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.
Thanks
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.
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.
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
Thank you both.
Yes, you are correct in assuming that the final temperature (T) and pressure (P) at equilibrium would be the same in both cases, whether there is a conductive barrier or a highly insulated barrier that is later removed.
To explain why this is the case, let's first understand what happens during the process of achieving equilibrium. When two samples of nitrogen at different pressure, volume, and temperature (PVTs) are separated by a conductive barrier, they can exchange heat and particles with each other.
In this situation, heat transfer between the samples occurs due to the conductive nature of the barrier. As a result, the samples will keep exchanging heat until their temperatures become equal. Meanwhile, particle exchange will occur until the pressure becomes equal.
Once the temperatures and pressures are equalized, equilibrium is reached. At this point, the system is in a state where no further changes occur.
Now, let's consider the scenario where the barrier is highly insulated and then removed to allow equilibrium to happen. This means that initially, there is no heat transfer occurring between the samples.
However, removing the barrier effectively opens up the system and allows heat and particles to freely move between the two samples. As a result, they will eventually reach equilibrium, just like in the previous case.
Since the equilibrium state is determined by the properties of the system (such as the number of particles and energy), and not by the nature of the barrier, the final temperature and pressure at equilibrium will be the same, regardless of whether the barrier was conductive or highly insulated.
In conclusion, you are right to assume that the final T and P at equilibrium would be the same, whether there is a conductive barrier or a highly insulated barrier that is later removed.