First slide
Law of equipartition of energy
Question

Two perfect monoatomic gases at absolute temperatures T1 and T2 are mixed. There is no loss of energy. The masses of the molecules are ml and m2.The number of molecules in the gases are n1 and n2.The temperature of the mixture is

Moderate
Solution

32kT1×n1+32kT2×n2 = (n1+n2)32kT

or T = n1T1+n2T2n1+n2

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