# # of gas particles hitting one face of a cubic container

• FranzDiCoccio
In summary, the conversation discusses estimating the number of collisions per second between atoms in a gas and one of the walls of a cubic container. Two different approaches are presented, with the book's formula using Avogadro's number, Boltzmann's constant, and the atomic mass, while the other approach considers the root mean square velocity and average velocity along one direction. There is also a discussion about the accuracy of the book's formula and a suggestion for an alternative formula based on the effusion rate for a collisionless gas.
FranzDiCoccio

## Homework Statement

[/B]
1 mol of gas at temperature $T$ is contained in a cubic container of side $L$.
Estimate the number of collisions per second between the atoms in the gas and one of the walls of the cubic container.

My book gives this formula for that quantity
$$\frac{N_A}{6L}\sqrt{\frac{3 k T}{m}}$$

Where $N_A$ is Avogadro's number, $k$ is the Boltzmann's constant and $m$ is the atomic mass.

I tried two different lines of reasoning, and came up with the same answer, which is different from that of the book
$$\frac{N_A}{2L}\sqrt{\frac{ k T}{m}}$$

## Homework Equations

a) Root mean square velocity

$$v_{rms}=\sqrt{\frac{3 k T}{m}}$$

b) average velocity along one direction:

$$\bar{v}_x=\frac{v_{rms}}{\sqrt{3}}=\sqrt{\frac{ k T}{m}}$$

## The Attempt at a Solution

[/B]
A) Via the momentum transferred to a wall in a time $\Delta t$.
The average number of atoms hitting a wall in a time $\Delta t$ is
$$\bar{N} = \frac{\bar{F} \Delta t}{2 m \bar{v}_x}=\frac{p L^2 \Delta t}{2 m \bar{v}_x}=\frac{p V \Delta t}{2 L m \bar{v}_x}=\frac{N_A k T \Delta t}{2 L m \bar{v}_x}=\Delta t \frac{N_A}{2L}\sqrt{\frac{kT}{m}}$$

B) Via the density of atoms.
The average velocity along direction x is $\bar{v}_x$. On average, half of the particles goes in the direction of the wall I'm interested in.
The average number of atoms hitting a wall in a time $\Delta t$ is half of the number of particles contained in a volume $\bar{V}=L^2 \Delta t \bar{v}_x$.
Thus
$$\bar{N} =\frac{1}{2} \bar{V} \frac{N_A}{V}=\frac{1}{2} L^2 \Delta t \bar{v}_x \frac{N_A}{L^3}= \frac{N_A}{2L}\sqrt{\frac{kT}{m}}$$

C) My book says "on average $N_A/3$ atoms hit the same face in the time $\Delta t$", and takes
$$\Delta t = \frac{2 L}{v_{rms}}$$.

The statement about 1/3 of the particles sounds wrong to me.
Can someone help with this? Shouldn't it take into account my formula b) above? Isn't the square velocity along one direction 1/3 of the square modulus of the velocity?

Thanks a lot
Franz

Last edited:
FranzDiCoccio said:

## Homework Statement

[/B]
1 mol of gas at temperature $T$ is contained in a cubic container of side $L$.
Estimate the number of collisions per second between the atoms in the gas and one of the walls of the cubic container.

My book gives this formula for that quantity
$$\frac{N_A}{6L}\sqrt{\frac{3 k T}{m}}$$

Where $N_A$ is Avogadro's number, $k$ is the Boltzmann's constant and $m$ is the atomic mass.

I tried two different lines of reasoning, and came up with the same answer, which is different from that of the book
$$\frac{N_A}{2L}\sqrt{\frac{ k T}{m}}$$

## Homework Equations

a) Root mean square velocity

$$v_{rms}=\sqrt{\frac{3 k T}{m}}$$

b) average velocity along one direction:

$$\bar{v}_x=\frac{v_{rms}}{\sqrt{3}}=\sqrt{\frac{ k T}{m}}$$

## The Attempt at a Solution

[/B]
A) Via the momentum transferred to a wall in a time $\Delta t$.
The average number of atoms hitting a wall in a time $\Delta t$ is
$$\bar{N} = \frac{\bar{F} \Delta t}{2 m \bar{v}_x}=\frac{p L^2 \Delta t}{2 m \bar{v}_x}=\frac{p V \Delta t}{2 L m \bar{v}_x}=\frac{N_A k T \Delta t}{2 L m \bar{v}_x}=\Delta t \frac{N_A}{2L}\sqrt{\frac{kT}{m}}$$

B) Via the density of atoms.
The average velocity along direction x is $\bar{v}_x$. On average, half of the particles goes in the direction of the wall I'm interested in.
The average number of atoms hitting a wall in a time $\Delta t$ is half of the number of particles contained in a volume $\bar{V}=L^2 \Delta t \bar{v}_x$.
Thus
$$\bar{N} =\frac{1}{2} \bar{V} \frac{N_A}{V}=\frac{1}{2} L^2 \Delta t \bar{v}_x \frac{N_A}{L^3}= \frac{N_A}{2L}\sqrt{\frac{kT}{m}}$$

C) My book says "on average $N_A/3$ atoms hit the same face in the time $\Delta t$", and takes
$$\Delta t = \frac{2 L}{v_{rms}}$$.

Can someone help with this? Shouldn't it take into account my formula b) above? Isn't the square velocity along one direction 1/3 of the square modulus of the velocity?

Thanks a lot
Franz
Both answers are reasonably good=I think yours is closer to being right, and I think the book's is somewhat inaccurate. Both are in the same ballpark. I think I can offer an improvement, but it is only exact for collisionless gases: The effusion rate out of a small aperture ##A ## is given by ## dN/dt=(n \, \overline{v}/4)A ## where ## n ## is the number of particles per unit volume. ## \overline{v}=((8/\pi)kT/m)^{1/2} ## (mean speed) for a Maxwellian distribution. This gives ## dN/dt=(N_A/4L)(8/ \pi)^{1/2} (kT/m)^{1/2} ##. (Note: I have previously derived the effusion rate formula ##( R=n \, \overline{v}/4) ##from first principles for a collisionless gas (by performing a double integral in real space as well as velocity space to find the number of particles that have the necessary velocity to reach the aperture in time ## \Delta t ##, etc.), and it is exact. A little arithmetic gives my answer is approximately ## dN/dt=(1/2.5)(kT/m)^{1/2} ##. The book's answer you quoted has ## \sqrt{3}/6=1/3.4 ## (approximately), compared to your ## (1/2) ## and my ## (1/2.5) ##.

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FranzDiCoccio
Hi Charles,

and thanks a lot for your insight.

So what you're saying is "since both answers rely on a large number of approximations, they are equally good".
I think I see what you mean.

But what I thought is that the average number of collisions with a wall should be connected with the pressure. Specifically the pressure is related to the net force (or rate of momentum transfer) normal to the wall. So I think one should use the velocity along one direction ##\bar{v}_x## and not ##v_{rms}##.
What I'm saying that, even at this approximate level, the book's solution does not seem to be consistent with the equation of state for an ideal gas...
There would be a coefficient ##\sqrt 3## that should not be there.
Right now I have no time to check that (school is starting in half an hour), I'll do that later.

As for your alternative, it is really interesting, and gives more insight, but it's definitely too difficult compared to the level of the book.

Thanks a lot again!
Franz

FranzDiCoccio said:
Hi Charles,

and thanks a lot for your insight.

So what you're saying is "since both answers rely on a large number of approximations, they are equally good".
I think I see what you mean.

But what I thought is that the average number of collisions with a wall should be connected with the pressure. Specifically the pressure is related to the net force (or rate of momentum transfer) normal to the wall. So I think one should use the velocity along one direction ##\bar{v}_x## and not ##v_{rms}##.
What I'm saying that, even at this approximate level, the book's solution does not seem to be consistent with the equation of state for an ideal gas...
There would be a coefficient ##\sqrt 3## that should not be there.
Right now I have no time to check that (school is starting in half an hour), I'll do that later.

As for your alternative, it is really interesting, and gives more insight, but it's definitely too difficult compared to the level of the book.

Thanks a lot again!
Franz
I didn't want to be too critical of the book, but I think a good textbook should provide the answer that I provided you. I think you could do an alternative derivation of the collision rate that gives the same answer I got by simply considering the x component of the velocity, but I'll need to give that one some thought. Meanwhile, the pressure is known to satisfy ## PV=NkT=(2/3)N(1/2)mv_{rms}^2 ##.This makes ##(1/2)mv_{x \, rms}^2=(1/2)kT ##.

Hi Charles,
your last sentence is what I'm referring to. I'm not sure the book's answer is consistent with that relation.

By the way, the book is for Italian high school students. That means 4th year, 17-18 year old people.
The book contains a standard derivation of kinetic theory, and what I'm discussing here is an exercise.
I found the derivation I'm referring to in the "cheatsheet" that the editor gives to the teachers.
It's just a few lines but, as I say, I'm not very convinced.

Thanks again, as soon as I have some time I'll also try to rethink the derivation to keep your suggestion into account.
Franz

FranzDiCoccio said:
Hi Charles,
your last sentence is what I'm referring to. I'm not sure the book's answer is consistent with that relation.

By the way, the book is for Italian high school students. That means 4th year, 17-18 year old people.
The book contains a standard derivation of kinetic theory, and what I'm discussing here is an exercise.
I found the derivation I'm referring to in the "cheatsheet" that the editor gives to the teachers.
It's just a few lines but, as I say, I'm not very convinced.

Thanks again, as soon as I have some time I'll also try to rethink the derivation to keep your suggestion into account.
Franz
Your ## v_x ## proof for the exact answer would proceed as follows: You have a velocity distribution ## g(v_x)=A(N_A/L^3)exp^{-v_x^2/(2kT/m)} ##. At a distance x, particles will make it to the wall (assumes collisionless) in time ## \Delta t ## if ## v_x>x/(\Delta t) ## . Thereby ## \Delta N ## in time ## \Delta t ## is ## \Delta N=L^2 \, \int\limits_{x=0}^{+\infty} \int\limits_{v_x=x/\Delta t}^{+\infty} g(v_x) \, dv_x \,dx ##. If you know how to evaluate the double integral, I think you should be able to get the same answer with this integral that I previously obtained. I'm also going to evaluate it myself and see what I get. editing.. ## A=\sqrt{m/(2 \pi kT)} ## and the order of integration can be changed on the double integral to get ## \ ## ## I=\int\limits_{v_x=0}^{+\infty} \int\limits_{x=0}^{v_x \Delta t} g(v_x) \, d x \, d v_x ##. The ## dx ## integral just gives ## v_x \Delta t ##. The integral ## \int\limits_{v_x=0}^{+\infty} v_x \, g(v_x) \, dv_x ## is straightforward. This results in ## \Delta N=(N_A/L) (\Delta t) \sqrt{kT/(2 \pi m)} ## ## \ ## which is in agreement with with my first result. Please let me know if you have further questions about evaluating this last integral.

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Ok, I see what you mean. Integrals are not a problem for me, but they are for the book's audience, for at least one more year or so.
Integrals are almost the last topic in 5th year mathematics, and even then most students struggle with the easy ones.
At this point the students in the Physics course have just met the concept of Maxwell distribution and its properties, but in general they are not given its explicit form, and anyway they won't make much of it, mathematically.

The exercise is meant to be solved without integrals. I do realize that, at that level, the approximation is so rough that one can't really say that one choice is better than another. Yet the book's derivation sounds wrong to me, even assuming that one really has a bunch of particles with velocity ##v_{rms}## and random directions.

Thanks again, this is useful.

## 1. How do gas particles hit one face of a cubic container?

Gas particles hit one face of a cubic container due to their random motion and collisions with the walls of the container. As gas particles move around, they collide with the walls and exert a force on them, resulting in a pressure on that particular face of the container.

## 2. What factors affect the number of gas particles hitting one face of a cubic container?

The number of gas particles hitting one face of a cubic container is affected by the temperature, volume, and pressure of the gas. An increase in any of these factors will result in more frequent collisions and therefore, a higher number of gas particles hitting the face of the container.

## 3. How is the number of gas particles hitting one face of a cubic container related to pressure?

The number of gas particles hitting one face of a cubic container is directly proportional to the pressure of the gas. As pressure increases, the frequency of collisions between gas particles and the walls of the container also increases, resulting in a higher number of gas particles hitting the face of the container.

## 4. Can the number of gas particles hitting one face of a cubic container be calculated?

Yes, the number of gas particles hitting one face of a cubic container can be calculated using the ideal gas law, which states that the pressure of a gas is directly proportional to the number of gas particles, temperature, and volume. By rearranging the formula, we can solve for the number of gas particles hitting one face of the container.

## 5. How does the shape of the container affect the number of gas particles hitting one face of a cubic container?

The shape of the container does not affect the number of gas particles hitting one face of a cubic container. As long as the other factors (temperature, pressure, and volume) are constant, the number of gas particles hitting the face of the container will remain the same regardless of the container's shape.

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