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When to apply the ideal gas law PV=nRT

  1. Oct 1, 2006 #1

    quasar987

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    Is the formula only applicable during quasi-static processes? In other words, is it only true for a gaz at equilibrium?
     
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  3. Oct 1, 2006 #2

    Clausius2

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    No, that equation holds even under translational, vibrational, rotational and chemical non equilibrium of the gas molecules. One can arrive to that equation from the Kinetic Theory and also from the Statistical Mechanics.

    BUT, have into account that only ideal gases yield that equation. With that I mean that only small perturbations off the equilibrium are allowed. For instance, strong vibrational non equilibrium can cause a coupling with the rotational non equilibrium of the molecules, leading to crossed terms in the partition functions and thererfore not yielding the equation of state. Also, ionized gases in which the electrostatic interaction is dominant, this equation does not hold.

    In the context of Gasdynamics, we usually admit that equation as valid in every point of the fluid domain for normal gases, if the temperature is not high enough (or low enough) for not having coupled effects (a cause of non ideality).
     
  4. Oct 1, 2006 #3
    Although this does not go at the heart of what your asking; Pv=RT is applicable when it is not near the saturation region and critical point on the phase diagram. If you are at those locations, you have to approximate it as Pv=ZRT, where Z is a correction factor.

    Now, if you have a piston that is moving at low velocities, then you have a quasi-equilibrium process. This means you can approximate the pressures inside the cylinder as the same throughout.

    On a real car engine where the velocities are very fast, the pressure is not going to be the same throughout, it will vary with location. However, you can use the pressure at the inner face of the piston, and find the work done at the boundary as:

    [tex] W_b = \int^2_1 P_i dV [/tex]

    Where [tex]P_i[/tex] is obviously the pressure at the inner face of the piston. This integral has to be evaluated experimentally and not analytically.

    Side: Properties are, by definition, taken for equilbrium states.
     
    Last edited: Oct 1, 2006
  5. Oct 1, 2006 #4

    quasar987

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    Thanks you both but I pretty much only understood one thing from your posts. That is, that it hold even when the macroscopic properties of the system are changing. This is surprising to me for the very simple reason that in PV=NRT, we assume that the pressure is constant at all points in the gaz.

    But when a piston abruptly compresses a gaz, pressure becomes non uniform in the gaz (it is higher near the piston surface since there is a net accumulation of particles there). So what is to be taken as P in the ideal gaz formula then? The higher P, the lower P? The average?
     
  6. Oct 2, 2006 #5
    Using this formula, you have to assume that the process consists of a 'continuum of equilibrium states'. To a reasonable approximation, this is true for most simple systems, like hand-pumped pistons and so on. The relationships break down where there is no well-defined set of properties at any given point - where there's an unquantified heat loss, pressure waves, chemical reactions going on, etc.
     
  7. Oct 2, 2006 #6

    Andrew Mason

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    For an ideal gas, I don't see how PV=nRT holds unless there is equilibrium. Pressure and temperature are defined only for equilibrium. The extreme case would be a free expansion of a ball of gas. While the gas is expanding, [itex]PV \ne nRT[/itex]. Pressure is non-uniform (greatest at the centre and 0 at the edges). Since the kinetic energy of the gas molecules does not follow the Boltzmann distrbution, temperature of the gas as a whole is not really defineable.

    AM
     
  8. Oct 2, 2006 #7
    [tex] P_i [/tex]

    <text>
     
  9. Oct 3, 2006 #8

    Clausius2

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    Maybe my assertion was too naive and vague. I should have complemented it with other corolary: the time of relaxation of the nonequilibrium process must be much shorter than the characteristic time of flow. That's what happens in a compressing cylinder.
     
  10. Oct 3, 2006 #9

    quasar987

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    The vocabulary you use is quite disarming to me but what you said sounds just like the definition of a quasi-static process.
     
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