Computation of 2nd and 3rd Virial Coefficients

[amattiol]

Homework Statement

Compute the 2nd and 3rd Virial coefficients at 300K of Ar. Experimental parameters are mentioned in xlist and ylist below :)

Homework Equations

My professor has given a hit, which is as follows: "Use the data to determine the (z) compressability (sp?) and plot Z as a function of the reiprocal of molar volume."

The Attempt at a Solution

In Maple this is where I am at:

xlist := [.4, .5, .6, .8, 1, 1.5, 2, 2.5, 3, 4];
ylist := [6.2208, 4.9736, 4.1423, 3.1031, 2.4795, 1.6483, 1.2328, .98357, .81746, .60998];
xylist := zip(proc (x, y) options operator, arrow; [x, y] end proc, xlist, ylist);
xylist:= [[0.4, 6.2208], [0.5, 4.9736], [0.6, 4.1423], [0.8, 3.1031], [1, 2.4795], [1.5, 1.6483], [2, 1.2328], [2.5, 0.98357], [3, 0.81746], [4, 0.60998]]

Then I have done non linear fitting for 7 Virial coefficients with the resulting equation of fit:
f(x) =0.733475918379429/x^6-5.86019434244588/x^5+18.1883313118305/x^4-27.6759879335724/x^3+21.3651589091868/x^2-5.25691168982818/x+1

 

[epenguin]

Why fit with 7 terms when you are asked for only 3?

Not strong on the statistical theory but I think you positively lose information by fitting to too many terms.

 

[amattiol]

epenguin, I fit 7 terms because it was such a good fit I couldn't resist keeping it fit to the 8th coefficient. If I reduce the fit eqn to z = 1 + B/Vbar + C/(Vbar^2) then B and C would be the 2nd and 3rd coefficients correct? Then I use those to calculate Z?
Thanks for the response...I realize I didn't format the question very well.

 

[Borek]

When fitting polynomials, the more coefficients you use, the better fit you will get. When number of coefficients equal number of data points the curve will go exactly through all the points (think why), but it will be useless between points.

 

[epenguin]

When fitting polynomials, the more coefficients you use, the better fit you will get. When number of coefficients equal number of data points the curve will go exactly through all the points (think why), but it will be useless between points.

Exactly what I was about to say. and fitting 10 points with 6 coefficients you are getting that way.