If 2 samples of nitrogen (different PVTs) are separated by a conductive barrier, they will eventually achieve equilibrium, I believe.
Am I right to assume that the final T and P at equilibrium in this case would be the SAME T and P if the barrier was highly insulated, then removed allowing equilibrium to happen.
Physics ideal gases - Damon, Saturday, January 3, 2009 at 5:24pm
In one case the process is at equilibrium at every step and reversible.
In the other case you have a sudden mixing. I would not assume the same U change for both cases.
Physics ideal gases - Damon, Saturday, January 3, 2009 at 6:24pm
But I would if the gases do no work in case 1. If no work is done and the system is totally isolated from the surroundings so that no heat can pass in or out, then the initial and final states will be the same.
Physics ideal gases - Count Iblis, Sunday, January 4, 2009 at 8:04am
If the pressures are not the same then when the barrier is removed, you get a free expansion of the gasses. When equilibrium is reached, the total internal energy will be the same as in the original state.
It follows from this that the final pressure is given by
P = (P1 V1 + P2 V2)/(V1+V2)
Here P is the final pressure, P1 and P2 are the initial pressures, V1 and V2 are the volumes of the two parts.
In case of a conductive barrier, only the temperatures will equalize. The total internal energy also stays the same. Equal final temperatures means that the final pressures are related according to:
P_fin1 V1 = P_fin2 V2
Conservation of energy then implies:
P1 V1 + P2 V2 = 2 P_fin1 V1 --->
P_fin1 = 1/2 (P1 V1 + P2 V2)/V1
P_fin2 = 1/2 (P1 V1 + P2 V2)/V2
Physics ideal gases - charlie, Sunday, January 4, 2009 at 10:37am
Thank you both.