I am learning to derive the equation of pressure of the gas which is given at the bottom of following photo.

enter image description here

Just want the explanation of the statement "Because there are many molecules and because they are all moving in random direction, the average values of the squares of their velocity components are equal, so that $(v_x)^2=\frac{v^2}{3}$"

I think author wants to say,

$$\frac{(v_{1x})^2 + (v_{2x})^2 + ... + (v_{Nx})^2}{N}=\frac{(v_{1y})^2 + (v_{2y})^2 + ... + (v_{Ny})^2}{N}=\frac{(v_{1z})^2 + (v_{2z})^2 + ... + (v_{Nz})^2}{N}$$

Which implies:

$$(v_{1x})^2 + (v_{2x})^2 + ... + (v_{Nx})^2 = (v_{1y})^2 + (v_{2y})^2 + ... + (v_{Ny})^2 = (v_{1z})^2 + (v_{2z})^2 + ... + (v_{Nz})^2 ---(A)$$

The problem is I can't understand how the statement is logical. How is it possible to assume and say something which is shown by equation A. I don't understand why the sums of the squares of the velocity components of all the molecules would be equal to each other. Or whether I have misunderstood the author.

  • $\begingroup$ Consider the statement "they are all moving in random direction". What does that mean about the expectation values of the different $v^2$'s? Why would one larger than the others? In a system that is the same in all directions, how could $\langle v_x^2 \rangle$ differ from $\langle v_y^2 \rangle$? $\endgroup$ – ZachMcDargh Nov 3 '16 at 16:33
  • $\begingroup$ @ZachMcDargh You all are using statments bounded by angular brackets. What do those stand for? $\endgroup$ – user104909 Nov 3 '16 at 17:45

You're making a simple matter complex. It's simply stated that average of each velocity component of the gas molecules are same with no preference to one component over the other. If you turn the gas container upside down all velocity components are on average same in equilibrium. I.E. $$\langle v_x\rangle =\langle v_y\rangle =\langle v_z\rangle $$ so $\langle v^2_x\rangle $ can be taken as 1/3rd of the total velocity squared.

  • $\begingroup$ Actually I am trying to replace $\bar{(v_x)^2}$ in $p=\frac{nM(v_x)^2}{V}$ with $\frac{\bar{v^2}}{3}$ because this is what author has done apparently. So I want to know if It would be correct to use here the fact that pressure is equal on each of walls. Because this would imply, $$\frac{nM\bar{(v_x)^2}}{V}=\frac{nM\bar{(v_y)^2}}{V}=\frac{nM\bar{(v_z)^2}}{V}$$ which further implies $$\bar{(v_x)^2}=\bar{(v_y)^2}=\bar{(v_z)^2}$$. $$v^2=(v_x)^2 +(v_y)^2 + (v_z)^2 \implies \bar{v^2}=\bar{(v_x)^2}+\bar{(v_y)^2}+ \bar{(v_z)^2} \implies \bar{v^2}=3 \bar{(v_x)^2}$$. $\endgroup$ – user104909 Nov 3 '16 at 16:38
  • $\begingroup$ In the first line, it is $\frac{nM \bar{v_x}}{V}$. I placed everywhere bar because I don't know how to make symbol of average in latex. $\endgroup$ – user104909 Nov 3 '16 at 16:56
  • $\begingroup$ Is it correct to use this fact in the derivation like I did in comment above? $\endgroup$ – user104909 Nov 3 '16 at 17:27
  • $\begingroup$ You're confusing average with root mean square. In your last equation to the first comment you write $\bar{v^2}=3 \bar{(v_x)^2}$ but it's actually $$\bar{v^2}=\sqrt{3} \bar{(v_x)^2}$$ $\endgroup$ – Weezy Nov 3 '16 at 18:56
  • $\begingroup$ When you take the square root you have to apply it to the "1/3" factor as well $\endgroup$ – Weezy Nov 3 '16 at 18:57

The author is saying that, $v^2 = v_x^2 + v_y^2 +v_z^2$, and because the velocities are random, then $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle$, and thus, $$v^2 = 3 \langle v_x^2 \rangle,$$ where the angled brackets denote averages.

The averages can be interpreted both as ensemble (over the different particles) and over time, i.e. both that $\langle v^2_{i,x}(t_1) \rangle = \langle v^2_{i,x}(t_2) \rangle$, and that $\langle v^2_{i,x}\rangle = \langle v^2_{j,x} \rangle = \langle v^2_{j,y} \rangle$. This follows from the premistes of equipartition (in dimensions and between particles) and an equilibrium configuration.

  • $\begingroup$ He has stated, "the average values of squares of ... " How do you interpret "Average" here? $\endgroup$ – user104909 Nov 3 '16 at 14:14
  • $\begingroup$ @SufyanNaeem Ah, gotcha, I edited my answer. Let me know if that is still unclear. $\endgroup$ – DilithiumMatrix Nov 3 '16 at 15:23
  • $\begingroup$ What do that left angle and right angle mean you used? Do they stand for "average"? $\endgroup$ – user104909 Nov 3 '16 at 17:38
  • $\begingroup$ Do you want to say that author is telling us that we can suppose that each molecule has a velocity v whose all three components are equal in magnitude. Is not it? $\endgroup$ – user104909 Nov 3 '16 at 17:58
  • $\begingroup$ @SufyanNaeem yes, the angle brackets are common symbols for averages --- sorry, I shouldn't have taken that for granted! And you're exactly right. There is no preferred direction, so on average, each component should have the same magnitude, for all particles! $\endgroup$ – DilithiumMatrix Nov 3 '16 at 19:11

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