The average translational kinetic energy of molecules of an ideal gas is (symbols have their usual meanings)

\frac{1}{2} k_{B} T

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\frac{3}{2} k_{B} T

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\frac{7}{2} k_{B} T

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\frac{5}{2} k_{B} T

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Solution

The correct option is B $\frac{3}{2} k_{B} T$
The average translational kinetic energy of the molecules of an ideal gas is given as $\frac{3}{2} k_{B} T$ .

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