The translatory kinetic energy of a gas per g | vedclass

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(B) The translatory kinetic energy of $n$ moles of an ideal gas is given by $E = \frac{3}{2}nRT$.For $1 \ gm$ of gas,the number of moles $n = \frac{1}

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$\frac{3}{2}\frac{RT}{M}$

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(B) The translatory kinetic energy of $n$ moles of an ideal gas is given by $E = \frac{3}{2}nRT$.For $1 \ gm$ of gas,the number of moles $n = \frac{1}{M}$,where $M$ is the molar mass of the gas.Substituting this into the energy equation,we get the kinetic energy per gram as $E_{gm} = \frac{3}{2} \left( \frac{1}{M} \right) RT = \frac{3}{2} \frac{RT}{M}$.

The translatory kinetic energy of a gas per $gm$ is

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