A mixture of ideal gases has 2 moles of He,4 | vedclass

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(C) The internal energy $U$ of a gas is given by $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom,$n$ is the number of moles,$R$ is the unive

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$16$

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(C) The internal energy $U$ of a gas is given by $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.For $He$ (monatomic gas),degrees of freedom $f_{He} = 3$.For $O_2$ (diatomic gas),degrees of freedom $f_{O_2} = 5$.For $O_3$ (non-linear polyatomic gas),degrees of freedom $f_{O_3} = 6$.Given:$n_{He} = 2$$n_{O_2} = 4$$n_{O_3} = 1$The total internal energy of the mixture is the sum of the internal energies of individual components:$U_{total} = U_{He} + U_{O_2} + U_{O_3}$$U_{total} = \left( n_{He} \cdot \frac{f_{He}}{2} + n_{O_2} \cdot \frac{f_{O_2}}{2} + n_{O_3} \cdot \frac{f_{O_3}}{2} \right) RT$$U_{total} = \left( 2 \cdot \frac{3}{2} + 4 \cdot \frac{5}{2} + 1 \cdot \frac{6}{2} \right) RT$$U_{total} = (3 + 10 + 3) RT$$U_{total} = 16 RT$.Thus,the internal energy is $16 RT$.

$A$ mixture of ideal gases has $2$ moles of $He$,$4$ moles of oxygen $(O_2)$,and $1$ mole of ozone $(O_3)$ at absolute temperature $T$. The internal energy of the mixture is ............. $RT$.

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