Three perfect gases at absolute temperatures | vedclass

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(C) The total internal energy of the mixture is conserved since there is no loss of energy.The internal energy of a gas with $n$ molecules at temperat

Vedclass · Accepted Answer

$\frac{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}{{n_1} + {n_2} + {n_3}}$

Vedclass · Answer

(C) The total internal energy of the mixture is conserved since there is no loss of energy.The internal energy of a gas with $n$ molecules at temperature $T$ is given by $U = \frac{f}{2} n k_B T$,where $f$ is the degrees of freedom and $k_B$ is the Boltzmann constant.For the mixture,the total energy is the sum of the energies of the individual gases:$U_{total} = U_1 + U_2 + U_3$$\frac{f}{2} (n_1 + n_2 + n_3) k_B T_{mix} = \frac{f}{2} n_1 k_B T_1 + \frac{f}{2} n_2 k_B T_2 + \frac{f}{2} n_3 k_B T_3$Canceling the common terms $\frac{f}{2} k_B$ from both sides,we get:$(n_1 + n_2 + n_3) T_{mix} = n_1 T_1 + n_2 T_2 + n_3 T_3$Therefore,the final temperature of the mixture is:$T_{mix} = \frac{n_1 T_1 + n_2 T_2 + n_3 T_3}{n_1 + n_2 + n_3}$

Three perfect gases at absolute temperatures $T_1, T_2$ and $T_3$ are mixed. The masses of molecules are $m_1, m_2$ and $m_3$ and the number of molecules are $n_1, n_2$ and $n_3$ respectively. Assuming no loss of energy,the final temperature of the mixture is

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