Two Perfect gases at absolute temperatures T1 and T2 are mixed. There is no loss of energy. Find the temperature of the mixture if the masses of the molecules are m1 and m2 and the number of molecules in the gases are $n_{1}$ and $n_{2}$ .

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Solution

The correct option is A
According the kinetic theory of gases, the kinetic energy of an ideal gas molecule at temperature T is given by KE = $\frac{3}{2}$ kT, And as there is no force of attraction among the molecules of a perfect gas, PE of molecules is zero. So the energy of a molecule of a perfect gas.
E = KE + PE = $\frac{3}{2}$ kT + 0 = $\frac{3}{2}$ kT
Now if T is the temperature of the mixture, by conservation of energy, i.e.,
$n_{1} E_{1} + n_{2} E_{2} = (n_{1} + n_{2}) E$
We have
$n_{1} (\frac{3}{2} k T_{1}) + n_{2} (\frac{3}{2} k T_{2}) = (n_{1} + n_{2}) \frac{3}{2} k T$
i.e., $T = \frac{(n_{1} T_{1} + n_{2} T_{2})}{(n_{1} + n_{2})}$

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