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:

Step-by-step solution

Let the final temperature of the mixture be T . Since, there is no loss in energy. Δ U = 0 ⇒ F 1 2 n 1 R Δ T + F 2 2 n 2 R Δ T = 0 ⇒ F 1 2 n 1 R T 1 - T + F 2 2 n 2 R T 2 - T = 0 ⇒ T = F 1 n 1 R T 1 + F 2 n 2 R T 2   F 1 n 1 R + F 2 n 2 R = F 1 n 1   T 1 + F 2 n 2   T 2   F 1 n 1 + F 2 n 2