Q.Calculate the total translational kinetic energy of $1\ \text{mol}$ of an ideal gas at $27\,^\circ\text{C}$. Take $k_B = 1.38 \times 10^{-23}\ \text{J/K}$ and Avogadro's number $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$.

The total kinetic energy of an ideal gas equals $\tfrac{3}{2}PV$, which can be evaluated directly at STP without knowing the number of moles.

For an ideal gas, total translational kinetic energy $K = \tfrac{3}{2}nRT$. Since $PV = nRT$,

$K = \dfrac{3}{2}PV$

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Each translational degree of freedom carries average energy $\tfrac12 K_BT$; a gas molecule has 3 translational degrees of freedom, giving $\tfrac32 K_BT$ per molecule.

From the kinetic theory of gases, the mean square speed of gas molecules is related to pressure and volume by $PV = \tfrac13 Nm\overline{c^2}$, and using the ideal gas law $PV = NK_BT$ (for $N$ molecules), we get

$\tfrac13 m\overline{c^2} = K_BT \;\Rightarrow\; \tfrac12 m\overline{c^2} = \tfrac32 K_BT.$ …