0
$\begingroup$

Consider two containers separated by a removable wall, each side of which is a perfect mirror for the gas in the respective container. Also the walls of the containers are ideal mirrors. In each container the gas is the same, it is ideal and in thermodynamic equilibrium, therefore the distributions of the energies of the molecules is the Maxwell-Boltzmann law. The gas in the container A has temperature $T_A$, and the gas in the container B has temperature $T_B$, with $T_A$ > $T_B$.

It is known that when the wall is removed, each gas expands into the other container and in the end, all the gas reaches thermodynamic equilibrium, at some intermediate temperature $T_C$, i.e. $T_A$ > $T_C$ > $T_B$.

In there a way to calculate the final parameters, temperature, pressure, entropy, from the two initial distributions of energies (velocities) + volumes of containers and initial pressures in them?

$\endgroup$
2
$\begingroup$

The final distribution is a function of the initial distributions in this sense: the initial distributions determine thermodynamic quantities - energy, temperature - which determine the thermodynamic quantity $T_C$ after the partition is removed, which finally determines the new statistical distribution. The best way to solve this problem is to find $T_C$ from conservation of energy, and use that to determine the new Boltzmann distribution.

$\endgroup$
  • $\begingroup$ This is indeed my problem. How to obtain $T_C$ . I went to the first law of thermodynamics, assuming that the total energy of the gas doesn't change because there is no communication with the environment. Am I wrong in this? Next, the change in entropy can be calculated (though it is very hard). But how changes the pressure? $\endgroup$ – Sofia Dec 2 '14 at 20:13
  • $\begingroup$ Correct - the system is completely isolated, so energy is conserved. The heat capacity of an ideal gas then gives you the temperature. Knowing the temperature, volume, and amount of the gas, the ideal gas law will give you the pressure. $\endgroup$ – user27118 Dec 3 '14 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.