Standard deviation of V_x of He gas

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

$\frac 1 2 m<v_x^2> =\frac 1 2 k_BT , \\ \sqrt{ <v_x^2>} = 556~ m/s$ So, I guess that the standard deviation should be less than rms speed.

So, the option is (a)

$\left< ax + b \right> = a\left<x\right> + b$

So, $\left< v_x^2 - <v_x^2> \right> = \left< v_x^2\right> - \left< v_x^2\right> = 0$

Is this correct?
[/B]

 

[Charles Link]

I think it your last statement, you are attempting to write $\sigma^2=E(X^2)-(EX)^2$. If that is the case, and you are saying $EX=0$ , yes, it is correct. Please put in the mass and temperature so that we can check your arithmetic. For the final answer, the standard deviation, $\sigma_{v_x}=\sqrt{E(v_x)^2}=v_{x \, rms}$. Also, the statement of the problem is incomplete. The available answers did not show up.

 

[Pushoam]

The question is not showing in the OP. So, posting it again.

I think it your last statement, you are attempting to write $\sigma^2=E(X^2)-(EX)^2$. I

I didn't understand what you mean by the above line. Please explain the symbol EX and so on.

 

[Charles Link]

The question is not showing in the OP. So, posting it again.
View attachment 217447

I didn't understand what you mean by the above line. Please explain the symbol EX and so on.

In probability theory, the $E$ means expectancy. $EX$ is the same thing as $<X>$.

 

[Pushoam]

So, $\left< v_x^2 - <v_x^2> \right> = \left< v_x^2\right> - \left< v_x^2\right> = 0$

In the above statement, as $< v_x ^2>$ is constant, I am taking it out of $\left< v_x^2 - <v_x^2> \right>$ and hence I got $\left< v_x^2 - <v_x^2> \right> = \left< v_x^2\right> - \left< v_x^2\right> = 0$.
But, this way I will always get 0 standard deviation. So, something is wrong with it. Right?

 

[Charles Link]

In the above statement, as $< v_x ^2>$ is constant, I am taking it out of $\left< v_x^2 - <v_x^2> \right>$ and hence I got $\left< v_x^2 - <v_x^2> \right> = \left< v_x^2\right> - \left< v_x^2\right> = 0$.
But, this way I will always get 0 standard deviation. So, something is wrong with it. Right?

The equation correctly reads $\sigma^2_{v_x}=<v_x^2>-<v_x>^2$. $\\$ Editing: This comes from $\sigma^2_{v_x}=<(v_x-<v_x>)^2>$. And I think your arithmetic in the OP is incorrect.

 

[Pushoam]

$\sqrt{ <v_x^2>} = \sqrt{ \frac { k_BT = 4.14 * 10^{-21}}{m=4* m_p = 6.69 * 10^{-27}}}= 786~ m/s$

Is this correct?

 

[Charles Link]

Yes. Very good. Round that off and (d) becomes the correct answer. See also my edited additions to post 6.

 

[Pushoam]

See also my edited additions to post 6.

I reloaded the page three times. Still, no edition to post #6 is shown.
What I get in post # 6 is:

The equation correctly reads $\sigma^2_{v_x}=<v_x^2>-<v_x>^2$. This comes from $\sigma^2_{v_x}=<(v_x-<v_x>)^2>$. And I think your arithmetic in the OP is incorrect.

 

[Charles Link]

I reloaded the page three times. Still, no edition to post #6 is shown.
What I get in post # 6 is:

I didn't use the word "edit". Let me put it in there for clarity. After I wrote the original post, I came back a couple of minutes later and added the last two sentences.

 

[Pushoam]

Thanks.

 

[Pushoam]

Now I understood.

$\sigma_{v_x} = \sqrt{ \left< (v_x - \left <v_x \right >)^2 \right> = \left < v_x^2 \right > – 2 \left < v_x \right >^2 + \left < v_x \right >^2 = \left < v_x^2 \right > – \left < v_x \right >^2 }$

Since , $\left < v_x \right > = 0$

$\sigma_{v_x} = v_{rms}$

 

[Charles Link]

Very good. :) (You need a parenthesis { } around v_x in the subscript of $\sigma_{v_x}$. Then your Latex will work.)