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The average kinetic energy of 28 g CO at, 300K is E kcal. The average kinetic of 2 g $H_{2}$ at the same temperature would be _______ k cal.

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14 E

$\frac{1}{14} E$

28 E

answer is A.

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Detailed Solution

28 g of $CO = \frac{28}{28} = 1$ mole; $2 g H_{2} = \frac{2}{2} = 1$ mole

Since the temperature and moles are the same, the average
kinetic energy will also be the same

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