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Virial equation, minimum pressure point

  1. Mar 18, 2014 #1
    Good evening PF! I'm having trouble figuring out how to attack this problem. I have tried two different ways but I don't know if either of them is correct.

    1. The problem statement, all variables and given/known data
    Using the provided virial coefficients, determine analytically the pressure at which the graph of PV versus P for N2 at -50° C, reaches a minimum point.
    Virial coefficients for N2 at -50° C:
    A = 18.31
    B = -2.88x10-2
    C = 14.98x10-5
    D = -14.47x10-8
    E = 4.66x10-11

    2. Relevant equations
    [tex]PV_m=A+BP+CP^2+DP^3+EP^4[/tex]

    3. The attempt at a solution
    So, in order to find the minimum point I need two differentiate the function, and I'm trying two ways of doing this. I hope at least one of them is correct.

    Option A:
    [tex]\frac {d(PV_m)}{dP} = B+2CP+3DP^2+4EP^3[/tex]
    In order to find the critical points I equate the above derivative to zero.
    [tex]B+2CP+3DP^2+4EP^3 = 0[/tex]
    Now I have to solve this cubic equation analytically, I could solve it with the help of a CAS, but the problem is asking for an analytic solution. This is as far as I can go. I did solve the equation with a software, and got two complex solutions and a real one. I assume the only solution that is relevant to me is the real one, right?

    Option B:
    I cleared Vm and expressed it as a function of P first, then differentiated.
    [tex]V_m(P) = \frac{A}{P} +B+CP+DP^2+EP^3[/tex]
    [tex]\frac{dV_m}{dP} = -\frac{A}{P^2} +C+2DP+3EP^2[/tex]
    Equating to zero.
    [tex]-\frac{A}{P^2}+C+2DP+3EP^2=0[/tex]
    I have to solve this equation analytically too, but I have no idea with this one.

    Well, I hope at least one of my procedures is right. Any help or insight will be greatly appreciated!
     
  2. jcsd
  3. Mar 18, 2014 #2

    haruspex

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    If PV is the quantity to be minimised then PV is what you must differentiate. dV/dP = 0 will find where V is minimised, which will be a different point.
    There is an equation for solving cubics (Google it), but I'd be surprised if you were expected to know it or use it.
     
  4. Mar 19, 2014 #3
    Yeah, option A seemed the most sound. I solved the equation with both WolframAlpha and Excel, the former gave me ~113 atm as a result, and the latter ~114 atm. Since this was the only real solution for the equation I'm assuming it is the minimum point I'm looking for and skipping the second derivative step. I also googled the equation you mentioned, I might try to solve it tomorrow. It's a LONG formula but it's just plug and chug. Thank you!

    Now, I have a minor concern with notation, how should I denote the derivative of PV? d(PV) (applying product rule), VdP or PdV?
     
  5. Mar 19, 2014 #4

    haruspex

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    I would just leave it as d(PV). PV is the entity to be minimised.
     
  6. Mar 19, 2014 #5
    "LONG" is way too mildly put. You will be surprised by how many pages you have to fill before you get to the solution (if ever).
    It may be that by "analytically" they mean exactly what you did. They did not say find an analytic solution. What would be the point in giving you the values of the parameters, if this were what they mean?
    But you could have just plotted the (PV) function and graphically estimate the minimum. This would be a graphical method.
     
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