Three ideal gases at absolute temperatures T_ | vedclass

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(D) For an ideal gas,the average kinetic energy per molecule is given by $K.E. = \frac{3}{2} k_B T$ (assuming monatomic gas for simplicity,though the

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$\frac{n_1T_1 + n_2T_2 + n_3T_3}{n_1 + n_2 + n_3}$

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(D) For an ideal gas,the average kinetic energy per molecule is given by $K.E. = \frac{3}{2} k_B T$ (assuming monatomic gas for simplicity,though the factor cancels out). The total internal energy of $n$ molecules is $U = n 	imes (\frac{3}{2} k_B T)$.Since there is no loss of energy,the total internal energy of the mixture is the sum of the internal energies of the individual gases:$U_{total} = U_1 + U_2 + U_3$$(n_1 + n_2 + n_3) 	imes (\frac{3}{2} k_B T) = n_1 (\frac{3}{2} k_B T_1) + n_2 (\frac{3}{2} k_B T_2) + n_3 (\frac{3}{2} k_B T_3)$Canceling the common factor $\frac{3}{2} k_B$ from both sides:$(n_1 + n_2 + n_3) T = n_1 T_1 + n_2 T_2 + n_3 T_3$Therefore,the final temperature $T$ is:$T = \frac{n_1 T_1 + n_2 T_2 + n_3 T_3}{n_1 + n_2 + n_3}$

Three ideal gases at absolute temperatures $T_1, T_2,$ and $T_3$ are mixed. The number of molecules are $n_1, n_2,$ and $n_3$ respectively. Assuming no loss of energy,what is the final temperature of the mixture?

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