Kinetic energy of one molecule of an ideal gas is

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$\frac{2}{3} K T$

$\frac{3}{2} K T$

$\frac{3}{2} R T$

$\frac{2}{3} R T$

answer is B.

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Detailed Solution

The K.E equation of a gas for n moles is

$K . E = \frac{3}{2} nRT$

The kinetic energy (K.E) for one mole of gas (𝑛 = 1 ) we can write as

$K . E = \frac{3}{2} RT$

Note: 1 mole of any gas contains N molecules

So, K.E per one molecule is

$K . E = \frac{3}{2} \frac{RT}{M}$

$K . E = \frac{3}{2} KT; (Since \frac{R}{N} = K)$

Therefore, the correct answer is option “B”.

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