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JEE Main 2019 (11 Jan, Shift 1) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2019 (11 Jan, Shift 1)

Q33JEE Main 2019 · 11 JAN · SHIFT 1Medium

Degrees of Freedom & Internal Energy

A gas mixture consists 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:

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Answer: (d)

Translational + rotational only: $\text{O}_2$ $f = 5$ , Ar $f = 3$ .
$U = 3 \times \frac{5}{2}RT + 5 \times \frac{3}{2}RT = 15RT$
So the answer is (d) $15RT$ .

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