Calculate the number of moles in a real gas

AI Thread Summary
To calculate the number of moles in a gas mixture under specific conditions, the ideal gas law (PV = nRT) is initially applied, yielding an incorrect result of 196.93 moles. The Van der Waals equation is then used, which accounts for real gas behavior, but it leads to a pressure discrepancy. An iterative approach is suggested to refine the calculation, starting with the ideal gas result and substituting it back into the Van der Waals equation. This method can provide a more accurate estimate, with a final approximation of around 174 moles. The discussion emphasizes the importance of recognizing the differences between ideal and real gas pressures.
larsb
Messages
3
Reaction score
0

Homework Statement


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 L2bar mol-2
b = 0,0388 L mol-1

Homework Equations


PV = nRT ( ideal gas law)

P = nRT (V-nb)-1 - n2a V-2

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 accommodate 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 - 0,00152n2

So please enlighten me, how can I proceed to calculate the value for n?
 
Physics news on Phys.org
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

n=\frac{PV}{RT}(1+\frac{n^2a}{PV^2})(1-\frac{nb}{V})

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
 
Char. Limit said:
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?

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"?
 
ehild said:
You can apply some iteration procedure. Rewrite the Van der Waals equation in the form

n=\frac{PV}{RT}(1+\frac{n^2a}{PV^2})(1-\frac{nb}{V})

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

Ok, that indeed helps me a lot! I do not have to be completely spot on, however the closer the better.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top