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Calculating term in Virial Expansion

  • #1

Homework Statement


...For each of the temperatures...compute the second term in the virial equation ##B(T)/(V/n)##, for nitrogen at atmospheric pressure.

Homework Equations


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

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.
 

Answers and Replies

  • #2
19,954
4,108
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).

Chet
 
  • #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.
 
  • #4
19,954
4,108
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.
Show us some of your work and maybe we can help.

chet
 
  • #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. :)
 
  • #6
19,954
4,108
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. :)
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?

Chet
 
  • #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?
 
  • #8
19,954
4,108
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?
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.

Chet
 
  • #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..
 
  • #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##
 
  • #11
19,954
4,108
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..
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.

Chet
 
  • #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?
 
  • #13
19,954
4,108
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?
There's a sign error in the 4ac term of your discriminant.

Chet
 
  • #14
Where exactly? I've been working on it so long it's harder and harder to find my mistakes.
 
  • #15
19,954
4,108
Where exactly? I've been working on it so long it's harder and harder to find my mistakes.
(-B)=+160
 
  • #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.
 
  • #17
19,954
4,108
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.
So, how come it doesn't match the solution you got by successive substitutions? They should give the same answer.

Chet
 
  • #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.
 
  • #19
19,954
4,108
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.
So you get the same answer doing it both ways, right? (I too get 8042 using successive substitutions).

Chet
 
  • #20
Yes, I get 8040 using the quadratic formula.
 

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