Average translational K.E. of one mole of helium gas at $273 K$ in calories is:

819

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81.9

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3412.5

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34.125

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Solution

The correct option is D $3412.5$
$Average translational k E of one mole He = \frac{3}{2} R T$
$\begin{matrix} Given T = 273 K K \cdot E & = \frac{3}{2} \times 8.314 \times 273 = 3402.945 J = 3403 J \end{matrix}$
Correct Answer $- C$

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