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Rotational kinetic energy of O₂

Find the average rotational kinetic energy of an oxygen molecule O₂ (diatomic) at T = 320 K. (k = 1.38×10⁻²³ J/K)

Given data

Diatomic molecule → 2 rotational d.o.f.T = 320 Kk = 1.38×10⁻²³ J/K

Steps

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By the equipartition theorem, each quadratic degree of freedom contributes on average ½kT. A diatomic molecule like O₂ rotates about 2 axes (not about its own axis) → 2 rotational d.o.f. → E_rot = 2·½kT = kT. Compute kT (in 10⁻²¹ J).
Write your working — the numerical result must appear in the text ×10⁻²¹ J

Full worked solution

By the equipartition theorem, each quadratic degree of freedom contributes on average ½kT. A diatomic molecule like O₂ rotates about 2 axes (not about its own axis) → 2 rotational d.o.f. → E_rot = 2·½kT = kT. Compute kT (in 10⁻²¹ J).
$E_{rot} = kT = 1.38\times10^{-23}\cdot 320$
$E_{rot} = kT = 1.38\times10^{-23}\cdot320 = \mathbf{4.42\times10^{-21}\,J}$ .

Result:

E_{rot} = 4.42\times10^{-21}

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