Translational Kinetic Energy — Question 2

Question 2

Equal masses at different temperatures can have equal total kinetic energy

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). Sample A has mass 4.00 g and molar mass 4.00 g/mol. Sample B has mass 4.00 g and molar mass 20.0 g/mol. A is at 100 K.

1. Find the temperature of B that makes total translational energies equal.

2. At that condition, compare mean energies per molecule and rms speeds. Explain how these comparisons coexist with equal total energy.

Question 2 - Solution

Equal masses at different temperatures can have equal total kinetic energy

(a) Method and calculation

$$n_A=1.00\,\mathrm{mol},\quad n_B=.200\,\mathrm{mol}.$$$$n_AT_A=n_BT_B\ \Rightarrow\ T_B=500\,\mathrm K,\quad U_A=U_B=1247.10\,\mathrm J.$$

(b) Interpretation and further calculation

$$\overline K_B/\overline K_A=T_B/T_A=5.$$$$\frac{v_{{\mathrm{rms}},B}^2}{v_{{\mathrm{rms}},A}^2}=\frac{T_B/M_B}{T_A/M_A}=1.$$B has one-fifth as many molecules, each with five times the mean energy; its molecular mass is also five times larger, giving the same rms speed.

Physical check

Total translational energy depends on molecule count, whereas equal total mass alone does not determine it.