Fluctuations in the ideal gas

The well-known equation of state for an ideal gas, P V = N k B T , PV = Nk_BT, is only true on average. In reality, the pressure fluctuates from moment to moment as more or fewer gas particles hit the wall in a given instant.

Suppose we measure the pressure on the walls of the container at m m different times P = { P t 1 , P t 2 , , P t m } . \mathcal{P} = \{P_{t_1}, P_{t_2},\ldots, P_{t_m}\}. We can measure the magnitude of the pressure fluctuations by finding the standard deviation σ P = P 2 P 2 . \sigma_P = \sqrt{\langle P^2\rangle - \langle P\rangle^2}.

How will the size of the fluctuations shrink relative to the mean as we increase the number of gas particles?

In other words, if N N denotes the number of particles, what is σ P P \dfrac{\sigma_P}{\langle P\rangle} proportional to?

1 N \frac{1}{N} 1 e N \frac{1}{e^N} 1 N 2 \frac{1}{N^2} 1 N \frac{1}{\sqrt{N}}

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2 solutions

Mark Hennings
Apr 2, 2018

This result was proved by R E Burgess, "Pressure fluctuations in an ideal gas", Physics Letters A , but note that there is a typo in the abstract on the Elsevier page (it should read P 2 / M \langle P\rangle^2/M and not just P 2 M \langle P\rangle^2 M . You have to pay Elsevier if you want to read it!

That's awesome, did not know about the Burgess paper. My derivation is much more cavalier.

Josh Silverman Staff - 3 years, 2 months ago

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I just took a calculated guess, what's the actual way to do it?

Sravanth C. - 3 years, 1 month ago

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There is some pretty hardcore statistical mechanics in the Burgess paper...

Mark Hennings - 3 years, 1 month ago

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@Mark Hennings I'll write up my back of the envelope solve. No intense stat mech involved.

Josh Silverman Staff - 3 years, 1 month ago

It's posted.

Josh Silverman Staff - 3 years, 1 month ago

For those not knowing, the quantity sought is coefficient of variance, a usual measure of dispersion in a distribution.

Venkata Karthik Bandaru - 3 years, 2 months ago

It seems just like the statistics and how a mean of samples are much tighter to the population mean. The variance of the spread of these means is inversely proportional to n.

Characterized by the equation z = x x ˉ σ / n z=\frac{x-\bar{x}}{\sigma / \sqrt{n}}

I'm not too familiar with statistical terminology, sorry for any confusion.

Jerry McKenzie - 3 years, 2 months ago
Josh Silverman Staff
May 8, 2018

This argument doesn't lean on any statistical mechanics, it just takes advantage of the lack of correlations in the ideal gas. As it turns out, that's the foundation for the statistical mechanics of the ideal gas, elements of surfactant chemistry, etc. but it also lets us skip the formalities.

The pressure reading on any wall is proportional to the number of particles that hit the wall in a given time. If that time is Δ T , \Delta T, then we care about the molecules inside the box of length v Δ T v\Delta T that's in front of the wall (where v v is the typical speed of gas molecules — we'll assume this is a constant, but it doesn't change anything).

For simplicity let's say that length is L T L_T and the length of the box is L box . L_\textrm{box}. Then the probability that a given gas molecule is in the box is ρ = L T / L box . \rho = L_T/L_\textrm{box}. Similarly, the probability that it isn't in the box is 1 ρ . 1 - \rho.

That means that expected number of particles in the box follows the binomial distribution, so that the mean of pressure P \langle P\rangle is proportional to N ρ N\rho and similarly the variance σ P 2 \sigma_P^2 is proportional to N ρ ( 1 ρ ) . N\rho\left(1 - \rho\right).

Putting those together, we get a scaling relationship noise-to-signal ratio and N N : σ P P N ρ ( 1 ρ ) ) N ρ , \dfrac{\sigma_P}{\langle P\rangle} \sim \frac{\sqrt{N\rho\left(1 - \rho\right))}}{N\rho}, which is proportional to 1 / N . 1/\sqrt{N}.

We also have P avg = m N v rms 2 3 V P_{\text{avg}}=\dfrac{mNv_{\text{rms}}^2}{3V}

So can we evaluate the average value of ρ \rho by equating P \langle P\rangle and P avg P_{\text{avg}} from the expression above?

Sravanth C. - 3 years, 1 month ago

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Mmm, no you can't and the reason is that the whole time I'm using scaling arguments. Since I just want to end up knowing how the noise to signal ratio varies with N , N, I dropped the the other details that aren't needed.

Josh Silverman Staff - 3 years, 1 month ago

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