Changing temperature in a perfect cylinder with piston

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Nemode
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Homework Statement



Perfect gas in an enclosed cylinder, the thermal capcity of the components in the cylinder and piston are small and they are perfect thermal insulators. The piston is pushed in slowly and allowed to return to its original position slowly.

How does the Temperature vary with the displacement of the piston?

Homework Equations



PV=nRT
P=(Nmv^2)/(3*V)

The Attempt at a Solution



Subbing the second equation into the first It would seem that the Temperature varies with the velocity, as would be expected. But I seem to conclude that velocity would remain constant and there would therefore be no change in temperature, which can't be right...
Though the particles are colliding more frequently, where does a change in velocity come into play, as the particles can't gain any energy can they?

Thanks
 
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Nemode said:
Subbing the second equation into the first It would seem that the Temperature varies with the velocity, as would be expected. But I seem to conclude that velocity would remain constant and there would therefore be no change in temperature, which can't be right...
Though the particles are colliding more frequently, where does a change in velocity come into play, as the particles can't gain any energy can they?
It takes work to push the piston in, so yes the particles can gain energy. In fact, since the piston is perfectly insulating all of that energy ends up in the gas (before the piston is released)
 
Oh dear, yes of course. Thank you. I'd decided the slowly part meant I could ignore the motion of the piston, but obviously that's not the case. (This isn't a homework question, just my own exam preparation)

So the work done on the gas by the piston is proportional to the displacement squared due to th increasing pressure as the volume decreases?

If so then as temperature is KE of particles, and KE = 0.5mv^2 the temperature increase is also linear as energy increase is exponential. ie. (where Z is a constant) Zs^2 = 0.5mv^2.

Meaning linear temperature increase with displacement?

Not so sure about this. I'm snowed in at home so I don't have any tutor I can run this past any time soon, sorry. Thanks.
 
I don't believe that is correct. See this link:
http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/adiab.html

If you think about it, if temperature increases linearly with displacement it implies some kind of upper bound on the temperature as you squeeze it towards zero volume, and conversely you'll have negative energy as it expands to a very large volume. Neither of those possibilities makes sense physically.