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If I have 2 ideal gases with different molecular masses, mixed together in a container and waited for a long time such that they are both in equilibrium. Suppose that in equilibrium the system is in temperature T.

Obviously the molecules with heavier molecules will move more slowly. However, will both gas individually follow a Maxwell-Boltzmann distribution of temperature T (and their corresponding mass)?

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Obviously the molecules with heavier molecules will move more slowly. However, will both gas individually follow a Maxwell-Boltzmann distribution of temperature T (and their corresponding mass)?

If the gas is collisionally dominated, then yes they will both relax to Maxwell–Boltzmann distributions with a single temperature, $T$, if given enough time. They will not, however, share the same thermal speeds, as those go as $V_{Ts} \propto \sqrt{\tfrac{T}{m_{s}}}$.

If I have 2 ideal gases with different molecular masses, mixed together in a container and waited for a long time such that they are both in equilibrium. Suppose that in equilibrium the system is in temperature T.

If the gases are ideal gases and in thermal equilibrium, then they should not change temperature in time and should share the same temperature, as you assume.

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I think that the answer is “Yes”.

The distribution of speeds of a gas $A$ alone is controlled by the mass of the molecules and the temperature of the gas.
As a result of collisions between the molecules on average there is no net transfer of kinetic energy between molecules of gas $A$.

The same is true of the molecules of gas $B$.

Assume that the two gases at the same temperature and pressure are in two containers and a tap is opened to allow the two gases to mix.
On average there will be no net transfer of kinetic energy between the two types of molecules and so on average there will be no change in the distribution of speed for each type of molecule.
Molecules of one type do not gain kinetic energy at the expense of the other.

When you bring two gases together which are at different temperature they reach an equilibrium state such that the temperature of the two gases is the same - on average there is no net transfer of kinetic energy between the molecules - so the distribution of speeds of each type of molecule as a mixture is the same as it would be if the gases were not mixed.

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