0
$\begingroup$

There is a problem given as -

N molecules of an ideal gas are in a container of volume $V_o$. If a molecule has a same probability of being at any point in the container. The variance of number of particles in a smaller volume $V$ is

We can see this situation as binomial distribution.
In binomial distribution we have $N$ trials and probability of success in each trial is, say $p$ with each trial being independent.
Here in our case, total trials is analogous to total number of particles, $N$ and the success in each trial corresponds to finding the particle in $V$.
As in container there is ideal gas thus the gas particles are not having any sort of interactions, so it is sort of equally likely outcomes case in continuum.
So, probability of finding particles in volume $V$ is $\frac{V}{V_o}$

Probability of finding $M$ particles in volume $V$ is analogous to $M$ successes in $N$ trials
So, probability of finding $M$ particles in volume $V$ is $P_{N_{o}}(M)=\begin{pmatrix}N_o \\ M\end{pmatrix}\Big(\frac{V}{V_o}\Big)^M\Big(1-\frac{V}{V_o}\Big)^{N-M}$

Average (Expectation) of finding the particles in volume $V$ is $E(n)=\sum_{i=1}^{N_o}iP_{N_o}(i)=\sum_{i=1}^{N_o}i\begin{pmatrix}N_o \\ i\end{pmatrix}\Big(\frac{V}{V_o}\Big)^i\Big(1-\frac{V}{V_o}\Big)^{N-i}$
This expectation value comes out to be $\boxed{E(n)=\Big(N_o\frac{V}{V_o}\Big)}$

Similarly variance comes out to be $\boxed{Var(n)=E((n-E(n))^2)=N_o\Big(\frac{V}{V_o}\Big)\Big(1-\frac{V}{V_o}\Big)}$
This is the answer.

The above analysis is clear to me.

But I get different result if I analyze it physically by not modelling it as a probability distribution.
We have an ideal gas thus it does not has any sort of interaction. So probability of finding particles in an elemental volume is same in all parts of container.
So we can use unitary method to find number of particles in volume $V$.
Number of particles in volume $V_o\;=\;N$
So, number of particles in volume $V\;=\;N\frac{V}{V_o}$
Thus average number of particles in volume $V$ is $\Big(N\frac{V}{V_o}\Big)$

With this reasoning there is always a uniform number of particles in a volume because the gas is ideal. So at any time, in volume $V$, the number of particles should be $\Big(N\frac{V}{V_o}\Big)$, it can't be distributed about this value otherwise the symmetry would be broken down. Because some excees or reduction of particles from the mean number in volume $V$ suggests that there is something special in this volume $V$ (say, the presence of interaction between particles) or their is some external influence which causes this change to happen, but no such influence is present.
Particles should enter or leave this volume in such a way that the mean number remains constant.

According to me, all the times, the number of particles in volume $V$ should be $\Big(N\frac{V}{V_o}\Big)$ otherwise the symmetry arises due to the fact that gas is ideal gets broken down. Thus the variance should be $0$.

So, the question is that the statistical description of this system is physically correct or not? As in statistical description we get non-zero variance. Expectation is in accordance to the physical intuition.
Please tell me the way I am physically analyze the system is correct or not.

$\endgroup$
0
$\begingroup$

The first statistical analysis is correct.

The symmetry argument only holds for the equilibrium macrostate and not for fluctuations about that state.

In your second analysis, you neglect the fact that any given particle in a non-interacting system has no knowledge of the existence of any other particle in the system and so their positions can have no impact on any physical properties of any other particle. For this reason, particles will move in and out of your volume $V$ due to their non-zero velocities, and the other particles are none the wiser and so will not redistribute due to this. It is only on average that the number of particles in $V$ is $N\frac{V}{V_0}$.

$\endgroup$
3
  • $\begingroup$ Thanks for the answer. But I have a doubt that suppose we made that system which is isolated. After sometimes, the equilibrium has been achieved like the temperature and pressure (macrovariables) becomes fixed. At that time shouldn't in the volume $V$, the number of particles has to be fixed? $\endgroup$
    – Iti
    Jun 23 '21 at 17:37
  • $\begingroup$ In short why can't we just use unitary method to find the number of particles in volume $V$ instead of statistical description? $\endgroup$
    – Iti
    Jun 23 '21 at 17:45
  • 1
    $\begingroup$ If the system is isolated then fluctuations are not due to external systems but you can think of dividing the system into $n$ volumes which now act as open systems and so these subsystems can exchange particles and energy. This means that the number of particles in one subsystem could decrease as long as another subsystem's particle number increases, that particle will also carry away energy. This leads to local fluctuations in pressure, temperature and particle number. However, if you average over the whole system then these properties will remain constant as the system is isolated. $\endgroup$
    – Chris Long
    Jun 23 '21 at 17:55

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.