A gaseous mixture consists of 2 moles of oxyg | vedclass

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(D) The total internal energy $U$ of a gaseous mixture is the sum of the internal energies of its individual components.For a gas with $n$ moles and d

Vedclass · Accepted Answer

$11$

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(D) The total internal energy $U$ of a gaseous mixture is the sum of the internal energies of its individual components.For a gas with $n$ moles and degrees of freedom $f$,the internal energy is given by $U = n \cdot \frac{f}{2} RT$.Oxygen $(O_2)$ is a diatomic gas. Neglecting vibrational modes,its degrees of freedom $f_1 = 5$.Internal energy of oxygen: $U_1 = n_1 \cdot \frac{f_1}{2} RT = 2 \cdot \frac{5}{2} RT = 5 RT$.Argon $(Ar)$ is a monatomic gas,so its degrees of freedom $f_2 = 3$.Internal energy of argon: $U_2 = n_2 \cdot \frac{f_2}{2} RT = 4 \cdot \frac{3}{2} RT = 6 RT$.Total internal energy $U = U_1 + U_2 = 5 RT + 6 RT = 11 RT$.

$A$ gaseous mixture consists of $2$ moles of oxygen and $4$ moles of argon at an absolute temperature $T$. Neglecting all vibrational modes,the total internal energy of the mixture of the gases is (in $RT$)

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