -P_{systA} dV [/tex] where [tex]P_{systA}[/tex] is a function of Volume and Temperature of the gas

    2. you can decrease the internal energy of the gas at constant Volume with the same change in temperature as #1. [tex] dU_{syst V} = C_{V}ndT [/tex]

    so, [tex] dU_{systA} = dU_{systV} = C_VndT [/tex]
    [tex]C_VndT = -P_{systA} dV[/tex]
    so
    [tex] C_VndT + P_{systA} dV = 0 [/tex]

    3. you can also take the isochoric then isobaric pathway to end up from T1 to T2 to T3. The total energy for this 2 step pathway is
    [tex] PdV + VdP = nRdT [/tex] where dT is the same dT as #1 and #2

    if you isolate dT from #3 and plug it in to #2, you obtain this expression.
    [tex] C_V VdP + C_V PdV + R P_{systA} dV = 0 [/tex]

    THIS IS WHERE I AM STUCK!!!!!!!!!!!!!!!!!!!!

    I DO NOT THINK THAT
    [tex] PdV = P_{systA} dV [/tex] where [tex] P_{systA} dV = [/tex] work involved to expand the gas ADIABATICALLY and [tex] PdV = [/tex] isobaric expansion of the gas to get to T3 from #3 above!!!!

    why are these two integrals the same??????????? they are not the same!!!!!!!!!!!!!!!!!!

    HELP HELP HELP!!!!
     
    Last edited: Mar 28, 2009
  2. jcsd
  3. Mar 28, 2009 #2
    does anybody have any clues please????
     
  4. Mar 28, 2009 #3

    Andrew Mason

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    The work done by the system and the work done on the surroundings are always the same magnitude (opposite sign) in a reversible process. It is only if the process is non-reversible (where there is more than an infinitessimal difference in internal and external pressure) that the two are different. [itex]PV^\gamma[/itex] is constant for a reversible adiabatic process only. For example, it does not hold if there is a free expansion

    AM
     
  5. Mar 28, 2009 #4
    thank you , but you haven't answered my question. why is the work done by an isobaric pathway equal to the work done by the adiabatic pathway for the expansion of an ideal gas????????
     
  6. Mar 28, 2009 #5

    Redbelly98

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    Those shouldn't be equal, should they?

    In general, the work done depends on the path taken between two points in a PV diagram. Since work is the integral of P·dV, and the area under the PV curve will be different for the different paths.
     
  7. Mar 28, 2009 #6
    yes, that's my point!!! but, according to the derivation of [tex] PV^{\gamma} [/tex] the work done isobarically and the work done adiabatically to expand the gas to T3 and V3 are the same!!!

    please look at the 3 ways you can expand the gas adiabatically above (#1,2,3)

    we are missing a CRUCIAL POINT here...!
     
  8. Mar 28, 2009 #7

    Redbelly98

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    I agree with that, except that I am going to write what happens when you actually do the integrals (because it's easier for me to think about it that way):

    Cv V1 ΔP + Cv P3 ΔV + R PsystA dV = 0 ​

    where V1 is initial volume, P3 is final pressure

    This is where I lose you. How does the previous equation imply what you are saying here? I don't see it.
     
  9. Mar 29, 2009 #8
    please note that the pressure in [tex] P_{systA} dV[/tex] is varying as the gas is being expanded adiabatically. However, there is a derivation in my textbook where the [tex] P_3 dV = P_{systA} dV [/tex]

    and when you take it out, you are left with [tex] C_v + R = C_p[/tex] that is substituted into the equation to get [tex] PV^{\gamma} = Constant_2 [/tex]

    if you differentiate the ideal gas law, you get
    [tex] PdV + VdP = nRdT [/tex] this is the total energy involved in the isochoric and then the isobaric pathway. Am I wrong????? While my textbook used this differential equation, I followed the pressures, volumes, and temperature as you go to the isochoric and then the isobaric pathway to finally expand the gas adiabatically. If I blindly accept the differential equation of the ideal gas, then there is no problem in getting to the final derivation. HOWEVER, if you take the isochoric and then the isobaric pathway, I get stuck at the above problem because the work involved in the Isobaric pathway is NOT EQUAL to the work involved in the adiabatic expansion.

    I hope I cleared it up some more.

    please try to derive the equation yourself....tell me if you come across the same problem.
     
  10. Mar 29, 2009 #9

    Mapes

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    This is not correct. Does this resolve the problem?
     
  11. Mar 29, 2009 #10
    The work done in the isochroic pathway is zero since V remains constant.
     
  12. Mar 29, 2009 #11

    Redbelly98

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    Which isn't correct, the equation itself or the statement that it gives the change in energy for the path?

    That pathway consists of an isochoric and isobaric section; work is done in the isobaric part.
     
  13. Mar 29, 2009 #12
    That pathway consists of an isochoric and isobaric section; work is done in the isobaric part.[/QUOTE](from Redbelly)

    And to get to the same final condition heat must enter the system during the isochoric part
     
    Last edited: Mar 29, 2009
  14. Mar 29, 2009 #13
    you guys are all missing the point. you must trust that the 3 equations i've listed are absolutely correct.

    what is ambiguous is the derivation on the [tex] PV^{\gamma} = constant_2 [/tex]
    because we must either assume that the work done in the isobaric pathway is equal to the work done in the adiabatic pathway OR
    we must blindly accept that if you differentiate the ideal gas law, you obtain a dT that is equal to the dT of the adiabatic pathway [tex] PdV + VdP = nRdT [/tex]

    PLEASE TRY THE DERIVATION ON YOUR OWN. NO MORE "WORK IS NOT DONE IN ISOCHORIC PATHWAY" RESPONSES...DUH!!!! WE ALL KNOW THAT.
     
  15. Mar 29, 2009 #14
    I think you will see it more clearly if you sketch out the PV graph, the adiabatic change rises fairly steeply to the final state and you need to integrate to find the work done.In effect you are finding the area under the graph.With the isobaric/isochoric change you have a section parallel to the V axis where work is done and a section parallel to the V axis where no work is done.The work done in the isobaric section is P*change in V.Clearly the areas are different so you need to recheck your maths and the assumptions you made. :grumpy:
     
  16. Mar 29, 2009 #15

    Mapes

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    The equation [itex]P\,dV+V\,dP=nR\,dT[/itex] is correct, but it's not correct that it's the "total energy," which you said twice above. This mistake may be connected to the contradiction you're finding.

    It should be clear that this isn't the total energy because the energy change for an ideal gas is always the integral of [itex]nc_V\,dT[/itex], which is different from [itex]nR\,dT[/itex].

    The work done in the isobaric pathway is not equal to the work done in the adiabatic pathway; one can see this by comparing the P-V curves, as Dadface says.

    Here's how I would derive the fact that [itex]PV^\gamma=\mathrm{constant}[/itex]:

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

    [tex]nc_V\,dT=-p\,dV[/tex]

    [tex]nc_V\,dT=-\frac{nRT}{V}\,dV[/tex]

    [tex]\frac{c_V}{R}\int^{T_2}_{T_1}\frac{dT}{T}=-\int^{V_2}_{V_1}\frac{dV}{V}[/tex]

    [tex]\frac{c_V}{R}\ln\left(\frac{T_2}{T_1}\right)=-\ln\left(\frac{V_2}{V_1}\right)[/tex]

    [tex]\left(\frac{T_2}{T_1}\right)^{c_V/R}=\left(\frac{V_1}{V_2}\right)[/tex]

    [tex]\left(\frac{P_2 V_2}{P_1 V_1}\right)^{c_V/R}=\left(\frac{V_1}{V_2}\right)[/tex]

    [tex]P_1 V_1^{c_P/c_V}=P_2 V_2^{c_P/c_V}[/tex]

    Does this resolve the issue?
     
  17. Mar 29, 2009 #16
    What is wrong with my method???? It should work! Three ways exist to expand this ideal gas to obtain same energy state
     
  18. Mar 29, 2009 #17

    Mapes

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    I've read your original post at least 10 times, and I'm with Redbelly98; I don't know what contradiction you're finding with approach #3. Can you show the steps in detail?
     
  19. Mar 29, 2009 #18
    how do you go from

    [tex] \left(\frac{P_2 V_2}{P_1 V_1}\right)^{c_V/R}=\left(\frac{V_1}{V_2}\right)[/tex]

    to
    [tex]
    P_1 V_1^{c_P/c_V}=P_2 V_2^{c_P/c_V}[/tex]
     
  20. Mar 29, 2009 #19

    2 STEP PROCESS TO GET TO SAME ENERGY STATE AS DIRECT ADIABATIC EXPANSION
    PdV + VdP = nRdT ought to be the total energy change for the isochoric then the isobaric pathway to expand the gas.

    VdP is the total energy change in the isochoric pathway from T1 to T2.

    PdV is the total energy change in the isobaric pathway from T2 to T3.

    the total energy change in this 2 step process would be nRdT where integral of dT = (T3 - T1)

    and we know this for a fact:
    Adiabatically, the temperature change would be T3 - T1. The change in Energy state from T1 to T3 would equal the work of the system to expand the gas adiabatically where T1 > T3.

    are these statements correct? if there are any flaws with these statements, then the contradiction is valid. However, if all these statements are correct so far, then I will tell you where the contradiction lies again.

    THANK YOU.
     
  21. Mar 29, 2009 #20

    Mapes

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    [tex] \left(\frac{P_2 V_2}{P_1 V_1}\right)^{c_V/R}=\left(\frac{P_2 V_2}{P_1 V_1}\right)^{c_V/(c_P-c_V)}=\left(\frac{V_1}{V_2}\right)[/tex]

    [tex] \left(\frac{P_2 V_2}{P_1 V_1}\right)=\left(\frac{V_1}{V_2}\right)^{(c_P-c_V)/c_V}=\left(\frac{V_1}{V_2}\right)^{c_P/c_V-1}[/tex]

    [tex] \left(\frac{P_2}{P_1}\right)=\left(\frac{V_1}{V_2}\right)^{c_P/c_V}[/tex]

    [tex]P_2 V_2^{c_P/c_V}=P_1 V_1^{c_P/c_V}[/tex]
     
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