Two monoatomic ideal gases at temperatures T_ | vedclass

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(D) Since there is no loss of energy and no external work is done,the total internal energy of the system is conserved.The internal energy of a monoat

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$\frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$

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(D) Since there is no loss of energy and no external work is done,the total internal energy of the system is conserved.The internal energy of a monoatomic gas is given by $U = \frac{3}{2} n R T$.Let $T$ be the final temperature of the mixture.The total initial internal energy is $U_i = \frac{3}{2} n_1 R T_1 + \frac{3}{2} n_2 R T_2$.The total final internal energy is $U_f = \frac{3}{2} (n_1 + n_2) R T$.By the law of conservation of energy,$U_i = U_f$.$\frac{3}{2} n_1 R T_1 + \frac{3}{2} n_2 R T_2 = \frac{3}{2} (n_1 + n_2) R T$.Dividing both sides by $\frac{3}{2} R$,we get $n_1 T_1 + n_2 T_2 = (n_1 + n_2) T$.Therefore,the final temperature is $T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.

Two monoatomic ideal gases at temperatures $T_1$ and $T_2$ are mixed. There is no loss of energy. If the masses of molecules of the two gases are $m_1$ and $m_2$ and the number of their molecules are $n_1$ and $n_2$ respectively,the temperature of the mixture will be:

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