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JEE Main 2026 (6 Apr, Shift 2) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2026 (6 Apr, Shift 2)

Q2JEE Main 2026 · 6 APR · SHIFT 2Medium

Degrees of Freedom & Internal Energy

If 2 mole of an ideal monoatomic gas at temperature $T$ , is mixed with 6 mole of another ideal monoatomic gas at temperature $2T$ then the temperature of mixture is:

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Answer: (d)

No energy lost and both gases monoatomic: $(n_1 + n_2)T_{mix} = n_1T_1 + n_2T_2$ .
$T_{mix} = \frac{2T + 6(2T)}{8} = \frac{14T}{8}$
So the answer is (d) $\frac{7}{4}T$ .

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