Three perfect gases at absolute temperatures | vedclass

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(C) The internal energy of an ideal gas is given by $U = \frac{f}{2} n R T$,where $f$ is the degrees of freedom,$n$ is the number of moles,and $T$ is

Vedclass · Accepted Answer

$\frac{n_1T_1 + n_2T_2 + n_3T_3}{n_1 + n_2 + n_3}$

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(C) The internal energy of an ideal gas is given by $U = \frac{f}{2} n R T$,where $f$ is the degrees of freedom,$n$ is the number of moles,and $T$ is the absolute temperature.Since the gases are mixed and there is no loss of energy,the total internal energy of the mixture is the sum of the internal energies of the individual gases.Total internal energy $U_{total} = U_1 + U_2 + U_3$.Using $n = \frac{N}{N_A}$ (where $N$ is the number of molecules and $N_A$ is Avogadro's number),the energy is $U = \frac{f}{2} \frac{N}{N_A} R T = \frac{f}{2} N k_B T$.Assuming the gases have the same degrees of freedom $f$,the total energy is $\frac{f}{2} k_B (n_1 T_1 + n_2 T_2 + n_3 T_3) = \frac{f}{2} k_B (n_1 + n_2 + n_3) T_{mix}$.Canceling the common terms $\frac{f}{2} k_B$,we get $n_1 T_1 + n_2 T_2 + n_3 T_3 = (n_1 + n_2 + n_3) T_{mix}$.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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