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karnten07
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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) [tex]\Delta[/tex]p per particle = -2mv
m=mass of particle and v =velocity
In time [tex]\Delta[/tex]t particles within a distance v[tex]\Delta[/tex]t can hit the wall but only half are traveling towards one wall at any instant.
So the number of collisions = (nV[tex]\Delta[/tex]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[tex]\Delta[/tex]t)/2 x (2mv)/[tex]\Delta[/tex]t = nmv[tex]^{}2[/tex]
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
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