Derive PV=nRT: Exploring the Ideal Gas Equation

[kidsasd987]

http://quantumfreak.com/derivation-of-pvnrt-the-equation-of-ideal-gas/



here is the link.


so we assume the particle hits two surfaces of the cube, thus pressure is 1/3.




combine the equation #11 and #12 we solve kinetic energy equation #12 for mv2.

13. mv^2=2E_{kinetic}\Rightarrow\frac{mv^2}{3}=\frac{2E_{kinetic}}{3}



My question is, I don't understand where 1/3 comes from. is this because the pressure is 1/3?


Thanks for all your help:)

 

[mfb]

Well at that point it is just the same equation written in a more complicated way (both sides get divided by 3). The definition of k is arbitrary here, it is chosen in such a way that the final equation does not have a numerical prefactor, and this requires the 1/3 in the intermediate steps.

 

[256bits]

Equation 8
N is the particles moving only in the x-direction.
It should have been written as Nx, to avoid confusion
and we take the velocity only in the x-direction Vx
We also have Ny = Nz = Nx
Similarily Vy = Vz = Vx


Equation 11
N is all the particles
( = Nx + Ny + Nz, see the previous logic in the statement before equation 7 )
The velocity is that of the particle along its path, and not parallel to any particular coordinate axis.
Here though, it is the velocity of the particle that is split into components Vx=Vy=Vz, rather than thinking about the number of particles that are moving in the x, y or z directions.

 

[nasu]

http://quantumfreak.com/derivation-of-pvnrt-the-equation-of-ideal-gas/

My question is, I don't understand where 1/3 comes from. is this because the pressure is 1/3?
Thanks for all your help:)

The 1/3 comes from expressing the average of the x component (squared) in terms of the total velocity squared.
You have
$$v^2=v_x^2+v_y^2+v_z^2$$
When we take the average
$$<v^2>=<v_x^2>+<v_y^2>+<v_z^2>$$
(I use brackets for "average")
For an isotropic gas, the average is the same for each component (the three terms on the right are equal). So any of the averages on the right hand side is 1/3 of the average of v^2.

 

[Delta2]

The 1/3 comes from expressing the average of the x component (squared) in terms of the total velocity squared.
You have
$$v^2=v_x^2+v_y^2+v_z^2$$
When we take the average
$$<v^2>=<v_x^2>+<v_y^2>+<v_z^2>$$
(I use brackets for "average")
For an isotropic gas, the average is the same for each component (the three terms on the right are equal). So any of the averages on the right hand side is 1/3 of the average of v^2.

Can u give a good explanation why for an isotropic gas those 3 averages are equal?

 

[mfb]

Symmetry. There is no reason why one coordinate should look different from the others (and the definition of those coordinates is arbitrary anyway).

 

[DrStupid]

Can u give a good explanation why for an isotropic gas those 3 averages are equal?

Because it is isotropic?

 

[dauto]

Can u give a good explanation why for an isotropic gas those 3 averages are equal?

That's what isotropic mean. Any quantity measured along any direction - say the x-axis, will give the same result as any other direction - say the y-axis or the z-axis.