Talk:Effusion

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!!!! There's an error in the page, or in Wikipedia !!!! --NOT REALLY.208.67.99.244 (talk) 20:05, 21 February 2009 (UTC)Reply

The example comparing the rates of deflation of a balloon filled oxygen and hydrogen seems to be misleading. A oxygen molecule is several times larger than a hydrogen molecule, and in this case with the pores in the membrane being on the molecular scale, would be an equally if not more important factor. Mentioning the use of effusion for enrichment of nuclear isotopes might provide a better example.

I've removed the balloon example. It is irrelevant to the topic of effusion, as Graham has shown in 1866 (https://doi.org/10.1080/14786446608644207) that the passage of gases through rubber membranes is by chemical diffusion, not by effusion through microscopic pores. Niel (talk) 14:07, 3 April 2021 (UTC)Reply

ACTUALLY: at the molecular scale gasses are interacting in such a way that the particle size can be essentially disregarded. Hence the whole, ideal gas law. For the size of the gas molecule to matter temperatures need to approach the condensation temperature of the gas.

The implication that two gases having the same temperature would have the same kinetic energy is false. Temperature is a measure of the change in entropy vs. the change in energy; two gases at the same temperature do NOT necessarily have the same amount of energy! To put it another way, recall the concept of specific heat. Specific heat is the amount of energy required to raise a given amount of a substance by 1 degree C. If one gas has a higher specific heat than another, it can end up containing more energy than the gas with the lower specific heat by the time it's raised to the same temperature. Because of this, I edited the text to contain "(and having the same specific heat)" to the sentence about energy and average velocity.

ACTUALLY: Temperatue is defined as a measure of the average transitional kenetic energy, or the energy of motion of the particles. It in no way maters how difficult it was to change the temperature (specific heat) here because the temperature value is comparing the current temperature and therefore current kenetic energy, which by definition will be the same.

I am adding a see also section at the bottom that links to the Graham's Law of Effusion

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"The average molecular speed is about 0.921 vrms." What is this supposed to mean? Average over what? For what molecules? Context is missing. — Preceding unsigned comment added by MrMischelito (talk • contribs) 09:04, 2 November 2016 (UTC)Reply

The average (or mean) molecular speed is just the ordinary arithmetic average of the speeds of all the molecules in the gas: add up all the speeds and divide by the number of molecules. And v_rms is the root-mean-square speed: add up the squares of all the speeds, divide by the number of molecules to obtain the mean-square, and then take the square root. For the usual distribution of molecular speeds, one can prove that v_avg = 0.921 v_rms.

What should be added to the article to clarify these points? It does seem to be hard to find an article which discusses the average speed. Dirac66 (talk) 18:48, 7 November 2016 (UTC)Reply

I have now clarified the reference to average molecular speed a little, and added a link to the article on the Maxwell velocity distribution which provides more detail. Dirac66 (talk) 00:56, 9 December 2016 (UTC)Reply

The rate equation clearly shows the rate as being inversely proportional to the square root of temperature which means inversely proportional to the rms velocity. I'm no expert but the units of the equation are 1/s so it appears correct. Later the article states "At a given pressure and temperature, the effusion rate is proportional to the root-mean-square speed" and "The effusion rate for a gas depends directly on the average velocity of its particles. Thus, the faster the gas particles are moving, the more likely they are to pass through the effusion orifice". Not only does this seem intuitively wrong but it's inconsistent with the equation. Could an 'expert' check this. 130.246.58.79 (talk) 15:43, 24 March 2017 (UTC)Reply

The key words here are "At a given pressure and temperature", which means that this paragraph is not considering the effect of T at all. Instead it is considering the effect of changing M at a constant temperature. If T is constant, the root-mean-square speed and the effusion rate are both inversely proportional to the square root of M. Historically the effect of M has been of greater interest than the effect of T, because of the connection to Graham's law.

However perhaps the article should also consider the effect of varying T. I will revise the article and try to clarify the distinction between the two effects. Thank you for raising this point. Dirac66 (talk) 15:52, 26 March 2017 (UTC)Reply

I will now consider here the effect of T. The equation is

${\displaystyle Rate={\frac {pAN_{A}}{\sqrt {2\pi MRT}}}}$, so the rate is inversely proportional to √T only provided that p is constant. In that case, n = pV/RT, so the reason the effusion rate decreases on heating is that there are less molecules to effuse, even though their root-mean-square speed does increase.

It seems more normal to consider the case when a given amount of gas is heated, so that n is constant and p = nRT/V increases. In that case the rate is in fact directly proportional to √T, since the p in the numerator increases faster than the √T in the denominator.

However I have not yet found a source which discusses this point, so for the moment it is "original research" on the talk page and I will not insert it into the article yet. Dirac66 (talk) 00:55, 31 March 2017 (UTC)Reply