Microscopic model of a physical system, ideal gas

[karnten07]

Homework Statement

A simple model of a gas in one-dimensional consists of N classical particles, each of mass m bouncing between the reflecting walls which are separated by a distance L.

i) Calculate the average pressure on a wall in terms of the mean kinetic energy per particle.
ii) L is varied by moving one wall at a slow constant speed. Calculate the change in the mean kinetic energy and compare it with the work done against the force on the wall due to the gas pressure.

[Note: usually, (in a three dimensional system) pressure is 'force per unit area.' In our one dimensional example we can equate pressure with the force on the wall.]

Homework Equations

The Attempt at a Solution

i) $$\Delta$$p per particle = -2mv
m=mass of particle and v =velocity

In time $$\Delta$$t particles within a distance v$$\Delta$$t can hit the wall but only half are traveling towards one wall at any instant.

So the number of collisions = (nV$$\Delta$$t)/2 where n = number of particles per unit area (not volume as this is a one dimensional example).

Force exerted on the wall is the rate of momentum transfer:

P=(nv$$\Delta$$t)/2 x (2mv)/$$\Delta$$t = nmv$$^{}2$$

This must be averaged because the particles have a spread of values of v. So we put a bar across the top of the v to denote the average value of v.

Total kinetic energy in the system , U = (nAmv^2)/2 where A is the area between the two walls.

This gives us, nmv^2 = 2U/A

So substituting into the equation of pressure, P = 2U/A which can be thought of mean kinetic energy of the system per unit area.

ii) How does the kinetic energy of the system vary when changing the area that the particles are occupied in? I assume when decreasing this area, more energy is imparted to the walls at a faster rate, but what equation can i use to show this?
I think i need to treat this as an isothermal system and i should find that the change in U=0 i have equation for the work done for a compression and expansion.

Am i on the right track here guys?, thanks

 

[anathema]

for any help.

Dear fellow scientist,

Thank you for your post and for sharing your thoughts on this interesting problem.

Firstly, your calculation for the average pressure on the wall in terms of mean kinetic energy per particle looks correct to me. It is a good starting point for understanding the behavior of this one-dimensional gas system.

For the second part of the problem, I agree with your approach of treating the system as isothermal and using the equation for work done during compression and expansion. In this case, we can use the equation W = -PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume.

Since we are only varying the length of the system, we can rewrite this equation as W = -FΔL, where F is the force on the wall due to the gas pressure and ΔL is the change in length. This is because the change in volume is equal to the change in length in this one-dimensional system.

Now, we can compare this work done with the change in mean kinetic energy of the system. As you mentioned, when the length is decreased, the particles will collide with the walls more frequently and with higher velocities, resulting in a higher pressure and thus a higher force on the wall. This means that more work is done against the force on the wall, resulting in a decrease in mean kinetic energy of the system. On the other hand, when the length is increased, the particles will collide less frequently and with lower velocities, resulting in a lower force on the wall and less work being done against it. This will result in an increase in mean kinetic energy of the system.

I hope this helps and provides some insight into the problem. Let me know if you have any further questions or if you would like to discuss this further.

 

[Moetasim]

for your help

Your approach seems to be on the right track. Here are some additional thoughts and suggestions:

i) Your calculation for the average pressure is correct. It is important to note that in this one-dimensional example, the pressure is the same on both walls since the particles are bouncing back and forth between them.

ii) You are correct in thinking that this can be treated as an isothermal system. In this case, the change in total kinetic energy will be equal to the work done on the system. The work done against the force on the wall due to gas pressure can be calculated using the formula W = PΔV, where P is the pressure and ΔV is the change in volume. In this case, ΔV is equal to the change in the area between the walls, since the system is one-dimensional. So, the change in kinetic energy will be equal to the work done, which means that the kinetic energy will increase as the walls move closer together and decrease as they move further apart.

You can also use the ideal gas law, PV = NkT, to relate the pressure, volume, and temperature of the system. Since the temperature is constant, this equation can be rearranged to solve for the change in pressure as the volume changes. This will give you another way to calculate the change in pressure and compare it to the change in kinetic energy.

Overall, your approach and thinking are correct. Keep up the good work!