Two ideal polyatomic gases at temperatures $T_{1}$ and $T_{2}$ are mixed so that there is no loss of energy. If $\mathrm{F}_{1}$ and $\mathrm{F}_{2}, \mathrm{~m}_{1}$ and $\mathrm{m}_{2}, \mathrm{n}_{1}$ and $\mathrm{n}_{2}$ be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :
Correct Option: , 2
(2)
Let the final temperature of the mixture be $\mathrm{T}$. Since, there is no loss in energy,
$\Delta U=0$
$\Rightarrow \frac{\mathrm{F}_{1}}{2} \mathrm{n}_{1} \mathrm{R} \Delta \mathrm{T}+\frac{\mathrm{F}_{2}}{2} \mathrm{n}_{2} \mathrm{R} \Delta \mathrm{T}=0$
$\Rightarrow \frac{\mathrm{F}_{1}}{2} \mathrm{n}_{1} \mathrm{R}\left(\mathrm{T}_{1}-\mathrm{T}\right)+\frac{\mathrm{F}_{2}}{2} \mathrm{n}_{2} \mathrm{R}\left(\mathrm{T}_{2}-\mathrm{T}\right)=0$
$\Rightarrow \mathrm{T}=\frac{\mathrm{F}_{1} \mathrm{n}_{1} \mathrm{RT}_{1}+\mathrm{F}_{2} \mathrm{n}_{2} \mathrm{RT}_{2}}{\mathrm{~F}_{1} \mathrm{n}_{1} \mathrm{R}+\mathrm{F}_{2} \mathrm{n}_{2} \mathrm{R}} \Rightarrow \frac{\mathrm{F}_{1} \mathrm{n}_{1} \mathrm{~T}_{1}+\mathrm{F}_{2} \mathrm{n}_{2} \mathrm{~T}_{2}}{\mathrm{~F}_{1} \mathrm{n}_{1}+\mathrm{F}_{2} \mathrm{n}_{2}}$
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