In: Chemistry
At 200 K, the compression factor of oxygen varies with pressure as shown in the table below. Evaluate the fugacity of oxygen at this temperature and 124 atm.
p/atm | 1.0000 | 4.0000 | 7.0000 | 10.000 | 40.00 | 70.00 | 100.0 |
Z | 0.9971 | 0.98769 | 0.9788 | 0.9695 | 0.8734 | 0.7764 | 0.6871 |
Theta = Z – 1/ p ‚ dp
which is the area under the curve of Z - 1 p vs. p,
where p is the x-value and Z - 1 p is the y-value, in an x, y coordinate system.
The fugacity coefficient f is then the antilog of the area under the curve.
The fugacity of oxygen, in atm = theta p, where p is the pressure in atm.
Make a list of the p and z values given in the problem.
Using the p and z lists, make a list for the Z - 1 p values. Finally, make a list of {p, Z - 1 p } values. Don't forget to transpose.
p = 81.0, 4.0, 7.0, 10.0, 40.0, 70.0, 100.0<;
z = 80.9971, 0.98796, 0.97880, 0.96956, 0.8734, 0.7764, 0.6871<;
yvalue = (z – 1)/ p;
points = [8p, y value]
data = [Transpose[ points]
area =1/2 * sum[ (y value[i+i]) + y value[i]) * ∗(p[][i + 1]] – p[[i]]),{I,1,6]]
The fugacity coefficient φ at 100 atm = 0.73315
−0.310405
Take the anti log of the area to find f:
fugcoeff = Exp[area];
Print["The fugacity coefficient φ at 100 atm = ", fugcoeff]
The fugacity of O2 at 200 K and 100 atm = 73.315 atm.