Two ideal polyatomic gases at temperatures T₁ and T₂ are mixed so that there is no loss of energy. If F₁ and F₂, m₁ and m₂, n₁ and n₂ be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :

Solution

Physics sir · 3 steps

Mixing keeps the total internal energy fixed, so add up F/2kT for every molecule before and after.

1. Energy per molecule. By equipartition, each degree of freedom carries 1/2kT. A molecule with F degrees of freedom has energy F/2kT, so n such molecules have nF/2kT.
2. Before and after. Before mixing, the total energy is n₁F₁/2kT₁ + n₂F₂/2kT₂. After mixing, every molecule is at the common temperature T, but each one keeps its own F, so the total is n₁F₁/2kT + n₂F₂/2kT.

   k/2(n₁F₁T₁ + n₂F₂T₂) = k/2(n₁F₁ + n₂F₂)T

3. So. Cancel k/2 and solve: T = (n₁F₁T₁ + n₂F₂T₂)/(n₁F₁ + n₂F₂), which is option (b). The masses m₁ and m₂ never enter, because the energy depends only on F and T.

The idea to keep. Final temperature after mixing is a weighted average of the starting temperatures. Each gas is weighted by its total number of degrees of freedom, nF, not by its mass.

Answer from the NTA final answer key. Solution by PrepWiser.