# Ideal gas pressure from Maxwell-Boltzmann distribution

Hi everyone

I'm having trouble with solving an exercise in statistical physics. I need to argue why the average number of particles with a velocity between ##v## and ##v+dv## that hit a surface area ##A## on the container wall in a time interval ##\Delta t## is $$N_{collision}=v_{x}A\Delta t f(\mathrm{\textbf{v}})dv_{x}dv_{y}dv_{z}$$ where ##f(\mathrm{\textbf{v}})## is the Maxwell-Boltzmann distribution. Consider the gas as an ideal gas.

I don't quiet know where to start so...

Thanks for your help!

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## Answers and Replies

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stevendaryl
Staff Emeritus
Science Advisor
Hi everyone

I'm having trouble with solving an exercise in statistical physics. I need to argue why the average number of particles with a velocity between ##v## and ##v+dv## that hit a surface area ##A## on the container wall in a time interval ##\Delta t## is $$N_{collision}=v_{x}A\Delta t f(\mathrm{\textbf{v}})dv_{x}dv_{y}dv_{z}$$ where ##f(\mathrm{\textbf{v}})## is the Maxwell-Boltzmann distribution. Consider the gas as an ideal gas.

I don't quiet know where to start so...

Thanks for your help!
Are you sure that there isn't an additional factor of density (number of particles per unit volume) involved?

What I would say is this: You have a wall that is oriented perpendicular to the x-axis. So for particles traveling at velocity $v_x$ in the x-direction, consider all particles that will hit the wall in the next $\delta t$ seconds. Clearly, for a particle to hit the wall in that time interval, it must be closer than $v_x \delta t$ in the x-direction. So there is a certain region of space that contains all the particles that could possibly hit the wall in the next $\delta t$ seconds (traveling at velocity $v_x$ in the x-direction). What is the volume of that region? The number of particles in that region is proportional to that volume.

H Psi equal E Psi
Are you sure that there isn't an additional factor of density (number of particles per unit volume) involved?

What I would say is this: You have a wall that is oriented perpendicular to the x-axis. So for particles traveling at velocity $v_x$ in the x-direction, consider all particles that will hit the wall in the next $\delta t$ seconds. Clearly, for a particle to hit the wall in that time interval, it must be closer than $v_x \delta t$ in the x-direction. So there is a certain region of space that contains all the particles that could possibly hit the wall in the next $\delta t$ seconds (traveling at velocity $v_x$ in the x-direction). What is the volume of that region? The number of particles in that region is proportional to that volume.
Thank you very much for your answer!
I guess the number of particle traveling in x-direction would be: $$n_{x}=\int_{0}^{\infty} f(\mathrm{\textbf{v}}) dv_{x}$$ right? But how do I include the infinitesimal time Intervall ##\Delta t##?

stevendaryl
Staff Emeritus
Science Advisor
Thank you very much for your answer!
I guess the number of particle traveling in x-direction would be: $$n_{x}=\int_{0}^{\infty} f(\mathrm{\textbf{v}}) dv_{x}$$ right? But how do I include the infinitesimal time Intervall ##\Delta t##?
Can you answer my question: what is the volume of the region of points such that a particle at that point will hit the wall in the next $\Delta t$ seconds?

Can you answer my question: what is the volume of the region of points such that a particle at that point will hit the wall in the next $\Delta t$ seconds?
You said that:
The number of particles in that region is proportional to that volume.
So by finding the number of particle which will hit the wall in the next ##\Delta t## seconds I can find the volume? The number of particle which will hit the wall should be equal to: $$dn_{x}=\Delta tv_{x}f(\mathrm{\textbf{v}})dv_{x}dv_{y}dv_{z}$$ with $$\mathrm{\textbf{v}}= \begin{bmatrix} v_{x}\\ 0\\ 0 \end{bmatrix}$$
Now i need to link this with the volume right?
I'm not that good in statistical physics so what I just stated could be completely wrong...
Thanks for your help!