Boltzmann distribution and ideal gases

[arnesmeets]

We know the Maxwell-Boltzmann distribution for the energy and the speed of a molecule of an ideal gas. Using derivatives it is easy to see that the most probable speed for a gas molecule is given by sqrt(2kT/m), which corresponds to kinetic energy kT. Calculating the most probable energy, we get kT/2. My question is: where did the factor 2 go to? I don't see any explanation.

 

[Hootenanny]

Here is a very good analysis, http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/molke.html#c1

~H

 

[arnesmeets]

I still don't understand

 

[Hootenanny]

What don't you understand?

~H

 

[arnesmeets]

I don't understand why the most probable energy and the energy corresponding to the most probable speed do not coincide, this seems completely illogical to me. I get the mathematical argument but I can't find and reasonable physical explanation. Also, I didn't find any explanation on the website you showed me.

 

[Hootenanny]

I think you missunderstand the meaning. $\frac{1}{2}kT$ is the average kinetic energy for one degree of freedom, i.e. if the particle only moves in one plane. Because the particle are assumed to follow Brownian motion (random paths), it is equally likely that they will be traveling in each of the three dimensions. Therefore, the average kinetic in three dimensions is given by;

$$3 \times \frac{1}{2}kT = \frac{3}{2}kT$$

Does that make sense? It is actually pointed out (but without any explanation) in the penultimate paragraph of that page I gave you.

~H

 

[arnesmeets]

No, this is not what I mean. We've got some Boltzmann distribution for the energy of a particles, something of the form $$dn = \frac{2\pi N}{\left(\pi kT\right)^{3/2}} W^{1/2} e^{-W/kT} dW.$$ Suppose we want to find the energy which appears more than any other energy - consider it as some kind of mode. Taking derivatives, we get that this energy equals kT/2. There is no mistake, it's in Alonso & Finn. The point is that this is only half of the energy corresponding to the most probable speed (this is NOT the average speed or anything else, this is simply the maximum/highest peak of the Maxwell Boltzmann speed distribution). And that doesn't make sense to me. If some energy appears more than any other energy, than the corresponding speed must surely appear more than any other speed?

 

[Andrew Mason]

No, this is not what I mean. We've got some Boltzmann distribution for the energy of a particles, something of the form $$dn = \frac{2\pi N}{\left(\pi kT\right)^{3/2}} W^{1/2} e^{-W/kT} dW.$$ Suppose we want to find the energy which appears more than any other energy - consider it as some kind of mode. Taking derivatives, we get that this energy equals kT/2. There is no mistake, it's in Alonso & Finn. The point is that this is only half of the energy corresponding to the most probable speed (this is NOT the average speed or anything else, this is simply the maximum/highest peak of the Maxwell Boltzmann speed distribution). And that doesn't make sense to me. If some energy appears more than any other energy, than the corresponding speed must surely appear more than any other speed?

The most probable energy has to correspond to the most probable speed. The most probable energy is kT. What is your source for the statement that the most probable energy is kT/2? This may be the source of the confusion. If you were to give us the Edition and page or type out the passage from Alonso and Finn we might be able to help you.

AM

 

[arnesmeets]

I have a Dutch translation of Alonso & Finn, 1984. I'll try to translate:

"Example 1.7. Calculate the most probable energy and speed of gas molecules at a given temperature; these values correspond to the maxima of dn/dW and dn/dv.

Solution: to find the maximum of dn/dW, given by equation (1.44) [which is the one I typed in my previous post, by the way], we need to find the maximum of $$y = W^{1/2}e^{-W/kT}$$. [they are writing W for energy] So we have $$\frac{dy}{dW} = \left(\frac{1}{2}W^{-1/2} - \frac{W^{1/2}}{kT}\right)e^{-W/kT} = 0$$ and hence W(most probable) = kT/2. At room temperature, we get W = 0.012 eV.
To find the maximum of dn/dv, given by equation (1.45), we need to find the maximum of $$y = v^2 e^{-mv^2/2kT}$$. This gives $$\frac{dy}{dv} = \left(2v - \frac{mv^3}{kT}\right)e^{-mv^2/2kT} = 0$$ and hence v(most probable) = sqrt(2kT/m). This speed corresponds to an energy of W = kT and hence this differs from W(most probable). Can the reader explain the reason for this difference?"

So, that's (almost) a literal translation. I don't think there can be a mistake.

 

[arnesmeets]

No-one? :(

 

[Andrew Mason]

I have a Dutch translation of Alonso & Finn, 1984. I'll try to translate:

"Example 1.7. Calculate the most probable energy and speed of gas molecules at a given temperature; these values correspond to the maxima of dn/dW and dn/dv.

Solution: to find the maximum of dn/dW, given by equation (1.44) [which is the one I typed in my previous post, by the way], we need to find the maximum of $$y = W^{1/2}e^{-W/kT}$$. [they are writing W for energy] So we have $$\frac{dy}{dW} = \left(\frac{1}{2}W^{-1/2} - \frac{W^{1/2}}{kT}\right)e^{-W/kT} = 0$$ and hence W(most probable) = kT/2. At room temperature, we get W = 0.012 eV.
To find the maximum of dn/dv, given by equation (1.45), we need to find the maximum of $$y = v^2 e^{-mv^2/2kT}$$. This gives $$\frac{dy}{dv} = \left(2v - \frac{mv^3}{kT}\right)e^{-mv^2/2kT} = 0$$ and hence v(most probable) = sqrt(2kT/m). This speed corresponds to an energy of W = kT and hence this differs from W(most probable). Can the reader explain the reason for this difference?"

So, that's (almost) a literal translation. I don't think there can be a mistake.

Let's try to work through it.

The number of molecules having energy in the interval E+dE is given by the distribution function:

$$n(E) = \frac{2N}{\pi^{1/2}\left(kT\right)^{3/2}} E^{1/2} e^{-E/kT} dE$$

Since $\frac{1}{2}mv^2 = E$, $dE = \frac{1}{2}m\left((v+dv)^2 - v^2 \right) = m(vdv)$, the velocity distribution is:

$$n(v) = \frac{2N}{\pi^{1/2}\left(kT\right)^{3/2}} \left(\frac{1}{2}m\right)^{1/2}v e^{-mv^2/2kT} m(vdv)$$

$$n(v) = \frac{2Nm^{3/2}}{(2\pi)^{1/2}\left(kT\right)^{3/2}}v^2 e^{-mv^2/2kT} dv$$

Take the derivative of the that function with respect to v and see what you get.

AM

 

[freemailsda]

I am facing the exact same dilemma. There are two scenarios for MB distribution:

Expression of the energy distribution: which while maximizing gives: Most probable energy = kT/2.

Expression of the speed distribution: which gives: most probable speed = 2kT/m

But the contradiction is that the above two quantities don't correspond to each other.. WHY??

 

[freemailsda]

@ Mason

The expression that you derived in your last post indeed gives 2kT/m as the most probable speed.

 

[Dickfore]

There is a Jacobian associated with transformation from speeds to energies that gives a different weighing factor for the speed and the energy distribution. Therefore, you need to convert to energy variables first and only then differentiate the probability density function to find its extremum.

 

[freemailsda]

I am so sorry I can't really make out the interpretation of Jacobian. I'll be thankful if you comment on the following in simple terms..

Does this all mean that both the values ie. sqrt(2kT/m) (for most probable speed) and kT/2 (for most probable energy) are right. But its just that we can't really relate the two in direct terms.

 

[Dickfore]

What I am saying is:

$$f(v) |dv| = g(E) |dE|$$

so

$$g(E) = f(v(E)) \, \left|\frac{d v}{d E}\right|$$

In case of a single variable, the Jacobian is:

$$\left|\frac{d v}{d E}\right|$$

 

[freemailsda]

Fine that's what I did..

See..

The number of molecules having energy in the interval E+dE is given by the distribution function:

$$n(E) = \frac{2N}{\pi^{1/2}\left(kT\right)^{3/2}} E^{1/2} e^{-E/kT} dE ... (1)$$

The above is the expression for the distribution for energy.

Now, Since $dE = \frac{1}{2}m\left((v+dv)^2 - v^2 \right) = m(vdv)$ the velocity distribution is:

$$n(v) = \frac{2N}{\pi^{1/2}\left(kT\right)^{3/2}} \left(\frac{1}{2}m\right)^{1/2}v e^{-mv^2/2kT} m(vdv)$$

$$n(v) = \frac{2Nm^{3/2}}{(2\pi)^{1/2}\left(kT\right)^{3/2}}v^2 e^{-mv^2/2kT} dv ... (2)$$

The above is the expression for the distribution of velocity.

Maximizing eq (1) gives the most probable energy (= kT/2) and maximizing eq (2) gives the most probable speed (= sqrt(2kT/m)). But these two values don't correspond to each other.

Where am I going wrong??

 

[Dickfore]

NOTHING! The most probable energy does not correspond to the particles moving with the most probable speed.

 

[freemailsda]

Thanks anyways.. !

 

[PeterDonis]

Moderator's note: This old thread is being reopened to allow further replies.

 

[PeterDonis]

Where am I going wrong??

You're not. The most probable speed will, in general, not be the same as the speed that gives the most probable energy. The basic reason is that the energy (or more precisely the kinetic energy) goes like the square of the speed, so the peaks of the distributions don't correspond. That is what the (correct) math in post #17 is telling you.

 

[Philip Koeck]

I believe the confusion might come from textbooks calling the maximum of f(v) "most probable speed" and likewise for f(E).
Neither f(v) nor f(E) are probabilities so I don't think one should call their maxima "speed or energy with highest probability".
To get a probability one has to multiply with dv or dE respectively, and dE is not proportional to dv!