Question on gases

Discussion in 'MCAT Study Question Q&A' started by SephirothXR, Jul 28, 2011.

1. SephirothXR 5+ Year Member

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What happens to the mean free path when you increase the temperature? The gas molecules travel faster so they'll hit each other more often, correct (thus reducing MFP) When you increase the pressure, they collide with walls more often, and thus with each other, so that should also reduce MFP right?

2. MD Odyssey 2+ Year Member

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Correct. The mean free path is the average distance that particles are able to travel before they collide with another particle. So, if two particles are headed towards each other and you double their speed, they will collide at a different position than they would have at the lower temperature and, on average, this distance will be shorter.

Think about how you increased the pressure. If you increase the pressure by heating the gas, obviously the mean free path will decrease, as you observed earlier. If one increases the pressure by decreasing the volume in an isothermal process, then the particles will still wind up closer together because, while their energy hasn't changed, their relative density has increased. This also decreases the mean free path.

So, in short, yes, you're correct. The mean free path increases for increased temperature and increased pressures, but for different reasons.

Edit: I should point out that the speeds I refer to are not, in general equal, which is what may have led to the OP confusion. See my next reply.

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Last edited: Jul 30, 2011
3. MT Headed snow, PBR, and bears Lifetime DonorVerified Account 2+ Year Member

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You conclusion is correct (http://en.wikipedia.org/wiki/Mean_free_path#Mean_free_path_in_kinetic_theory) , but I'm not buying your justification. If two cars approach each other at 10mph or 100mph, they are going to collide at the exact same point. The time is reduced, but the distances will be the same.

Now I'm curious, what is the justification for the mean free path being shorter as temps go up? In any thought experiment I can come up with, it seems like the mean free path ought to be the same. I would have gotten this wrong on the MCAT

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4. MD Odyssey 2+ Year Member

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I figured that you might say this. Your example is correct, but assumes that they're headed directly towards each other at the same speed. In general, there is a distribution of energies and therefore speeds as well. Let's say that you have two particles that are 1000 m apart. Particle A is at the origin and has a velocity of 10 m/s. Particle B is at the other end and has a velocity of 20 m/s.

It can be shown, with a little bit of algebra that a general equation for the mean free path of particle A in this system is the following:

$image=http://latex.codecogs.com/gif.latex?%5Ctext%7BMean%20Free%20Path%7D%20%3D%201000%5Cleft%20%28%201%20+%20%5Cfrac%7BV_B%7D%7BV_A%7D%20%5Cright%20%29%5E%7B-1%7D&hash=ba4fc7801bb2f80a0cc0e988bbf5d967$

So, if particle B has a velocity that is twice that of particle A, then the distance that A travels prior to the collision is ~ 333 m. Particle B obviously travels 666 m. If particle B is nine times faster than particle A, the mean free path for A is only 100 m and for particle B it's 900 m. And clearly, in the simple case of equal and opposite speeds, the mean free path for both is 500 m, which is the center, just like you said.

I can post a derivation if you like - I doubt that you'll find it anywhere on the internet - but it's rather straightforward. Also, notice that I have explicitly described it with the assumption that the two velocities in the equation are actually speeds. In the derivation, one must remember that velocities are actually vector quantities.

The derivation I just gave you should be justification enough. Also, an additional complication would be that the particle directions are random, although as you can see, one can deduce some features of the mean free path with a simple gas of two molecules.

Hope this helps.

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