Translational Kinetic Energy — Question 1

Question 1

Per-molecule energy and sample energy scale differently

For equilibrium translational motion use $\overline K=\tfrac32k_BT$ per molecule and $U_{\mathrm{tr}}=\tfrac32nRT$. Take $k_B=1.381\times10^{-23}$ J/K and $R=8.314$ J/(mol K). A .200 mol ideal gas is at 400 K. Use $1\,\mathrm{eV}=1.602\times10^{-19}$ J.

1. Find mean translational energy per molecule in joules and electronvolts, and total translational energy.

2. If the sample amount doubles at unchanged temperature, state how the mean energy, total energy and rms speed change for the same species.

Question 1 - Solution

Per-molecule energy and sample energy scale differently

(a) Method and calculation

$$\overline K=\tfrac32(1.381\times10^{-23})(400)=8.286\times10^{-21}\,\mathrm J.$$$$\overline K=\frac{8.286\times10^{-21}}{1.602\times10^{-19}}\,\mathrm{eV}\approx.0517228\,\mathrm{eV},\quad U_{\mathrm{tr}}=\tfrac32(.200)(8.314)(400)=997.68\,\mathrm J.$$

(b) Interpretation and further calculation

$$\overline K_{\mathrm{new}}=\overline K,\quad U_{{\mathrm{tr,new}}}=1995.36\,\mathrm J.$$$$v_{\mathrm{rms}}=\sqrt{2\overline K/m}\ \Rightarrow\ v_{{\mathrm{rms,new}}}=v_{\mathrm{rms}}.$$Temperature fixes energy per molecule, while sample amount fixes how many such contributions are added.

Physical check

An electronvolt is an energy unit; a small energy per molecule can sum to a macroscopic total.