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PV^(gamma) = C

  1. Aug 27, 2007 #1
    Hi guys,


    I have some questions about this equation:

    P.V^(gamma) = constant, gamma = C_p / C_v

    Is it only valid for a adiabatic process, plus ideal gas? I thought it was at the first place, as I saw its derivation uses adiabatic properties(dQ=0) and assumes the gas is ideal (PV=mRT).
    But when I read my textbook, it doesn't mention anything about adiabatic. It says those processes follow:

    P.V ^ n = constant (n is another constant)

    are called 'polytropic'. If the gas is ideal, then the process is isothermal.

    I become more confuse when I saw 'isothermal', as I thought it should be adiabatic.

    And, is the equation only valid for a closed system?

    Last question, is 'TV^(gamma - 1) = constant' valid only for adiabatic process and ideal gas?

    I like thermodynamics, but I find some parts of it quite confusing, hope someone can clear my doubts.

    thank you.
     
  2. jcsd
  3. Aug 27, 2007 #2

    siddharth

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    These are two different process. In an adiabatic reversible process for an ideal gas, you'll get [tex]PV^{\gamma}=c[/tex] where [tex] \gamma [/tex] is the ratio of C_p and C_v.

    A polytropic process is generally different from an adiabatic process, where n can take any value depending on the system.

    I think that's wrong. When the value of n is 1, only then will the reversible polytropic process be isothermal.


    Yes, most of the equations you use are derived for a closed system. For an open system, you'd use the general form of the first law,
    [tex] dU = TdS - PdV + \mu dn[/tex]
    where [tex]\mu[/tex] is the chemical potential.

    Yes. Can you derive this from PV^gamma is constant? Hint: Use the ideal gas law to substitute for P.


    Hope this helped. If you have any further questions, feel free to ask. Thermodynamics can indeed be quite interesting :smile:
     
  4. Aug 27, 2007 #3

    Andrew Mason

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    It need not be reversible. Not every isothermal process is reversible but, for an ideal gas, PV=constant where T is constant.

    AM
     
  5. Aug 27, 2007 #4

    siddharth

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    Yeah, that's true. However, for a process which is irreversible and not in equilibrium, the idea of a macroscopic pressure and temperature don't hold, do they?
     
    Last edited: Aug 27, 2007
  6. Aug 27, 2007 #5

    olgranpappy

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    As long as the initial state and the final state are macroscopic equilibrium states then it can certainly be said that they possess an internal Energy (U) and, for an ideal gas this obeys U=3NT/2 where T is the temperature and N is the number of particles. So that the initial and final states both have some well-defined temperature... even thought there might not be any meaningful temperature for the in-between non-equilibrium.
     
  7. Aug 27, 2007 #6

    rbj

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    that ideal gas equation (we are assuming that units are chosen so that Boltzmann = 1) is also only valid for monatomic particles (like inert gasses) with 3 degrees of freedom (the x, y, z axis). if you have diatomic particles (like N2 and O2, together comprising 99% of air), the idealized equation for mean energy is 5/2 T per particle. this difference is important in getting the theoretical speed of sound in normal dry air to agree with experiment.
     
  8. Aug 27, 2007 #7

    olgranpappy

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    true.
     
  9. Aug 29, 2007 #8
    thanks siddharth. You are right, for the process to be isothermal, n has to be equal to 1. I checked my textbook again, and I realized I missed that part last time.

    Today I just learnt that PV^n = constant is also called path equation, all process in thermodynamics can be represent by this equation. (Correct me if I am wrong.)

    Thanks to others also, although I don't understand what some of you said. lol.
     
  10. Aug 29, 2007 #9

    Andrew Mason

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    A reversible process is, by definition, one in which the system is in constant equilibrium (or, to be more precise, one in which equilibrium can be restored by an infinitessimal change in conditions). Temperature is only defined for a system in equilibrium. So, you make a good point. A true isothermal process, in which temperature is defined and constant at all times during the process, must be reversible.

    AM
     
  11. Sep 17, 2007 #10
    v^(gamma)P=c

    heey, can you explain how to come from Vftf^f/2 = ViTi^f/2 to VT^f/2 and then to

    PV^gamma = c

    thanks
     
    Last edited: Sep 17, 2007
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