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Pressure, Volume, Temperature and Collisions

Pressure = P1
Volume = V1
Temperature = T1

Pressure remian the same
We increase the Volume to V2, so V2>V1
So Temperature must increase to T2 > T1 (to keep pressure the same)

But what does this tell us about the number of collisions of the molecules - are they the :

1. same
2. more
3. less
4. can't tell

Reply 1

SDA

I am assuming we are interested in the number of collisions in a given time, so my advice is to think about the effect the changes you described have on the distance between molecules, the speed with which they are moving and how these affect the average time between collisions and see whether the time gets longer or shorter.

Reply 2

Charries

Original post by SDA

Well I don't know because the the space is bigger, so the molecules are further apart, but the temerature rises, so they are moving faster - hence my question. But since the pressure is the same my inclination is to say no change in the number of collisions - but to keep the pressure the same with the bigger surface area of the container surely there should be more collisions with the container walls?

(edited 11 years ago)

Reply 3

SDA

Ok so my thinking is this:

From the rms speed equation we know:

v \propto \sqrt{T}

From the gas law we know:

V \propto T

and so

v \propto \sqrt{V}

From the equation for mean free path we see

d \propto V

So using

t = d/v

we see

t \propto V/V^{\frac{1}{2}} = V^{\frac{1}{2}}

So what this tells us is that as the volume increases the time between collisions also increases and so the collision rate should decrease.

I would be interested to get someone else's opinion on this also as this is by no means my area of expertise but it is how I understand it.

(edited 11 years ago)

Reply 4

Charries

1 molecule in box

Scenario 1
volume of box = 1x1x1 = 1 sqm
temperature = 1k

Scenario 2
Volume of box = 1.26 x 1.26 x 1.26 = 2 sqm
temperature = 2k (same pressure)

As velocity is proportional to the root of temperature, the velocity increases by root 2 = 1.41 m/s times the velocity before.

But since each side of the container has only increased by x1.26, this x1.41 increase in velocity means there must be more collisions with the container and more collisions with each other.

Reply 5

SDA

Original post by Charries

But since each side of the container has only increased by x1.26, this x1.41 increase in velocity means there must be more collisions with the container and more collisions with each other.

The problem I can see with this statement is that you have said the side of the container increases by x1.26 which is correct, however the mean free path between molecules increases proportional to the volume and so the distance of interest increases by x2.

Reply 6

Charries

Thanks SDA - we don't do the mean free path, so I have to bow to your argument - I can't see an "easy" way to do this without the maths, but it's a question from an example paper, and it a worded answer, like explain two difference before and after (2 marks) - so I'd have though this would be an obvious answer, but to me it isn;t "obvious". The first difference is easy though - increase in KE, the second not so obvious for me, although the mark scheme agreed with your answer.

Reply 7

SDA

I realise my argument is largely based on the maths, if I can think of a more descriptive way of saying it then I will post it here later. But for now I hope you are comfortable with the basic concepts and at least you will be able to answer future questions :smile: