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Pure Substances and Van der Waals

  1. Apr 5, 2008 #1
    1. The problem statement, all variables and given/known data

    For any Pure Substance, use the relation (1) to show that for a Van der Waals gas, over a range of conditions where c[tex]_{v}[/tex] is effectively constant, that equation (2) gives equations (3) and (4)


    2. Relevant equations

    (1) ~ Tds = du + pdv

    (2) ~ c[tex]_{v}[/tex]dT + T [tex]\frac{\partial p}{\partial t}[/tex][tex]_{v}[/tex]dv = 0

    (3) ~ T(v-b)[tex]^{\alpha}[/tex] = constant

    (4) ~ p + [tex]\frac{a}{v^{2}}[/tex][tex]\approx[/tex]T[tex]^{\frac{1+\alpha}{\alpha}}[/tex]

    (5) ~ [tex]\alpha[/tex]=[tex]\frac{R}{c_{v}}[/tex]

    (6) ~ (p + [tex]\frac{a}{v^{2}}[/tex])(v-b)=RT

    3. The attempt at a solution

    I really havent got a clue where to go with this one, this question obviously has something to do with c[tex]_{v}[/tex].

    I was thinking for the first part since its equal to a constant then maybe integration or diferrentiation will come into this at some point. As for the second part i haven't got the slightest clue where to go so any tips would be nice.

    I've written down all the equation i think could possibly come into play here, and have tried rearranging them but have not got anywhere with this yet. I can derive equation (2) from equation (1) but i can't see that being much use

    I'm new round here and would like to apologise for any thing that i've done wrong

    Thanks alot
     
    Last edited: Apr 5, 2008
  2. jcsd
  3. Apr 5, 2008 #2
    i have managed to show that

    p+[tex]\frac{a}{v^{2}}[/tex]=[tex]\frac{RT}{v-b}[/tex]=T([tex]\frac{\partial p}{\partial T}[/tex])[tex]_{v}[/tex]

    and that

    [tex]\frac{a}{v^{2}}[/tex] = ([tex]\frac{\partial u}{\partial v}[/tex])[tex]_{T}[/tex]

    but am now unsure where to go or if this is even the right path to take
     
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