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## Q. 9.1

Compute the mean translational kinetic energy of (a) a single ideal gas molecule in eV, and (b) a mole of ideal gas in J, both at room temperature 293 K.

Strategy The mean kinetic energy of a single molecule is given by Equation (9.8). For an entire mole, it’s necessary to multiply the result by Avogadro’s number $N_{A}$, the number of atoms in a mole.

$\overline{K}=\frac{\overline{1m v^{2}} }{2}=\frac{1}{2} m\left(\overline{v_{x}^{2}}+\overline{v_{y}^{2}}+\overline{v_{z}^{2}}\right)=\frac{1}{2} m\left(\frac{3 k T}{m}\right)=\frac{3}{2} k T$ (9.8)

## Verified Solution

(a) For a single molecule, Equation (9.8) gives

$\bar{K}=\frac{3}{2} k T=\frac{3}{2}\left(1.38 \times 10^{-23} J / K \right)(293 K )=6.07 \times 10^{-21} J$

Because of the small size of this energy, it may be useful to convert to units of electron volts.

$\bar{K}=6.07 \times 10^{-21} J \times \frac{1 eV }{1.60 \times 10^{-19} J }=0.038 eV$

The mean molecular energy at room temperature is about 1/25 eV. This is a good number to remember, because this computation applies for any ideal gas at room temperature.

(b) For an entire mole, the mean kinetic energy is a factor of $N_{A}$ larger:

\begin{aligned}\bar{K} &=6.07 \times 10^{-21} J \times N_{ A } \\&=\left(6.07 \times 10^{-21} J \right)\left(6.02 \times 10^{23}\right)=3650 J\end{aligned}

for one mole.