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Calculating term in Virial Expansion

  1. Jan 21, 2015 #1
    1. The problem statement, all variables and given/known data
    ...For each of the temperatures...compute the second term in the virial equation ##B(T)/(V/n)##, for nitrogen at atmospheric pressure.

    2. Relevant equations
    ##PV = nRT(1+\frac{B(T)}{V/n}+\frac{C(T)}{(V/n)^2}+...)##

    3. The attempt at a solution
    I'm given a list of temperatures and values for B(T), for starters, Let T=100K giving B(T)= -160 (cm^3/mol).
    I think I'm misunderstanding the question maybe. I'm confused on how I'm supposed to calculate the second term when I need the volume, but to find volume I need the second term. I'm also assuming I can just consider 1 mol of nitrogen.
  2. jcsd
  3. Jan 21, 2015 #2
    You know P, R, T, and B(T), and you need to solve for V/n. Substitute y = V/n, and then solve for y. It will be a quadratic equation in y, and you will need to use the quadratic formula (or you can probably solve it iteratively by using successive substitutions).

  4. Jan 21, 2015 #3
    Ok, thats what I tried doing. I did get an answer. It was close'ish to the answer I got when calculating V just from the ideal gas law, but off by a couple of factors of 10. I must have made some sort of conversion mistake.
  5. Jan 21, 2015 #4
    Show us some of your work and maybe we can help.

  6. Jan 21, 2015 #5
    I'm going to try it a couple more times carefully before I subject my self to typing the quadratic formula with units and what not. :)
  7. Jan 21, 2015 #6
    If you are using the quadratic formula, you may be encountering a roundoff problem. Is the 4ac term very small compared to the b2 term in the discriminant?

  8. Jan 21, 2015 #7
    Ok, I found my problem. I left out T^2 when I substituted numbers into the variables on my white board. So I kept making the same mistake over and over.

    Here's what I got, V=98.13cm^3 using the virial expansion and V=82.03cm^3 using the ideal gas law.

    Does this seem like a reasonable difference at 100K?
  9. Jan 21, 2015 #8
    Well, if B(T) is negative, wouldn't you expect V from the virial equation to be less than V from the ideal gas law? And, in addition, from the ideal gas law, I get 8203 cc/mole.

  10. Jan 21, 2015 #9
    I wasn't converting one of the constants, so I'm in the right degree of 10 now.

    It would make sense that V from the virial equation should be less than V from the ideal gas law. But now I've got 8358 cc/mole. Not really sure where I'm going wrong so quadratic formula madness is coming..
  11. Jan 21, 2015 #10
    ##P-Pressure=10.13 \frac{N}{cm^2}; R-constant= 8.31 \frac{J}{mol*K}=831\frac{N*cm}{mol*K}##

    ##PV^2 - nRTV - B(T)n^2RT = 0##
    ##V = \frac{-(-nRT) + \sqrt{(nRT)^2 - 4(P)(B(T)n^2RT)}}{2P}##
    ##V = \frac{(831*100) + \sqrt{(831*100)^2 - 4(10.13)(-160)(831*100)}}{2*10.13}##
    ##V = 8358 cm^3##
  12. Jan 21, 2015 #11
    Just do it by successive substitutions. Start by substituting the result from the ideal gas law into the virial term, and then resolving for V/n again on the left side. Keep substituting the previous solution into the virial term and then resolving until the answer stops changing.

  13. Jan 21, 2015 #12
    Ok, cool, now I've got V/n = 8147cc/mol. Which seems much more reasonable. Thank you!

    On the other hand, though, why would the quadratic formula fail to give an accurate V?
  14. Jan 21, 2015 #13
    There's a sign error in the 4ac term of your discriminant.

  15. Jan 21, 2015 #14
    Where exactly? I've been working on it so long it's harder and harder to find my mistakes.
  16. Jan 21, 2015 #15
  17. Jan 21, 2015 #16
    Ah ok, thank you. And jsut for completeness' sake, I found another error, I wasn't dividing by P and was misreading my calculator, my actual V/n was 8042 cc/mol.
  18. Jan 21, 2015 #17
    So, how come it doesn't match the solution you got by successive substitutions? They should give the same answer.

  19. Jan 21, 2015 #18
    No, that is the answer I get from successive substitution, the previous 8147 that I posted, was actually 8.147x10^4, and I still needed to divide by P to get my final answer.
  20. Jan 21, 2015 #19
    So you get the same answer doing it both ways, right? (I too get 8042 using successive substitutions).

  21. Jan 21, 2015 #20
    Yes, I get 8040 using the quadratic formula.
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