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Two perfect gases having masses m(1) and...

Two perfect gases having masses `m_(1)` and `m_(2)` at temperature `T_(1)` and `T_(2)` are mixed without any loses of internal kinetic energy of the molecules. The molecular weights of the gases are `M_(1)` and `M_(2)`. What is the final temperature of the mixture?

Text Solution

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According to the kinetic theory of gases, the kinetic energy of an ideal gas molecule at temperature .T. is given by KE (3/2)KT. And as there is no force of attraction among the molecules of a perfect gas, PE of the molecule is zero. So the energy of a molecule of perfect gas, `E=3/2 KT`.
Now if T is the temperature of the mixture, by conservation of energy,i.e, `n_1E_1+n_2E_2=(n_1+n_2)E`
`n_(1)(3/2 kT_1) +n_2 (3/2kT_2) =(n_1+n_2) (3/2kT)`
i.e, `T=((n_(1)T_(1)+n_(2)T_(2)))/((n_(1)+n_(2))` .
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