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Kinetic theory and gas pressure

Hi all, I was looking through my A2 Physics textbook and the textbook didn't seem too clear on how the gas pressure equation in terms of density and mean velocity was derived. I had a go myself and I just wanted to see if this was the right way to go about it. Would appreciate it if you could take a look at it.

Thanks,

M
2014-11-03 20.38.20.jpg294.1KB
Original post by Malgorithm
Hi all, I was looking through my A2 Physics textbook and the textbook didn't seem too clear on how the gas pressure equation in terms of density and mean velocity was derived. I had a go myself and I just wanted to see if this was the right way to go about it. Would appreciate it if you could take a look at it.

Thanks,

M



The change in momentum of 2mv is not due to a "collision with the opposite wall".
How do you justify v2 = mean square speed.
Reply 2
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Malgorithm
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I guess I worded the reasoning behind 2mv badly. I meant that because there would be a total of two collisions in each 'cycle', the total change in momentum would be mv+mv=2mv.
As for the mean square speed, I wasn't entirely sure how that was substituted in as the book was quite vague at that part. It said that because each individual particle in the system had a different velocity, a mean average of the speed squared had to be taken. It also mentioned that the mean square speed was different to the mean speed squared? I assume that it means c2ˉ≠cˉ2\bar{c^2}\not=\bar{c}^2
Original post by Malgorithm
I guess I worded the reasoning behind 2mv badly. I meant that because there would be a total of two collisions in each 'cycle', the total change in momentum would be mv+mv=2mv.
As for the mean square speed, I wasn't entirely sure how that was substituted in as the book was quite vague at that part. It said that because each individual particle in the system had a different velocity, a mean average of the speed squared had to be taken. It also mentioned that the mean square speed was different to the mean speed squared? I assume that it means c2ˉ≠cˉ2\bar{c^2}\not=\bar{c}^2


The change is 2mv because the particle has momentum mv before it collides with the wall and -mv after. The change is 2mv on a single collision with the wall.
I can't now but I'll post a better explained proof here tomorrow scanned from a good text book.
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Malgorithm
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Original post by Stonebridge
The change is 2mv because the particle has momentum mv before it collides with the wall and -mv after. The change is 2mv on a single collision with the wall.
I can't now but I'll post a better explained proof here tomorrow scanned from a good text book.

OK, thank you :smile:
Original post by Malgorithm
OK, thank you :smile:


Here are two pages showing a proof.
The bit missing before the 1st scan reads:

A gas is in a cubical box of side l and contains N molecules each having mass m. Consider one molecule moving with velocity c as shown in Fig 19.6. We can resolve c into 3 components u, v and w in the three directions Ox, Oy and Oz respectively, parallel to the..

___________________________________________________________________________
___________________________________________________________________________


Let me know if there is anything you don't understand.
Reply 6
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Malgorithm
OP
6
Original post by Stonebridge
Here are two pages showing a proof.
The bit missing before the 1st scan reads:

A gas is in a cubical box of side l and contains N molecules each having mass m. Consider one molecule moving with velocity c as shown in Fig 19.6. We can resolve c into 3 components u, v and w in the three directions Ox, Oy and Oz respectively, parallel to the..

___________________________________________________________________________
___________________________________________________________________________


Let me know if there is anything you don't understand.

This is really great, thanks! We just went over this in lesson, and had to derive it for homework. This has really cleared things up for me, particularly with the c2ˉ\bar{c^2}

Also, which brand of textbook was that from? It definitely goes a lot more in depth that my current OCR one.
(edited 9 years ago)

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