Ideal gas pressure from Maxwell-Boltzmann distribution

[H Psi equal E Psi]

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!

 

[stevendaryl]

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]

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?

 

[H Psi equal E Psi]

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!