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In Principles of Physics by Resnick,Halliday,Walker, when evaluating the $v_{rms}$ ,they first found $p = \frac{nM(v_x)^2}{V}$ & then substiuted $(v_{avg})^2 = \frac{1}{3} .v^2$ where $v^2 = (v_x)^2 + (v_y)^2 + (v_z)^2 $ saying that there are many molecules and all are moving in random directions hence average values of the square of the velocity components are equal and wrote $v^2= \frac{1}{3} (v_x)^2$. But i couldn't understand their reasoning. How have they came to this conclusion that the square of the components of the velocity are equal? Can anyone explain me their reasoning why they have taken the square of the components of velocity be equal? Help.

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  • $\begingroup$ We assume non-relativistic limit and dynamical equilibrium, which lead us to the statement that speed distribution of gas is isotropic. $\endgroup$ – Andrew McAddams Aug 23 '14 at 13:19
  • $\begingroup$ The velocity components of individual particles at any given time are generally not equal. An appeal to an average (either across many particles or in time) is essential to the argument. Notice the use of angle brackets ($\langle \rangle$) in some of the answers---they denote averages. $\endgroup$ – dmckee --- ex-moderator kitten Aug 23 '14 at 23:43
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The main assumption used here is that the direction of the velocity is distributed uniformly. This means that if I take a particle at random, the probability of its velocity pointing in any particular direction is the same for every direction; that is, no direction is more likely than the rest.

Now suppose I take a whole bunch of particles and measure $\langle v_x^2\rangle$, $\langle v_y^2\rangle$, and $\langle v_z^2\rangle$. Ignore the $z$ component for a moment. If it were to happen that $\langle v_x^2\rangle$ > $\langle v_y^2\rangle$, it would mean that on average, the $x$ component of the velocities is greater (in absolute value) than the $y$ component. But this would mean that there are more velocities pointing in the general area of the $x$ direction (both towards $+x$ and $-x$, because we are taking squares) than there are pointing in the general area of the $y$ direction. But this is unacceptable, because we said that we want the same amount of velocities pointing in any direction. The same argument shows that we cannot have $\langle v_x^2\rangle < \langle v_y^2\rangle$, so $\langle v_x^2\rangle = \langle v_y^2\rangle$ is the only option. The same holds for $\langle v_z^2\rangle$.

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The key step in this derivation is the assumption of isotropy of particle velocities. Think of the velocity vector of each particle as a randomly selected direction in space, in combination with a randomly selected speed from a specified distribution of speeds. The result of such a random sampling is that on average $\langle v_x\rangle=\langle v_y\rangle=\langle v_z\rangle=0$ and $\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle=\frac13 \langle v^2\rangle$.

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In essence, there is no reason why the gas should have more speed in one direction than in the other. So, we can say that the speed in each component is the same.

So, $v(x)=v(y)=v(z).$ Using the Pythagorean Theorem, $v^2(xy)=v^2(x)+v^2(y),$ where $v(xy)$ is the speed in the $xy$ plane. Using the Pythagorean theorem again, $$v^2(xyz)=v^2(xy)+v^2(z)=v^2(x)+v^2(y)+v^2(z)$$ where $v(xyz)$ is your regular three dimensional speed, denoted in your book simply as $v.$ Since $$v(x)=v(y)=v(z),$$ $$v^2(xyz)=3\cdot v^2(x)$$

In reality, I gravity on earth will play a role. But since we are dealing with an idealisation, the above is valid enough.

So the takeaway lesson is: there is no reason why the speed should be more in a particular direction. In an ideal situation, if you claim one direction has more speed, I can use the same logic to claim that another direction has the same speed.

I hope that helped

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  • $\begingroup$ Can you plz use latex to write the equation? $\endgroup$ – user36790 Aug 23 '14 at 14:00
  • $\begingroup$ Can you tell me why they are taken equal? What is the cause? $\endgroup$ – user36790 Aug 23 '14 at 16:41
  • $\begingroup$ See, there is no reason why the gas should have a preferred direction. Since we're dealing with ideal gases, we make an idealisation: the gas prefers all directions equally. This is very close to what we see on Earth, and therefore a very good approximation. If the gas did have higher speed in one direction, you would expect more pressure in that direction. In most situations, the pressure is the same, so we can use the ideal gas approximation. If we find that there is different pressure, we'll have to start from scratch for that particular condition. $\endgroup$ – user46268 Aug 24 '14 at 4:05
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It is just splitting of vectors component wise: $$v_{i}= v_{x}^{2}+v_{y}^{2}+v_{z}^{2} \,.$$ Like when you split the velocity of a projectile in to two components. We know from the molecular model of ideal gas: $$ v_{x}=v_{y}=v_{z} \,,$$ squaring all sides we get $$ v_{x}^{2}=v_{y}^{2}=v_{z}^{2} \,.$$ Thus $$ v_{i}= v_{x}^{2}+v_{x}^{2}+v_{x}^{2}=3v_{x}^{2} \,.$$

I am guessing you meant $ v_{i}= 3v_{x}^{2}$, instead of $ v_{i}= \frac{1}{3}v_{x}^{2}$

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  • $\begingroup$ One of the assumptions of the molecular model is that the gases move at random, in that they move in all direction at the same rate. $\endgroup$ – Derg Aug 23 '14 at 18:54
  • $\begingroup$ :plz can you tell me about molecular model of ideal gas? $\endgroup$ – user36790 Aug 23 '14 at 18:58
  • $\begingroup$ Oh,this is an assumption? Or actually it happens? $\endgroup$ – user36790 Aug 23 '14 at 19:03
  • $\begingroup$ my book says they are taken equal due to random motion in all directions. Can you plz justify? $\endgroup$ – user36790 Aug 23 '14 at 19:05
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    $\begingroup$ @user36790: it's an assumption, but if it were false, the gas would have a preferred direction, and you'd expect bulk motion, or different pressures depending on the orientation of matter, or something like that. As is, this assumption, when fed into equations, gives a very, very close approximation to what you get when you measure real gasses. $\endgroup$ – Jerry Schirmer Aug 23 '14 at 20:29

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