Gas in a box with Maxwell-Boltzmann distribution

In summary: The velocity would be proportional to the surface area of a sphere of radius v around the particle. How would this expression be linked to the final answer that I'm told to prove?Thanks!In summary, the probability distribution for a particle colliding against a wall at ##z=L## is given by ##\rho_{collision}(\vec{v})=2\rho_v(\vec{v})\theta(v_z)##, where ##\rho_v(\vec{v})## is the Maxwell-Boltzmann distribution and ##\theta(v_z)## is the Heaviside step function. This can be interpreted either as the distribution for a group of particles over a certain time interval or for
  • #1
rogdal
14
2
Homework Statement
Given a box of dimensions ##V=L \times L \times L## in the ##xyz## space which has ##N \gg1## particles of a certain gas, the velocity of a random particle is given by the Maxwell-Boltzmann distribution:
$$\rho_v(\vec{v})=\frac{1}{\sqrt{(2\pi \beta)^3}}e^{-\frac{v^2}{2\beta}},$$
where ##\beta = k_BT/m##

With this data, prove that the probability distribution function with which a particle chosen at random collides against the wall located in the plane ##z = L## is:
$$\rho_{collision}(\vec{v})=2\rho_v(\vec{v})\theta(v_z),$$
where ##\theta(v_z)## is the Heaviside theta function.
Relevant Equations
$$\rho_v(\vec{v})=\frac{1}{\sqrt{(2\pi \beta)^3}}e^{-\frac{v^2}{2\beta}}$$
I have considered two scenarios:

1) A particle that has just collided with the wall at ##z=L## is moving with a velocity ##v_z<0## moving away from the wall. Hence, the probability that this particle has of colliding again is ##0##, so its distribution is also ##0##.

2) A particle moving with positive ##v_z##. Since the particles are constrained to move in that box, it will necessarily collide against the wall, and assuming a completely elastic collision, the velocity distribution in this case would be ##2\rho_v(\vec{v})## because after the collision the particle emerges with the same velocity it had before.

Would these arguments be correct?

Thanks in advance!
 
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  • #2
rogdal said:
2) A particle moving with positive ##v_z##. Since the particles are constrained to move in that box, it will necessarily collide against the wall, and assuming a completely elastic collision, the velocity distribution in this case would be ##2\rho_v(\vec{v})## because after the collision the particle emerges with the same velocity it had before.
Hello, @rogdal. I don't understand your argument. The particle does not emerge from the collision with the same velocity it had just before the collision. Since ##v_z## changes in the collision, the final velocity vector is not the same as the initial velocity vector. If ##v_z## just changes sign in the collision with no change in ##v_x## or ##v_y##, then the initial and final speeds would be the same. But I don't see why this would give the factor of 2 in the answer.

I'm not completely sure of the interpretation of the statement of the problem:

prove that the probability distribution function with which a particle chosen at random collides against the wall located in the plane ##z = L## is:
$$\rho_{collision}(\vec{v})=2\rho_v(\vec{v})\theta(v_z),$$

Two possible interpretations come to my mind:

(1) Consider some arbitrary interval of time that is long enough for many particles to strike the wall. Imagine making a list of the incoming velocity ##\vec v## for each particle that strikes the wall during the time interval. What is the probability distribution ##\rho_{\rm collision}(\vec v)## associated with this list?

(2) Pick an arbitrary particle of the gas. Wait until it collides with the wall. (It could be a long wait, but that's ok.) Write down the incoming velocity ##\vec v## for this collision. Repeat this many times. What is the probability distribution ##\rho_{\rm collision}(\vec v)## associated with this list?

I favor the second interpretation. It seems to be more in line with the way I would literally interpret the wording of the problem statement. Also, I think the second interpretation will give the answer that's stated in the problem; whereas, I don't believe the first interpretation would.
 
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  • #3
Hello @TSny. Many thanks for your answer. Yes, obviously, after thinking it a bit, the argument I used in my 2nd scenario for justifying the origin of the 2 extra factor makes no sense.

Let me try to think about the probability distributions from your interpretations:

(1) In this case, we would obtain exactly the Maxwell-Boltzmann distribution ##\rho_{collision}(\vec{v}) = \rho_v(\vec{v})##. However, since we are only considering positive ##v_z## from, let's say, 0 to infinity, in this particular case, ##\rho_v(\vec{v})## wouldn't be normalized. Integrating, instead of 1 one would obtain 0.5, which is not correct. But if we normalize it by adding a factor of 2, we would get exactly ##\rho_{collision}(\vec{v}) = 2\rho_v(\vec{v})\theta(v_z)##. Does this make sense?

(2) Instead the Maxwell-Boltzmann distribution for a group of particles, this would be the Maxwell-Boltzmann distribution just for a single particle, right? This distribution would be given by:

$$\rho_{collision}(\vec{v}) = \left(\frac{m}{2kT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$
The factor of 4πv^2 would arise from the fact that the velocity distribution is for a single particle rather than for a group of particles.The velocity would be proportional to the surface area of a sphere of radius v around the particle. How would this expression be linked to the final answer that I'm told to prove?

Thanks!
 
  • #4
rogdal said:
Hello @TSny. Many thanks for your answer. Yes, obviously, after thinking it a bit, the argument I used in my 2nd scenario for justifying the origin of the 2 extra factor makes no sense.

Let me try to think about the probability distributions from your interpretations:

(1) In this case, we would obtain exactly the Maxwell-Boltzmann distribution ##\rho_{collision}(\vec{v}) = \rho_v(\vec{v})##. However, since we are only considering positive ##v_z## from, let's say, 0 to infinity, in this particular case, ##\rho_v(\vec{v})## wouldn't be normalized. Integrating, instead of 1 one would obtain 0.5, which is not correct. But if we normalize it by adding a factor of 2, we would get exactly ##\rho_{collision}(\vec{v}) = 2\rho_v(\vec{v})\theta(v_z)##. Does this make sense?

(2) Instead the Maxwell-Boltzmann distribution for a group of particles, this would be the Maxwell-Boltzmann distribution just for a single particle, right? This distribution would be given by:

$$\rho_{collision}(\vec{v}) = \left(\frac{m}{2kT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$
The factor of 4πv^2 would arise from the fact that the velocity distribution is for a single particle rather than for a group of particles.The velocity would be proportional to the surface area of a sphere of radius v around the particle. How would this expression be linked to the final answer that I'm told to prove?

Thanks!
I think your argument for the first interpretation is actually the one that should be used for the second interpretation.

We can consider a simple model to illustrate the difference between the first and second interpretations. Imagine a box of gas for which the molecules are uniformly spread in space and the molecules move only parallel to the z-axis. The molecules do not collide with themselves. So, each molecule just bounces back and forth between two walls and never changes its speed. Suppose half the molecules have a speed of 500 m/s and the other half a speed of 100 m/s. Then the velocity distribution would be 25% with velocity +500 m/s, 25% with - 500 m/s, 25% +100 m/s, and 25% -100 m/s. Or, if you consider molecules with ##v_z >0##, then 50% of those will have velocity +500 m/s and 50% +100 m/s.

(1) If you watch the wall at ##z = L##, then you will see molecules of velocity + 500 m/s and velocity +100 m/s strike the wall. But the faster molecules hit the wall more often than the slow molecules (just because they are moving faster). So, the velocity distribution for the molecules that hit the wall during some arbitrary interval of time will be more than 50% +500 m/s. That is, this velocity distribution ##\rho_{\rm collision}(\vec v)## of the molecules striking the wall is different from the velocity distribution of the molecules in the box that have ##v_z > 0##.

(2) However, if you pick a molecule at random in the box that has ##v_z > 0##, there is a 50% chance that it has a velocity of +500 m/s and a 50% chance of +100 m/s. Thus, a molecule picked at random will have a 50% chance of hitting the wall at +500 m/s and a 50% chance of hitting the wall at +100 m/s. So, for particles picked randomly, the velocity distribution ##\rho_{\rm collision}(\vec v)## is the same as the velocity distribution of the molecules in the box that have ##v_z >0##.

I think similar arguments hold for the actual gas which has the Maxwell-Boltzmann distribution.

In discussing interpretation (2), you wrote
This distribution would be given by:

$$\rho_{collision}(\vec{v}) = \left(\frac{m}{2kT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$
The right-hand side is the Maxwell-Boltzmann probability distribution for speeds rather than velocity. But, the question is asking for the velocity probability distribution.
 
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  • #5
rogdal said:
(2) Instead the Maxwell-Boltzmann distribution for a group of particles, this would be the Maxwell-Boltzmann distribution just for a single particle, right? This distribution would be given by:

$$\rho_{collision}(\vec{v}) = \left(\frac{m}{2kT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$
The factor of 4πv^2 would arise from the fact that the velocity distribution is for a single particle rather than for a group of particles.The velocity would be proportional to the surface area of a sphere of radius v around the particle. How would this expression be linked to the final answer that I'm told to prove?
The different distributions aren't for single particle vs. many particles. It's for Cartesian vs. spherical. You have ##dv_x\,dv_y\,dv_z = v^2 \sin\theta\,dv\,d\theta\,d\phi##. That's where the ##v^2## comes from. When you integrate out the angles to get a distribution in terms of the speed only, you get the factor of ##4\pi##.
 
  • #6
Oh, I see... Many thanks @TSny and @vela for your help. I understand it now!
 

FAQ: Gas in a box with Maxwell-Boltzmann distribution

What is the Maxwell-Boltzmann distribution?

The Maxwell-Boltzmann distribution is a statistical distribution of speeds for particles in a gas that is in thermal equilibrium. It describes the probability of finding a particle with a certain speed in a gas where the particles are not interacting with each other, aside from elastic collisions. The distribution is derived from the principles of classical mechanics and statistical thermodynamics.

How is the Maxwell-Boltzmann distribution related to temperature?

The Maxwell-Boltzmann distribution is directly related to the temperature of the gas. Higher temperatures result in a broader distribution of particle speeds, meaning that particles are more likely to have higher speeds. Conversely, at lower temperatures, the distribution is narrower, indicating that most particles have lower speeds. The distribution's shape and peak are determined by the temperature, with the peak shifting to higher speeds as the temperature increases.

What are the key assumptions behind the Maxwell-Boltzmann distribution?

The key assumptions behind the Maxwell-Boltzmann distribution are: (1) the gas consists of a large number of identical, non-interacting particles, (2) the particles are in thermal equilibrium, meaning their energy distribution does not change over time, (3) the particles obey classical mechanics, and (4) the collisions between particles are elastic, conserving both kinetic energy and momentum.

How does the Maxwell-Boltzmann distribution change with different gases?

The Maxwell-Boltzmann distribution changes with different gases primarily due to differences in molecular mass. For a given temperature, lighter gas molecules will have a broader distribution of speeds, with a higher average speed, compared to heavier gas molecules. This is because the distribution depends on the mass of the particles; lighter particles move faster on average at the same temperature.

Can the Maxwell-Boltzmann distribution be applied to real gases?

While the Maxwell-Boltzmann distribution provides a good approximation for the behavior of ideal gases, real gases can deviate from this distribution due to factors such as intermolecular forces and quantum effects. At high pressures and low temperatures, real gases exhibit behaviors that differ from the idealized assumptions of the Maxwell-Boltzmann distribution. However, under conditions where the gas behaves approximately ideally (low pressure and high temperature), the Maxwell-Boltzmann distribution is a useful and accurate model.

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