The gas mixture constists of $3$ moles of oxygen and $5$ moles of argon at temperature $T$ . Considering only translational and rotational modes, the total internal energy of the system is:

12 R T

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20 R T

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15 R T

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4 R T

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Solution

The correct option is C $15 R T$
$U = \frac{f_{1}}{2} n_{1} R T + \frac{f_{2}}{2} n_{2} R T$
$= \frac{5}{2} (3 R T) + \frac{3}{2} \times 5 R T$
$U = 15 R T$

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The gas mixture constists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is:

The correct option is C 15 RTU=f12n1RT+f22n2RT=52(3RT)+32×5RTU=15RT

The gas mixture constists of $3$ moles of oxygen and $5$ moles of argon at temperature $T$ . Considering only translational and rotational modes, the total internal energy of the system is:

The correct option is C $15 R T$
$U = \frac{f_{1}}{2} n_{1} R T + \frac{f_{2}}{2} n_{2} R T$
$= \frac{5}{2} (3 R T) + \frac{3}{2} \times 5 R T$
$U = 15 R T$