# Year 12: Cambridge Physics Problem (Distance of O atom in atmosphere)

1. Jun 29, 2012

### johnconnor

At an altitude of about 100km above the Earth's surface, the density and temperature of the atmosphere are about 10^-14 kg/m^3 and 2000K, respectively. At this altitude, the major constituent of the atmosphere is atomic oxygen. Use this information to estimate the pressure of the atmosphere at an altitude of 1000km.

On average, how far apart are the oxygen atoms? What is their root mean square speed? Discuss whether the temperature of a gas at this pressure can be interpreted in the same way as the temperature of a gas at normal atmospheric pressure.

Attempt:

The first thing that crossed my mind is whether the pressure of the atmosphere oxygen changes linearly with the height of the atmosphere. But how do I know whether it's a linear variation or not?

What other factors should I consider when attempting the question? P=ρhg doesn't really help, I think.

Thank you!

2. Jul 1, 2012

### Infinitum

I don't think it is linear. The pressure depends on the temperature, and density of oxygen molecules, and they are both not related linearly to the height from sea level. Aren't you given the temperature at 1000km height?

3. Jul 1, 2012

### Staff: Mentor

I think the 1000 km must be a typo. I think it was meant to read 100 km. The problem asks you to calculate the pressure at the altitude, which you can get from the ideal gas law. If you want to get the parameters at 1000 km, you need to know the temperature profile, and you also need to take into account the changes in g with radial distance. I don't think this is what was intended by the question.

4. Jul 2, 2012

### johnconnor

You're right. The guide didn't mention anything about a variation with height at ten times the height too. Thanks. Will solve it now.

p = (density)(rms square speed)/3 = (density)(temperature)/3
(using the proportion that rms speed is proportional to temperature)
p = 6.67E-12

But the guide given squares the temperature, giving the pressure as 1.33E-8. Why so?

Last edited: Jul 2, 2012
5. Jul 2, 2012

### johnconnor

You're welcome! I'd appreciate yr participating in the discussions to solve the questions as a sign of thanks!

Last edited by a moderator: May 6, 2017
6. Jul 2, 2012

### johnconnor

As for the final part, the guide says
What other comments have you got? I can't think of any... :P

7. Jul 3, 2012

### Infinitum

Isn't the given explanation clear? Since the pressure and molecule density is very low, there are lesser molecules that collide with the thermometer, and this makes attaining the same temperature as that of the atmosphere, take a lot of time. It implies you would need to keep the thermometer in there for a longer time than you normally keep in the 'near-earth' atmosphere to measure temperature.

As for your work, I think you should instead consider the ideal gas equation over the mean square speeds. <C> is proportional to T, but it is unknown what the proportionality constant is...Rearranging the ideal gas equation will give you,

$$PV = \frac{m}{M}RT \\ PM = \rho RT$$

Where M is the molecular mass of oxygen.