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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:

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

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

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

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

answer is D.

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Detailed Solution

$C h a n g e i n I n t e r n a l e n e r g y, △ U = n (\frac{F R}{2}) (T_{f} - T_{i})$

Loss in internal energy of hot portion of gas is equal to gain in internal energy of colder portion of gas

$\frac{1}{2} n_{1} R F_{1} (T - T_{1}) = \frac{1}{2} n_{2} R F_{2} (T_{2} - T)$
Solving

$T = \frac{n_{1} F_{1} T_{1} + n_{2} F_{2} T_{2}}{n_{1} F_{1} + n_{2} F_{2}}$

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