Translational Kinetic Energy — Question 5

Question 5

Opposing bulk flows can become random translational motion

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). Two equal monatomic ideal gas portions each have temperature 300 K and molar mass .0200 kg/mol. Their center-of-mass velocities are $+600$ and $-600$ m/s. They mix in a rigid insulated enclosure and equilibrate with zero final bulk velocity. Neglect other energy stores and wall energy uptake.

1. Find the initial bulk kinetic energy per mole of total gas and the final temperature.

2. If both portions initially moved at $+600$ m/s instead and the enclosure moved with them, would their common drift produce the same heating during internal equilibration?

Question 5 - Solution

Opposing bulk flows can become random translational motion

(a) Method and calculation

$$E_{\mathrm{bulk}}/n=\tfrac12Mu^2=\tfrac12(.0200)(600)^2=3600\,\mathrm{J/mol}.$$$$\tfrac32R(T_f-300)=3600\ \Rightarrow\ T_f=300+\frac{2400}{8.314}\approx588.670\,\mathrm K.$$

(b) Interpretation and further calculation

For common drift, the center-of-mass frame has no relative bulk motion:$$\mathbf v_i=\mathbf u+\mathbf c_i,\quad K=\tfrac12M_{\mathrm{total}}u^2+K_{\mathrm{relative}}.$$$$\Delta K_{\mathrm{bulk,relative}}=0\ \Rightarrow\ T_f=300\,\mathrm K.$$Uniform translation of the entire isolated apparatus is not a source of internal heating.

Physical check

Heating follows loss of relative organized motion, not a change of reference frame.