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Calc 3/Thermo differentiation

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  1. Nov 4, 2014 #1
    1. The problem statement, all variables and given/known data
    pFwiQ16.png

    2. Relevant equations
    au/as=T
    au/av=p
    S/R=ln[(v-b)(u+a/c)^2]

    3. The attempt at a solution
    1/T=1/au/as=as/au
    S=ln[(v-b)(U+a/v)^2]R
    as/au=[(v-b)2(U+a/v)(1)]R/[(v-b)(U+a/v)^2]=2R/(U+a/v)=1/T
    T=(U+a/v)/2R
    U=2RT-a/v
    au/av=-P
    au=-Pav
    integrate au to get u=-pv+c

    u=-pv+c=2RT-a/v
    -p=2RT-a/v-C
    p=a/v-2RT+c

    7uylucs.png uploaded my work in image format.

    One of the main problem I have is that we haven't learned multivariable integration(or partial integration I don't know what it's called.)

    edit:
    more attempts
    zQrph03.png

    2M8nJNS.png
     
    Last edited: Nov 4, 2014
  2. jcsd
  3. Nov 4, 2014 #2

    vela

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    You found an expression for U as a function of V, and you're told ##P = -\frac{\partial U}{\partial V}##. Why are you integrating? Just differentiate U with respect to V to find the pressure.
     
  4. Nov 4, 2014 #3
    Hi vela, I need to express pressure as a function of temperature and volume. I need to get both T and V in the expression. Which is the problem I'm having.

    taking the au/av would give me a/v^2
     
  5. Nov 4, 2014 #4
    kBIawEe.png
    Another attempt.
     
  6. Nov 4, 2014 #5
    You already correctly obtained ##\left(\frac{\partial U}{\partial S}\right)_V=T=\frac{(U+\frac{a}{V})}{2R}##. Now please show us what you get for ##\left(\frac{\partial U}{\partial V}\right)_S=-P##. You will need to combine the two relationships.

    Chet
     
  7. Nov 4, 2014 #6
    Hey Chet. In my previous posts I listed how I derived au/av from the chain rule in implicit differentiation. They might not be correct but
    au/av=e^(S/R) (-2/(u+a/v))^3 is what I got for fourth attempt

    au/av= 2R[((u+a/v)(-a/v^2)]/[(v-b)(u+a/v)^2]/[2R((v-b)(u+a/v))/((v-b)(u+a/v)^2)] is what I got for my second attempt

    au/av=-2R/(v-b)(u+a/v)(v^2)/[(2R/(u+a/v))+1/T] is what I got on the third attempt

    my derivation are all in the photos.(Not that I think they are right :(. So far unfortunately only my last attempt actually gave me a expression with T in it, T always somehow cancels out in the end in other attempts.
     
  8. Nov 5, 2014 #7
    I took a simpler/different approach

    28OftM6.png the answer looks appropriate but I'm not sure if I messed up somewhere.
     
  9. Nov 5, 2014 #8
    I get:
    [tex]\frac{S}{R}=ln(V-b)+2ln(U+\frac{a}{V})[/tex]
    So,
    [tex]\frac{dS}{R}=\frac{dV}{V-b}+\frac{2}{(U+\frac{a}{V})}(dU-\frac{a}{V^2}dV)[/tex]
    At constant S, this becomes:
    [tex]0=(\frac{1}{V-b}-\frac{2a/V^2}{(U+\frac{a}{V})})dV
    +\frac{2}{(U+\frac{a}{V})}dU[/tex]
    Solving for ##\partial U/\partial V## gives:
    [tex]\left(\frac{\partial U}{\partial V}\right)_S=-\frac{(U+a/V)}{2(V-b)}+a/V^2=-P[/tex]
    Does this match any of the results from your attempts?

    Chet
     
  10. Nov 5, 2014 #9
    Hi Chet, yes it matches the results from my last attempt. If I sub in U=2RT-a/v for U in your answer I would get p=RT/(v-b)-a/v^2

    [tex]\left(\frac{\partial U}{\partial V}\right)_S=-\frac{(2RT-a/v+a/V)}{2(V-b)}+a/V^2=-P[/tex]

    [tex]\left(\frac{\partial U}{\partial V}\right)_S=\frac{(RT)}{(V-b)}-a/V^2=P[/tex]
     
  11. Nov 5, 2014 #10
    I forgot to mention. Thank you for the help Chet! I really appreciate it.
     
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