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I Internal Energy of a real gas

  1. Nov 4, 2016 #1
    I am looking over the kinetic theory of gases. It is most commonly described as
    U = (3/2)*N*k*T = (3/2)*mass*R*T
    for a monatomic gas, assuming the gas is ideal. This is based on the derivation, where ultimately
    (3/2)*P*V = N*K = total kinetic energy of particles.

    My question, for a real gas, such as following Van der Waal, or even a liquid, is it reasonable to assume:
    U = (3/2)*P*V
    for the internal energy, regardless of the phase? As far as I can derive, this should be valid, but I welcome other responses. Original references would be much appreciated as well.
     
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  3. Nov 4, 2016 #2

    Andrew Mason

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    No. First of all, U = (3/2)*P*V is true only for monatomic ideal gases. The ideal gas model assumes that there is no inter-molecular forces and all collisions are elastic. Where there are inter-molecular forces, one has to take into account the potential energies of the molecules. Potential energies of the molecules contribute to internal energy but not to pressure or volume. When the gas is polyatomic, the degrees of freedom (e.g. vibration, rotation) must be taken into account. The kinetic energies associated with those degrees of freedom do not contribute to pressure or volume but do contribute to internal energy. One also has to take into account the fact that different modes may not be fully active at a given temperature due to quantum effects, so the equipartition theorem does not apply until those modes are fully active.

    Welcome to PF Marko!

    AM
     
  4. Nov 4, 2016 #3
    Andrew,

    I appreciate the answer, but then the question is, if I am trying to predict the internal energy of a real monatomic gas, what equation should I use? The value of P would obviously be affected by the intermolecular attractive and repulsive forces, resulting in a lower pressure for a given temperature. But that being said, how else can one calculate it?

    Often I see for a real gas the equation:
    dU = Cv*dT + (T*dP/dT - P)*dV

    My problem with this equation is that it assumes a constant specific heat, which in reality is never the case for real gases. Plus, it assumes a normalized zero internal energy.

    Ultimately, I want to be able to figure out what the changes in internal energy are, so I can figure out what the heat inputs and outputs of a given thermodynamic cycle are.
     
  5. Nov 4, 2016 #4

    Andrew Mason

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    You may have to use tables and do a numerical integration of the equation :
    [tex]dU = C_VdT + \left[T\left(\frac{\partial{P}}{\partial{T}}\right)_V - P\right]dV[/tex]

    if there is a significant change in ##C_v## or ##\left(\frac{\partial{P}}{\partial{T}}\right)_V## over the temperature range you are dealing with.

    AM
     
  6. Nov 4, 2016 #5
    As Andrew Mason indicates, the equation does not assume constant specific heat, and it assumes no such thing as a normalized zero internal energy.
     
  7. Nov 4, 2016 #6
    Thank you, I understand your answer with regard to the equation for dU under discussion.

    Looking at the derivation for the kinetic energy:
    http://dev.physicslab.org/Document.aspx?doctype=3&filename=IdealGases_KineticTheory.xml

    it is clear that
    P = (2/3)*(N_particles / V)*(KE_particle)
    for a monatomic fluid, and this is derived without any reference to the ideal gas law. The ideal gas law doesn't come into play until the above equation is plugged into the ideal gas law to find the relationship between the temperature versus the internal energy.

    I agree the kinetic theory only applies to molecules bouncing back and forth on the boundary, but in the absence of any molecule rotations / vibrations, etc, shouldn't the internal energy be:
    (N_particles*KE_particles) = (3/2)*P*V ?

    i acknowledge that this only applies to the kinetic energy, but is it not a reasonable assumption?
     
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