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

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Correct answer: B. n _ 1 F _ 1 T _ 1 + n _ 2 F _ 2 T _ 2 n _ 1 F _ 1 + n _ 2 F _ 2

Let the final temperature of the mixture be $T .$

Since, there is no loss in energy.

$\DeltaU = 0$

$⇒ \dfrac{{F}_{1}}{2}{n}_{1}R\DeltaT + \dfrac{{F}_{2}}{2}{n}_{2}R\DeltaT = 0$

$⇒ \dfrac{{F}_{1}}{2}{n}_{1}R\left({T}_{1} - T\right) + \dfrac{{F}_{2}}{2}{n}_{2}R\left({T}_{2} - T\right) = 0$

$⇒ T = \dfrac{{F}_{1}{n}_{1}R{T}_{1} + {F}_{2}{n}_{2}R{T}_{2}}{{F}_{1}{n}_{1}R + {F}_{2}{n}_{2}R} = \dfrac{{F}_{1}{n}_{1}{T}_{1} + {F}_{2}{n}_{2}{T}_{2}}{{F}_{1}{n}_{1} + {F}_{2}{n}_{2}}$

Two ideal polyatomic gases at temperatures T _ 1 and T _ 2 are...

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