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Two monoatomic ideal gas at temperature $T_1$ 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 number of their molecules are $n_1$ and $n_2$ respectively. The temperature of the mixture will be :
$\frac{{{T_1} + {T_2}}}{{{n_1} + {n_2}}}$
$\frac{{{T_1}}}{{{n_1}}}\, + \,\frac{{{T_2}}}{{{n_2}}}$
$\frac{{{n_2}{T_1} + {n_1}{T_2}}}{{{n_1} + {n_2}}}$
$\frac{{{n_1}{T_1} + {n_2}{T_2}}}{{{n_1} + {n_2}}}$
Solution
Net $\mathrm{P}-\mathrm{V}$ work done $=0$ as no net change in volume takes place.
Also net heat absorbed $=0$
Thus by first law, net internal energy change $=0$
$\Rightarrow n_{1} C_{v{1}} \Delta T_{1}+n_{2} C_{\mathrm{v} 2} \Delta T_{2}=0$
Both gases are monoatomic $\Rightarrow C_{v_{1}}=C_{v_{2}}=\frac{3}{2} R$
Let final temperature of mixture at equilibrium be $\mathrm{T}$
Then $\frac{3}{2} R\left(n_{1}\left(T-T_{1}\right)+n_{2}\left(T-T_{2}\right)\right)=0 \Rightarrow T=\frac{n_{1} T_{1}+n_{2} T_{2}}{n_{1}+n_{2}}$
