There is no equilibrium state in an isolated system, and the second and 0th laws of thermodynamics fail.
2. Nonequilibrium fluctuation test
The helium and carbon dioxide in the container are mainly carbon dioxide in the lower part of the container, and are near its critical point, so the fluctuation energy of the gas is large. The upper half of the container is mainly composed of helium, which is a conventional gas. The fluctuation energy is small. The fluctuation energy of the lower part will be transferred to the upper part, which will destroy the thermodynamic equilibrium probability distribution of the upper and lower parts of the container. The average internal energy of the lower gas is converted into fluctuation energy, while the fluctuation energy of the upper gas is converted into average internal energy, resulting in an increase in the temperature of the upper helium and a decrease in the temperature of the lower carbon dioxide. The thermal equilibrium cannot exist, and the 0th law and the 2nd law of thermodynamics fail at the same time.
Complete stupidity.
Thermodynamic equilibrium - it either exists or it doesn’t. Whatever fluctuations there may be in a mixture of two gases, the temperature of the mixture will sooner or later become the same throughout its volume.
I agree with Mikhail N Dulin. If one waits long enough, the gases will mix, and the concentrations and the temperature will be the same verywhere in the vessel.
On the way to equilibrium, however, a temperature deviation may occur (diffusion thermo effect, Dufour effect). This is well known and does not violate any laws of thermodynamics.
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