Two ideal gases at absolute temperatures T_1 | vedclass

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(C) According to the law of conservation of energy,the total internal energy of the mixture is equal to the sum of the internal energies of the indivi

Vedclass · Accepted Answer

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

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(C) According to the law of conservation of energy,the total internal energy of the mixture is equal to the sum of the internal energies of the individual gases.The internal energy of an ideal gas is given by $E = \frac{f}{2} n k T$,where $f$ is the degrees of freedom,$n$ is the number of molecules,$k$ is the Boltzmann constant,and $T$ is the absolute temperature.For the mixture:$E_{total} = E_1 + E_2$$\frac{f}{2} (n_1 + n_2) k T = \frac{f}{2} n_1 k T_1 + \frac{f}{2} n_2 k T_2$Canceling the common terms $\frac{f}{2} k$ from both sides:$(n_1 + n_2) T = n_1 T_1 + n_2 T_2$Therefore,the temperature of the mixture is:$T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$

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

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