Two ideal polyatomic gases at temperatures T_{1}  and T_{2} are mixed so that there is no loss of energy. If F_{1} and F_{2} , m_{1} and m_{2}, n_{1} and 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 :
   
Option: 1 \frac{n_{1}F_{1}T_{1}+n_{2}F_{2}T_{2}}{n_{1}F_{1}+n_{2}F_{2}}
Option: 2 \frac{n_{1}T_{1}+n_{2}T_{2}}{n_{1}+n_{2}}
Option: 3 \frac{n_{1}F_{1}T_{1}+n_{2}F_{2}T_{2}}{n_{1}+n_{2}}  
Option: 4 \frac{n_{1}F_{1}T_{1}+n_{2}F_{2}T_{2}}{F_{1}+F_{2}}

Answers (1)

Let the final temperature of the mixture be T.

Since, there is no loss in energy.
\Delta \mathrm{U}=0

\begin{array}{l} \Rightarrow \frac{F_{1}}{2} n_{1} R \Delta T+\frac{F_{2}}{2} n_{2} R \Delta T=0 \\ \\ \Rightarrow \frac{F_{1}}{2} n_{1} R\left(T_{1}-T\right)+\frac{F_{2}}{2} n_{2} R\left(T_{2}-T\right)=0 \\ \\ \Rightarrow T=\frac{F_{1} n_{1} R T_{1}+F_{2} n_{2} R T_{2}}{F_{1} n_{1} R+F_{2} n_{2} R} \Rightarrow \frac{F_{1} n_{1} T_{1}+F_{2} n_{2} T_{2}}{F_{1} n_{1}+F_{2} n_{2}} \end{array}

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