darkalex wrote:So due to its molecular weight, O2 has a greater probability of exiting the keg. How does this probabilty change with each subsquent venting and purging of the keg? Is it simply multiplied by the number of times purged? Or does the O2 loss depend on the concentration of O2 and/or other gasses in the head space? If so, does the probability decrease, with each purging, since there is less of it around? Describe what you mean by 2nd order phenomena.
Let's see if I can answer both questions at once. For an ideal gas P*V = n*R*T (P = pressure, V = volume, n = the number of gas molecules, R is a constant and T is the absolute temperature). Rearranged n = P*V/R*T i.e. the number of molecules of a gas depends on its pressure. So if I have n_O molecules of oxygen the pressure from that at temperature T in volume V is P_O = n_O*R*T/V and if I have n_N = 4*n_O molecules of nitrogen the pressure of the nitrogen will be P_N = n_N*R*T/V = 4*P_O and the total pressure will be 5*P_O. Air which is 20% oxygen. At a pressure of 1 atmosphere 0.2 atm is attributable to nitrogen and 0.8 atm to nitrogen. If we close some air up in a Cornelius keg and then push in twice as many carbon dioxide molecules as there are nitrogen plus oxygen molecules, 2*(5*n_O), I will have a total of 3 atmospheres pressure. Turning it around the other way if I pump in CO2 until the the pressure in the keg is 3 atmospheres (absolute equal to 2 atmospheres gauge or about 30 psig) I will have n_O molecules of oxygen, 4*n_O molecules of nitrogen and 10*n_O molecules of CO2 for a total of 15*n_O molecules of gas of which n_O are oxygen. The percentage of oxygen molecules is 100*1/15 = 6.66667%. In air, the percentage was 20% so the concentration of oxygen molecules was reduced by a factor of 3. If I bleed the tank down to atmospheric pressure 2/3 of the molecules will escape. Assuming they all escape at the same rate (forgetting Grahams law for the moment) n_O/3 molecules of oxygen will remain. If I now repeat the process by pumping in CO2 to the point where the number of molecules is tripled the number of oxygen molecules as a percentage will decline by a factor of 3 and upon bleed down n_O/3/3 will remain. Thus each dilution with CO2 results in the same reduction factor. After m dilutions by a factor of 3 the remainging percentage of oxygen is 20/3^n.
This assumes 3 things
1) The number of molecules is proportional to the pressure
2) The molecules are uniformly mixed
3) The molecules are equally likely to leave the keg when the valve is opened.
None of these assumptions is true. The molecules interract with each other so that the pressure relationship is more complicated than P = n*R*T/V. In fact it is more like P = R*T*( (n/V) + B*(n/V)^2 + C*(n/V)^3 + ...). The first term, the linear term, describes ideal gas behaviour. B and C are called 'virial coefficients' and depend on temperature and the particular gas. The terms which contain them also contain powers of (n/V) and represent deviations from the ideal gas law. In a polynomial like this the power to which the independent variable (n/V) is raised is called the order of the term. Thus P = R*T*(n/V) is a first order approximation to the actual behaviour, P = R*T*(n/V) + R*T*B*(n/V)^2 a second order approximation and so on. If an engineer says a gas is ideal to first order that is equivalent to saying that the second order and higher terms in the expansion are not significant. Clearly if there are few molecules in a large volume (n/V <<1) then the first order approximation is valid as for example (n/V)^2 = 0.000001 is much smaller than (n/V) = 0.001. At higher gas concentrations (pressure) the B, C.. terms become significant. Calculations done on the basis of an ideal gas will not be exact. But clearly there is strong motivation to accept the first order model (ignore the 2d and higher order terms) as the solution becomes much messier.
Put much more simply, when an engineer says "it's a second order effect" he is saying that it, whaterver it is, does have a recognized effect but not a major one (and, frequently, that he does not want to be bothered with accounting for it). Hops additions at knockout have a second order effect on beer bitterness. Hops additions 1 hour before knockout have a second order effect on hops aroma.