Gas expansion temperature change

[ronthepon]

I had approached a thermodynamics(?) related excercise some time ago, and don't really feel sure about the way I approached it.

Suppose we had two gas filled chambers, perfectly isolated from the surroundings reigion, both separated with a valve.

The thermodynamic state of the chambers (chambers 1 and 2) are charecterised by gaseous pressures, Volumes and Gaseous temperatures as-

Pressures and

Chamber volumes and

Temperatures and

(Subscripts denote the chamber number.)

Supposing the valve was suddenly opened. Assuming ideal conditions, We've got to find out the final state the system reaches. After it reaches equilibrium that is.

[Drum]

Originally Posted by ronthepon

Pressures and

Chamber volumes and

Temperatures and

(Subscripts denote the chamber number.)

Is this a trick question ??

It looks too simple. Wouldn't 'equilibrium' be maximum entropy ?? Order (lower entropy) would only be possible if a greater amount of dis-order was released. As 'equilibrium' has already been reached, this is not possible.

My intuition tells me that if you have reached 'Maximum Entropy' in an isolated system, then all movement would have ceased and it would be about as cold as it is possible to get. The temperature would be falling towards Absolute Zero like a Stuka with a broken wing ....


Kind regards .... Drum

[snark1100]

the ideal gas law gives us PV = nrT
kinetic theory gives us the average kinetic energy = 3/2 kT

initially we have 2 boxes -


P1V1 = n1rT1 and P2V2 = n2rT2,

when we break the barrier between these, we have a total volumn of
V1+V2, and a total number of particles, n1+n2, and some final temperature
Tf, and pressure Pf:

A) Pf(V1+v2) = (n1+n2)rTf

now the total energy in box 1 will be the average * n1, simarly for box 2,
and kinetic energy will be conserved -
(total) KE = (3/2)n1kT1 + (3/2)n2kT2 = 3/2(n1+n2)kTf
and so Tf = (n1T1 +n2T2)/(n1+n2), and we can substitute this in equation
A to get the final pressure.

[Qfwfq]

Originally Posted by ronthepon

Supposing the valve was suddenly opened. Assuming ideal conditions, We've got to find out the final state the system reaches. After it reaches equilibrium that is.

Trouble is the question isn't quite perfectly defined, because there will be a very slow exchange through the connection, even after the pressures are equal and. Therefore saying "after it reaches equilibrium" doesn't unequivocally define the problem unless you specify better.

[Erasmus00]

It depends on how the divider is removed. Was it slow and gradual (adiabatic) or was the valve suddenly turned full open?

If we do the expansion adiabatically, we know that no new states in phase space become accessible to the particles, so this becomes the starting point of our analysis.

Now, what if, instead, we take the volume V goes to 2V instantly. What happens then? Well, now much of the new phase space of the 2V box is accessible to our particles.
-Will