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Thermodynamics: Defining Pressure

  1. Jun 21, 2012 #1
    I'm trying to understand the relation:

    P=T (∂S/∂V)U,N.

    I understand the definition of temperature as:

    1/T =(∂S/∂U)V,N.

    The relation for pressure has a similar derivation, but how can T be thrown in there? In my textbook, the author derives the relation for two systems in thermal equilibrium. Are we supposed to only compare the pressures of systems in thermal equilibrium? I'm clearly a little confused -- any help would be appreciated.
  2. jcsd
  3. Jun 21, 2012 #2
    Pressure as a single value is only well defined for a system in equilibrium, because if the system were not in equilibrium then the pressure would generally vary throughout the system. So unless you are studying non-equilibrium thermodynamics you will only compare pressures of systems in equilibrium
  4. Jun 21, 2012 #3


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    As far as fluids and other such things, you consider them as a collection of many subsystems, each of which have a well defined pressure. in the continuous limit, you get a continuous variation of pressure.

    As was said before, to consider the behavior of such systems in time, we need to look at non equilibrium thermodynamics, namely kinetics and transport theory in general. That will teach you about diffusion and the forces that drive the system as a whole toward equilibrium. You may want to look into the Boltzmann transport equation in particular.
  5. Jun 22, 2012 #4
    Thank you.

    Perhaps I'm thinking about this wrong. But:

    Say we have two systems that can exchange energy only. Their respective volumes are held constant, as well as number of particles, and total energy between them.

    The definition of temperature gives us a basis for seeing which way energy will flow in this system, knowing entropy of the total system to increase. Compare the derivatives for different distributions of energy.

    I'm having a hard time extending this intuition to pressure, given the extra T. (Or likewise, chemical potential.) I'm not sure what systems could have constant energy and number of particles and exchange volume only. Nonetheless, the situation seems akin to multiplying (∂S/∂V)U,N by P, in which case I'm not sure about its comparative value.
  6. Jun 22, 2012 #5
    Alternatively you can define pressure as the change in energy of the system with respect to volume.
  7. Jun 24, 2012 #6

    Andrew Mason

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    It follows from the first law, which leads to the fundamental equation of thermodynamics: TdS = dU + PdV

    If you start with TdS = dU + PdV, and keep U and N constant (dU = 0), then:

    [tex]T\left(\frac{\partial S}{\partial V}\right)_{U, N} = P\left(\frac{\partial V}{\partial V}\right)_{U, N} = P[/tex]

    This means that pressure is the rate of heat flow with respect to change in volume in a reversible process where U and N are held constant. For an ideal gas, of course, holding U constant means T is constant. So this is simply saying that the work done in a quasi-static isothermal process is equal to the heat flow.

    Last edited: Jun 24, 2012
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