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(B) The internal energy $U$ of a gas mixture is given by the sum of the internal energies of its components: $U = U_1 + U_2$.For an ideal gas,the inte

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$15$

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(B) The internal energy $U$ of a gas mixture is given by the sum of the internal energies of its components: $U = U_1 + U_2$.For an ideal gas,the internal energy is $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom.Oxygen $(O_2)$ is a diatomic gas. Since the bond is rigid,its degrees of freedom $f_1 = 5$ ($3$ translational + $2$ rotational).Argon $(Ar)$ is a monatomic gas,so its degrees of freedom $f_2 = 3$.Given $n_1 = 3$ moles of oxygen and $n_2 = 5$ moles of argon.Total internal energy $U = \frac{f_1}{2} n_1 RT + \frac{f_2}{2} n_2 RT$.$U = \left( \frac{5}{2} 	imes 3 	imes RT ight) + \left( \frac{3}{2} 	imes 5 	imes RT ight)$.$U = \frac{15}{2} RT + \frac{15}{2} RT = \frac{30}{2} RT = 15 RT$.Thus,the total internal energy in units of $RT$ is $15$.

$A$ gas mixture consists of $3$ moles of oxygen and $5$ moles of argon at temperature $T$. Assuming the gases to be ideal and the oxygen bond to be rigid,the total internal energy (in units of $RT$) of the mixture is

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