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 o

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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:

TestHub · Accepted Answer

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

Since, there is no loss in energy.

$Δ U = 0$

$\Rightarrow \frac{F_{1}}{2} n_{1} R Δ T + \frac{F_{2}}{2} n_{2} R Δ T = 0$

$\Rightarrow \frac{F_{1}}{2} n_{1} R (T_{1} - T) + \frac{F_{2}}{2} n_{2} R (T_{2} - T) = 0$

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

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