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PHY2 Thermal Physics help!

I am finding it difficult to understand the kinetic theory of gases. At first I understand the assumptions made by the theory: i.e. the particles don't exert forces upon each other, and the collisions are elastic and the volume ocupied is negligible.
Then it somehow comes to the fac that the pressure of an ideal gas is proportional to the mean of the squares of the particle speeds:
pV=1/3Nm<c^2>

I get that the come to the point that the pressure is equal to the p=1/3p<c^2>

p=density of gas.

Then they also lead to the fact that:

T is directly proportional to ½ m<c^2>

Now I get confused when they try the Boltzmann constant.
They make pV=nRT and pv=pV=1/3Nm<c^2> equal and come to ½ m<c^2>=3/2RT/N

PLS EXPLAIN
Reply 1
You've got the second bit right (almost). You just need to note that R/N = k (where N is avagadro's number). so pV = nRT = nNkT (nN is the number of gas molecules). and p = 1/3nm<c^2> = nkT (n is now the number density). so you get what you wrote in the end.

The first part is more interesting in my opinion, i will outline a derivation here but it's best if you google it since it will have pictures.

Start with a particle hitting a wall at velocity v_x. The change in momentum of the particle Delta_p = 2mv_x
consider a volume element at the wall with length v_x.Delta_t. The number of particles in this element is n.v_x.Delta_t (n is the number density). The fraction of these particles with velocity in range v_x -> v_x+dv_x (i.e. the fraction in this volume that will have enough velcoity to hit the wall) is given by the velocity distribution: f(v_x)dv_x

so the number of particles hitting the wall with velocity v_x in time Delta_t is:

n.v_x.Delta_t.f(v_x)dv_x

and the total change in momentum: Delta_p = 2mv_x.n.v_x.Delta_t.f(v_x)dv_x

The force is: Delta_p/Delta_t and we have already assumed the volume element to have unit cross sectional area, so we have the fractional pressure:

dp = 2mn.v_x^2.f(v_x)dv_x

integrate over this and by definition of <v_x^2> we have:

p = mn<v_x^2>

we have lost the 2 because the <v_x^2> is an integral over all velocity (both positive and negative) but we must only consider positive velocities (those going towards the wall). So you just divide by two to remove the half travelling away from the wall.

And the final thing is just to note that: <v_x^2> = <v_y^2> = <v_z^2> with <c^2> = <v_x^2> + <v_y^2> + <v_z^2>

so: p = 1/3 nm <c^2>

it's tough but I hope that gives you a flavour
Reply 2
Looking at the above confused me quite a lot, but then when i rewrote it, it made sense. Thanks a lot, for the explanation now can you please guide me further on how they come to the Boltzmann constant. I don't understand that too well.
Reply 3
you've pretty much got it there. The boltzmann constant is just k = R/N (N is avagadro's number).

p = (1/3)nm<c^2>

pV = hRT (where h is the number of moles - i've run out of n's)
pV = f/N * RT (where f is the number of molecules and N is avagradro's number)
pV = fkT (since R/N = k)
p = nkT (n is the number density of molecules = f/V)

so (1/3)nm<c^2> = nkT
but we want the energy of a single molecule = (1/2)m<c^2> = 3/2 kT

get it?
Reply 4
Okkk, thanks a lot.
Reply 5
is this A level physics? =.=;
Reply 6
This is AS physics, on PHY2 section of Thermal Physics. :smile:

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