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Van der waal's gas equation

  1. Sep 28, 2011 #1
    http://en.wikipedia.org/wiki/Van_der_Waals_equation

    I understand the first equation where a is the measure of attraction between particles, and b is the volume of one particle.

    The second equation includes 'n', the number of moles of gas.

    1) v-nb is simply the volume occupied by the gas minus the volume of the particles (usually one atomic diameter)

    why is the must the n be squared; n2a/V2 and not simply n?

    My general idea is that n is proportional to the volume, so whenever there is a V2 or even V3 n must sort of 'follow' the power?
     
  2. jcsd
  3. Oct 5, 2011 #2
  4. Oct 8, 2011 #3
    bumpp~
     
  5. Oct 8, 2011 #4
    I don't know why you have been unable to look this up for yourself, but since no one else is prepared to provide an answer and that is what PF is for:

    The correction term to the pressure component is due to molecular cohesion (forces of attraction between molecules). The effect is similar to surface tension and is derived like this.

    Molecules within the body of the gas are surrounded by other molecules and so (on average) experience no net force in any direction from the attractive forces between the molecules.

    Molecules in the outer layer have molecules on the gas side but none on the container side so experience a net force inward towards the body of the gas.

    But the pressure exerted by the gas on the container is made up by summing all the outward forces from the molecules. These are reduced by the net inward cohesive forces.

    To derive a mathematical expression note that the inward forces must be proportional to

    1) ζ1 - The number of molecules in the outer layer

    2) ζ2 - The number of molecules in the layer immediately next to the outer layer


    Since both layers are very thin Van Der Waals made the assumption that

    ζ1 = ζ2 = ζ

    Now the number of molecules in any given space (part of the volume) is proportional to the total number of molecules divided by the volume

    So the force responsible for the reduction of pressure [itex]\propto[/itex] ζ1ζ1 [itex]\propto[/itex] ζ2

    Since ζ [itex]\propto[/itex] n/V, ζ2 [itex]\propto[/itex] (n/V)2

    Introducing the constant of proportionality, a, we arrive at the pressure correction term, which we must add back to the pressure to get the true pressure.

    [tex]\left( {P + \frac{{a{n^2}}}{{{V^2}}}} \right)[/tex]

    go well
     
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