1. The problem statement, all variables and given/known data I would like to know how to calculate how much moles of gas I have in the following in a cylinder with a certain volume and pressure. The gas in the cylinder is a mixture of air, with added oxygen and helium, the mixture is 18% Oxygen, 36,6% Nitrogen, 45% Helium and 0,4% Argon. The cylinder is 24 liters big, the pressure in the cylinder is 200bar. I can calculate the a and b values for the gas mixture. P = 200 bar V = 24 L T = 293,15 K R = 0,083145 L bar K^{-1} mol ^{-1} a = 0,8746 L^{2}bar mol^{-2} b = 0,0388 L mol^{-1} 2. Relevant equations PV = nRT ( ideal gas law) P = nRT (V-nb)^{-1} - n^{2}a V^{-2} 3. The attempt at a solution According to the ideal gas law this should be 196,93 moles, but that is not right, since if I use the Vanderwaals equation I end up at a pressure of 234,6 bar to accomodate 196,93 moles of this gasmixture. Using the Vanderwaals equation I can't calculate the exact number of moles, this is where is end up: 200 = 24,37n (24 - 0,0388n)^{-1} - 0,00152n^{2} So please enlighten me, how can I proceed to calculate the value for n?
First, remember that the Van Der Waals pressure doesn't have to match the ideal pressure: in fact, it won't, unless a is 0. With an a as large as yours, the pressure can actually have quite a bit of variation between ideal and Van Der Waals. For that last equation, have you considered quadratics?
You can apply some iteration procedure. Rewrite the Van der Waals equation in the form [tex]n=\frac{PV}{RT}(1+\frac{n^2a}{PV^2})(1-\frac{nb}{V}) [/tex] Plug in the data, start with n you got for the ideal gas, substitute for n at the right side, and you get a new n. Continue the procedure with the new n... You get a value near 174 mol, if I am not mistaken. ehild
Thanks, I know that it doesn't match the actual pressure and the ideal pressure do not have to match... I know for sure that the pressure in the cylinder is 200 bar, or actually 201 bar, because I am reading the pressure from a gauge. What do you mean by "considering quadratics"?
Ok, that indeed helps me a lot! I do not have to be completely spot on, however the closer the better.