The translational kinetic energy of gas molec | vedclass

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Question

(A) The translational kinetic energy per molecule of an ideal gas is given by $K.E._{molecule} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann const

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$\frac{3}{2}RT$

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(A) The translational kinetic energy per molecule of an ideal gas is given by $K.E._{molecule} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.For one mole of gas,the number of molecules is equal to Avogadro's number,$N_A$.Therefore,the total translational kinetic energy for one mole is $K.E._{molar} = N_A 	imes (\frac{3}{2} k_B T)$.Since the universal gas constant $R = N_A k_B$,we substitute this into the equation.Thus,$K.E._{molar} = \frac{3}{2} RT$.

The translational kinetic energy of gas molecules for one mole of the gas is equal to

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