Three ideal gases at absolute temperatures T_ | vedclass

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Question

(A) For an ideal gas,the internal energy $U$ is given by $U = \frac{f}{2} nRT = \frac{f}{2} NkT$,where $f$ is the degrees of freedom,$N$ is the number

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For an ideal gas,the internal energy $U$ is given by $U = \frac{f}{2} nRT = \frac{f}{2} NkT$,where $f$ is the degrees of freedom,$N$ is the number of molecules,and $k$ is the Boltzmann constant.Since the gases are mixed and there is no loss of energy,the total internal energy of the mixture is equal to the sum of the internal energies of the individual gases.$U_{total} = U_1 + U_2 + U_3$$\frac{f}{2} (n_1 + n_2 + n_3) kT = \frac{f}{2} n_1kT_1 + \frac{f}{2} n_2kT_2 + \frac{f}{2} n_3kT_3$Canceling the common terms $\frac{f}{2} k$ from both sides:$(n_1 + n_2 + n_3) T = n_1T_1 + n_2T_2 + n_3T_3$Therefore,the final temperature $T$ is:$T = \frac{n_1T_1 + n_2T_2 + n_3T_3}{n_1 + n_2 + n_3}$

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

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