# Calculating the root mean square speed from pressure and density.

1. Jan 7, 2014

### Jamesey162

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

These are the equations I believe to be relevant:

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

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

$$p = \frac{1}{3}ρ<c^2>$$

3. 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?

Last edited: Jan 7, 2014
2. Jan 7, 2014

### SteamKing

Staff Emeritus
If c^2 = 225000, then the units can't be m/s

3. Jan 8, 2014

### Jamesey162

Sorry yes <c^2> should be in m^2s^-2 shouldn't it?

4. Jan 8, 2014

### BruceW

Are you sure there should be an $N$ here?

5. Jan 8, 2014

### Jamesey162

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

6. Jan 8, 2014

### ehild

How does your book define <c^2>?

ehild

7. Jan 8, 2014

### Jamesey162

It defines it as:

$$<c^2> = \frac{{c_1}^2 + {c_2}^2 + {c_3}^2 + ....{c_N}^2}{N} = \frac{{c_i}^2}{N}$$

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

8. Jan 8, 2014

### ehild

It should be

$$<c^2> = \frac{{c_1}^2 + {c_2}^2 + {c_3}^2 + ....{c_N}^2}{N} =\frac{\sum{{c_i}^2}}{N}$$

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

ehild

9. Jan 8, 2014

### BruceW

yeah, I think what ehild said is right. The equation
$$c_{rms} = \frac{\sqrt{<c^2>}}{N}$$
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.

10. Jan 8, 2014

### Jamesey162

Okay, I've got it now, thanks for the explanation.