Two perfect gases at absolute temperatures $T_{1}$ and $T_{2}$ are mixed. There is no loss of energy. The temperature of mixture, if masses of molecules are $m_{1}$ and $m_{2}$ and the number of molecules in the gases are $n_{1}$ and $n_{2}$ respectively

\frac{T_{1} + T_{2}}{2}

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\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}

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\frac{3 n_{1} T_{1} + 5 n_{2} T_{2}}{3 n_{1} + 5 n_{2}}

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\sqrt{T_{1} T_{2}}

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Solution

The correct option is B $\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$
$\begin{matrix} T h e a v e r a g e k i n e t i c e n e r g y p e r m o l e c u l e o f a p e r f e c t g a s = \frac{3}{2} K T A v e r a g e k i n e t i c e n e r g y o f t h e m o l e c l e s o f t h e f i r s t g a s = \frac{3}{2} n_{1} K T_{1} A v e r a g e k i n e t i c e n e r g y o f t h e m o l e c u l e s o f t h e sec o n d g a s = \frac{3}{2} n_{2} K T_{2} T h e r e f o r e, t h e t o t a l k i n e t i c e n e r g y o f t h e t w o m o l e c u l e s o f g a s b e f o r e t h e y a r e m i x e d K = \frac{3}{2} n_{1} K T_{1} + \frac{3}{2} n_{2} K T_{2} = \frac{3}{2} (n_{1} T_{1} + n_{2} T_{2}) K - - - - (1) I f T i s t h e t m e p e r a t u r e o f t h e m i x t u r e, t h e n t h e k i n e t i c e n e r g i e s o f t h e m o l e c u l e s (n_{1} + n_{2}) i s K^{'} = \frac{3}{2} (n_{1} + n_{2}) K T - - - (2) sin c e, t h e r e i s n o l o s s o f e n e r g y K - K^{'} E q u a t i n g (1) a n d (2) w e g e t n_{1} T_{1} + n_{2} T_{2} = (n_{1} + n_{2}) T T = \frac{n_{1} T_{1} + n_{2} T_{2}}{(n_{1} + n_{2})} H e n c e, t h e o p t i o n B i s t h e c o r r e c t a n s w e r . \end{matrix}$

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