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Van de Waals equation

  1. Apr 22, 2010 #1
  2. jcsd
  3. Apr 22, 2010 #2
    There is theory, but it only leads to approximate results.

    The equation models the gas as a system of incompressible spheres, of diameter [tex]\sigma[/tex], that experience an attractive force inversely proportional to the fourth power of their distance apart.

    This attractive pressure therefore varies as the square of the concentration. This is subtracted from the overall pressure seen by the outside world and is therefore negative.

    i.e. [tex]\frac{{a{'^2}N}}{{{V^2}}}[/tex]

    Secondly we must use the 'free volume' which is the total volume minus four times the total volume of the spheres

    [tex]N\frac{2}{3}\pi {\sigma ^3}[/tex]

    Substituting these into the gaw law yields VDW equation.

    [tex]P = \frac{{Nkt}}{{V - N\frac{2}{3}\pi {\sigma ^3}}} - \frac{{a{'^2}N}}{{{V^2}}}[/tex]

    The constants can also be derived from the critical points as below

    [tex]\begin{array}{l}
    b = \frac{{{V_c}}}{3} \\
    {P_c} = \frac{a}{{27{b^2}}} \\
    {T_c} = \frac{{8a}}{{27bR}} \\
    \frac{{R{T_c}}}{{{P_c}{V_c}}} = \frac{8}{3} \\
    \end{array}[/tex]
     
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