Calculating the root mean square speed from pressure and density.

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Homework Help Overview

The discussion revolves around calculating the root mean square speed of gas molecules in a tire, given the pressure and density of the gas. The relevant equations and concepts from kinetic theory are being examined.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between pressure, density, and the root mean square speed, questioning the necessity of the variable N in the equation. There is also confusion about the definition of and its implications for calculating the rms speed.

Discussion Status

Some participants have offered clarifications regarding the definition of and its calculation, suggesting that the original equation involving N may be incorrect. The discussion is exploring different interpretations of the equations and definitions provided in the textbook.

Contextual Notes

There is uncertainty regarding the definitions and assumptions in the textbook, particularly concerning the variable N and the calculation of . Participants are also noting discrepancies between their calculations and the answers provided in the textbook.

Jamesey162
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Homework Statement



A tyre contains gas at a pressure of 150 kPa. If the gas has a density of 2.0 kg m-3, find the root mean square speed of the molecules.

Homework Equations



These are the equations I believe to be relevant:

c_{rms} = \frac{\sqrt{<c^2>}}{N}

pV = \frac{1}{3}Nm<c^2>

p = \frac{1}{3}ρ<c^2>

The Attempt at a Solution



\frac{3p}{ρ} = <c^2>
\frac{3 * 150000}{2} = <c^2> = 225000ms^{-1}

But I'm not sure how to work out N as I don't any volume or temperature. I'm not quite sure how they get their answer in the back of the book which is 474 ms-1 (which is the square root of 225000) meaning that N = 1? How can that be?

Thank you for reading!
 
Last edited:
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If c^2 = 225000, then the units can't be m/s
 
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SteamKing said:
If c^2 = 225000, then the units can't be m/s

Sorry yes <c^2> should be in m^2s^-2 shouldn't it?
 
Jamesey162 said:
c_{rms} = \frac{\sqrt{&lt;c^2&gt;}}{N}
Are you sure there should be an ##N## here?
 
BruceW said:
Are you sure there should be an ##N## here?

I'm sure, atleast that's how they have quoted it in my textbook:

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How does your book define <c^2>?

ehild
 
ehild said:
How does your book define <c^2>?

ehild

It defines it as:

&lt;c^2&gt; = \frac{{c_1}^2 + {c_2}^2 + {c_3}^2 + ...{c_N}^2}{N} = \frac{{c_i}^2}{N}

I'm not exactly sure what the right-hand most part means with the subscript i.
 
Jamesey162 said:
It defines it as:

&lt;c^2&gt; = \frac{{c_1}^2 + {c_2}^2 + {c_3}^2 + ...{c_N}^2}{N} = \frac{{c_i}^2}{N}

I'm not exactly sure what the right-hand most part means with the subscript i.

It should be

&lt;c^2&gt; = \frac{{c_1}^2 + {c_2}^2 + {c_3}^2 + ...{c_N}^2}{N} =\frac{\sum{{c_i}^2}}{N}

i is the summation index, and √<c2> itself is the rms speed. No need to divide it by N. \frac{3p}{ρ} = &lt;c^2&gt;

ehild
 
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yeah, I think what ehild said is right. The equation
c_{rms} = \frac{\sqrt{&lt;c^2&gt;}}{N}
is most likely a mistake in the book. It doesn't make sense, if you think about it. If you had a bunch of atoms all moving at the same speed, then this equation would give an rms value that is smaller for a larger population of these identical atoms. Which doesn't make sense at all.
 
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ehild said:
It should be

&lt;c^2&gt; = \frac{{c_1}^2 + {c_2}^2 + {c_3}^2 + ...{c_N}^2}{N} =\frac{\sum{{c_i}^2}}{N}

i is the summation index, and √<c2> itself is the rms speed. No need to divide it by N. \frac{3p}{ρ} = &lt;c^2&gt;

ehild

BruceW said:
yeah, I think what ehild said is right. The equation
c_{rms} = \frac{\sqrt{&lt;c^2&gt;}}{N}
is most likely a mistake in the book. It doesn't make sense, if you think about it. If you had a bunch of atoms all moving at the same speed, then this equation would give an rms value that is smaller for a larger population of these identical atoms. Which doesn't make sense at all.
Okay, I've got it now, thanks for the explanation.
 

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