Average Kinetic Energy Per Mole

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An ideal monatomic gas is held at a uniform temperature of 350 K; calculate the average translational kinetic energy associated with one mole of the gas.

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Depending only on temperature, the average translational kinetic energy of one mole of an ideal gas equals $\overline{KE}=\frac{3}{2}RT$, since the three translational degrees of freedom each contribute $\frac{1}{2}RT$ per mole. Inserting $R=8.314$ J/mol\cdotK and $T=350$ K yields $\overline{KE}=\frac{3}{2}	imes8.314	imes350\approx4.36	imes10^{3}$ J. Importantly, this is the translational part only, and it is identical for a monatomic gas and for any other ideal gas at the same temperature, even though diatomic and polyatomic gases additionally store energy in rotation and vibration. The value $2.91	imes10^{3}$ J uses $RT$ without the factor $3/2$. The value $7.27	imes10^{3}$ J doubles the answer by misreading the temperature as 700 K. The value $1.45	imes10^{3}$ J halves the correct coefficient. This is the molar form of the NCERT kinetic interpretation of temperature, which establishes translational kinetic energy as a universal function of temperature alone. The result also explains why two gases at the same temperature exchange no net translational energy on contact, which is precisely the condition that defines thermal equilibrium. As a magnitude check, multiplying the per-molecule energy $\frac{3}{2}k_BT$ by Avogadro's number reproduces $\frac{3}{2}RT$, confirming internal consistency and the value $4.36	imes10^{3}$ J.

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