9
$\begingroup$

Among the five postulates or assumptions of kinetic molecular theory of gases (KMT), only the assumption that a gas is made up of large number of small particles doesn't seem to make any sense to me. The only reason I could of think of was due to lack of negligible pressure. I reasoned that if there aren't enough molecules in the container there will be no evidence of pressure, since most of collision would be within molecules and not between the wall of the container and the molecules. But there is something missing in this explanation or it is completely wrong, as it fails to adequately explain the above stated question. Why do we need this assumption?

$\endgroup$
  • $\begingroup$ What is the sound of one hand clapping? What is the temperature of a single atom of monoatomic gas? $\endgroup$ – Curt F. Oct 3 '16 at 5:53
  • 5
    $\begingroup$ Surely there have only to be enough particles for the average values of quantities to be meaningful so that the random nature of single molecule events is not observed, and in fact fluctuations due to this becomes quite negligible. Thus 'large' is determined by this consideration. Clearly Avogadro's number of particles is going to be more than enough $\endgroup$ – porphyrin Oct 3 '16 at 7:01
  • 1
    $\begingroup$ I'd assume that the notion of a large number of particles comes from statistics. By assuming that the use of statistics is a valid way to analyze the phenomena then as the number of particles that you're averaging over gets larger, then the statistical values calculated will be better estimators. $\endgroup$ – MaxW Oct 3 '16 at 19:34
3
$\begingroup$

To echo and expand on Porphyrin's comment, the idea of having a large number of particles shows up in KMT because this is a statistical theory. That is, everything KMT predicts is some average property, or, to be rigorous, the expectation value of a distribution.

Thus, if one only had a very small number of particles, you would never arrive at something like the Maxwell-Boltzmann distribution because you would have to treat each particle individually rather than averaging over a phase-space.

So, to answer your actual question, we need this assumption because it justifies us in doing things like the derivation in this link where they take the derivative of the unknown function (eq. 9.8). That assumes this function is continuous over all space because no constraints are put on the resultant function. The only way we can do that and expect to have our theory match reality is if reality has an approximately continuous distribution of particle speeds, i.e. a lot of particles present.

As it happens, $10^{23}$ particles really isn't that many in practice, so when you get down to very low pressures and temperatures (that is the number density $\frac{N}{V}$ is small) the theory still works because something like $10^6$ probably still makes a pretty continuous function.


As a totally unrelated aside, you might also wonder if the MB distribution is wrong because it allows particles to have speeds ranging from $0\rightarrow \infty > c$... I computed the special relativistic correction one time and it's so laughably small as to actually annoy you that you even took the time to do it.

I know this answer focused mostly on the MB distribution, but that is very central to KMT so I thought it would make for a representative though not exhaustive answer.


EDIT:

To be critical of my answer, however, I'm almost certain that much of KMT is derivable while only thinking about a single particle sampling its whole phase-space (perhaps all at once) and once could simply argue that the above function I mention taking the derivative of is a hypothetical distribution in which each of the infinite states need not be occupied at any given time. So basically taking a time average of the phase-space over infinite time gives this continuous distribution. That idea also lines up well with ideas to do with the partition function where every state could be equally likely but we get a non-even distribution because some states are more likely to be sampled.

Then perhaps it is just more physically instructive to think about a large number of particles.

$\endgroup$
  • $\begingroup$ Nice way of looking at it :) +1 $\endgroup$ – getafix Oct 4 '16 at 2:23
1
$\begingroup$

I don't remember the "large" in the assumptions. I'm with you on questioning why.

Possibly the writer of this assumption was taking the perspective that because atoms are so small, it takes a lot of them to effect a measurment. That is, all the meaniful quantities of gases we typically deal with are composed of large enough numbers for humans to manipulate in the lab or commercially.

But, as far as I know, the kinetic theory does not break down at low particle concentrations. All the gaw laws should still hold. Two particles in a container will still collide with the walls exerting pressure (assuming the other assumptions are in play). Good luck measuring it. I've never seen asterisks next to the gas laws admonishing you to not use these at low particle concentrations.

edit: I see someone has downvoted this answer, without a courtesy comment. For lack of a better focus, let me address current comments to the OP's post.

Temperature is a measure of the average kinetic energy of the prticles in a substance. This is the definition inherent in assumptions of KMT. It's the one I taught for 24 years and has appeared in every textbook I've used. This, the most useful, and accepted, definition with regard to KMT and the gas laws, does not rely on being able to use a thermometer to measure it. The average kinetic energy of the particles is equal to $3RT/2$, where $T$ is the Kelvin temperature and $R$ the gas constant. There is no imposition on the number of particles. And it is a average being used; a fraction of the number of particles, with the same average KE, still has the same temperature.

As to pressure and the implication of "fluctuations" at low particle numbers. Let's put ten particles in a container. With this small a number, there will be moments when none of those particles act to generate force on a particular chosen area. But at other moments all ten might be acting. Over very small intervals of time, the pressure at this locus will vary wildly. But over meaningful spans of time the pressure values will average out. I can't imagine why that time-averaged pressure would be any different than predicted by KMT.

I would agree that relying on "time averaging pressure" and considering the temperature of one particle are not practically useful concepts in the application of gas laws. I just checked my copy of Zumdahl (AP Chemistry), no use of "large". I just checked my Gialcolli (AP Physics), oops! He uses "large". I would postulate Zumdahl is a theoretical kind of guy and Giancolli is practical.

$\endgroup$
  • $\begingroup$ If it's worth anything, I agree with you. The temperature as defined in kinetic theory is simply related to $\langle v^2 \rangle$ which is a well-defined property even for a single molecule. There is no mention of such an assumption in Atkins' Physical Chemistry 9th ed. (Chapter 20). $\endgroup$ – orthocresol Oct 3 '16 at 17:45
  • $\begingroup$ It's worth my sanity. Thanks for the formatting edit too, I'm not yet conversant with equation formatting here. $\endgroup$ – bpedit Oct 3 '16 at 18:41
  • 1
    $\begingroup$ It would seem that you have invoked the ergodic hypothesis which allows the time average to be the same as the ensemble average, which seems reasonable. $\endgroup$ – porphyrin Oct 3 '16 at 21:16
  • $\begingroup$ @porphyrin Thanks, I'll look that up. I've always been a bit stymied explaining to students how gas pressure arises when there is no time element in the definition of pressure yet the gas pressure depends on particle velocities. $\endgroup$ – bpedit Oct 3 '16 at 22:52
  • $\begingroup$ Ah, @porphyrin I think you hit the nail on the head here. :) $\endgroup$ – orthocresol Oct 5 '16 at 5:14
1
$\begingroup$

Short answer: It is not absolutely necessary, the theory and the results derived from it still hold at low particle concentrations but they cease to be meaningful. It is not so much a theoretical requirement, but a statistical one (for the results to make sense)

The kinetic theory of gases attempts to describe macroscopic properties like pressure and temperature based on a microscopic particulate picture of matter.

Pressure, for instance, is described as the force per unit area exerted by the gas molecules on the walls of the container. So, if you have a large number of particles in a container, whizzing about with some velocity and bumping into the container walls all the time. The collisions exert a force, which we measure as pressure.

Now thing of a macroscopic container with just one gas molecule in it. It will still whizz about inside the container with some velocity, and will also collide with the container walls every now and then exerting a force on them. If we invoke the ergodic hypothesis, which means that the time spent by a molecule in region of space is related to the volume of said region, it would reasonable to say that the particle spends more time in the bulk volume of the container than near its walls.

Thus, we will not have the constant bombardment of particles on the container walls like the many molecule case, and thus we cannot obtain a meaningful measurement of pressure.

Similarly, temperature is related to the rms velocity in the kinetic theory. That quantity can be calculated for a single lone molecule, however, if you stick a thermometer in the container you wouldn't be able to get a meaningful measurement.

$\endgroup$
1
$\begingroup$

The kinetic theory seems to implicitly assume 'large' numbers of molecules in the sense that any one molecule is in thermal equilibrium with a heat reservoir at some temperature T, which is formed of all the other molecules in the gas. This is called a 'canonical ensemble'.

In the calculation of the Maxwell-Boltzmann distribution it is assumed that the state of any molecule can be defined by position r and momentum p. The state of a molecule is defined to be in the range $r \rightarrow r+dr$ in the volume element $d^3r=dxdydz$ and have momentum in the range $p \rightarrow p+dp$ in momentum space volume $d^3p=dp_xdp_ydp_z$. The molecule is in weak interaction with other molecules which then act as a heat reservoir, and as we assume a dilute gas we can think of this molecule as a distinct small system in contact with a heat reservoir, and thus having a canonical distribution.

As a molecule's position can be anywhere in the volume of gas and its speed from $-\infty$ to $+\infty$ this leads to the probability of being at $r\rightarrow dr$ and $p\rightarrow dp$ $$P(r,p)d^3rd^3p \approx exp(-\beta p^2/(2m)~)d^3rd^3p$$ with $\beta =(k_BT)^{-1}$. The exponential is the canonical distribution.

A consequence of being in contact with a heat reservoir is that there is a range of molecular speeds (and hence momenta). If there are $dN$ molecules with speed $v\rightarrow v+dv$ then $dN$ remains constant at equilibrium, even though the speeds of molecules are continually changing due to collisions. This also implies that a large number of molecules are colliding.

$\endgroup$
0
$\begingroup$

You're trying to describe a uniform gas. You want a large number of particles so that the gas appears to be uniform. If it's not uniform, it's hard to describe properties of the gas. As an extreme case, what is the pressure exerted by a single gas particle and how much volume (of the entire container) does it take up?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.