Translational Kinetic Energy — Question 6

Question 6

Selective removal of energetic particles cools what remains

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 monatomic gas initially has $N$ particles at temperature $T_i$. An ideal selector removes a fraction $f=.100$ of the particles, whose mean kinetic energy is three times the original mean $\overline K_i$. No work or other energy transfer occurs. The remaining gas then re-equilibrates at fixed volume.

1. Derive the final temperature ratio and evaluate it for $T_i=300$ K.

2. Find the final-to-initial pressure ratio and the largest mathematically allowed $f$ for the stated removed-particle energy. Explain the physical meaning of that limit.

Question 6 - Solution

Selective removal of energetic particles cools what remains

(a) Method and calculation

$$U_f=N\overline K_i-fN(3\overline K_i)=(1-3f)N\overline K_i.$$$$N_f=(1-f)N,\quad\frac{T_f}{T_i}=\frac{1-3f}{1-f}=\frac79\ \Rightarrow\ T_f=233.333\,\mathrm K.$$

(b) Interpretation and further calculation

$$\frac{p_f}{p_i}=\frac{N_fT_f}{NT_i}=1-3f=.700.$$$$U_f\ge0\ \Rightarrow\ f\le\frac13.$$The boundary value would leave no translational energy in the remainder; an actual distribution must supply the assumed energetic subset, so the algebraic bound is necessary rather than a realizability guarantee.

Physical check

Random removal with the original mean energy would preserve temperature; preferential energy removal changes it.