Translational Kinetic Energy — Question 3

Question 3

Thermal contact conserves the sum of molecular energy contributions

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 monatomic ideal gas samples occupy separate rigid compartments joined by a conducting wall. Amounts are .300 mol and .700 mol; initial temperatures are 500 K and 200 K. The outer enclosure is insulated, and the wall has negligible heat capacity. No molecules cross the wall.

1. Derive the final equilibrium temperature from molecular translational energy conservation.

2. Find the energy transferred from the initially hot sample and verify the gain of the cold sample.

Question 3 - Solution

Thermal contact conserves the sum of molecular energy contributions

(a) Method and calculation

$$\tfrac32R(n_AT_A+n_BT_B)=\tfrac32R(n_A+n_B)T_f.$$$$T_f=\frac{.300(500)+.700(200)}{1.00}=290\,\mathrm K.$$

(b) Interpretation and further calculation

$$Q_{A\to B}=\tfrac32(.300)(8.314)(500-290)=785.673\,\mathrm J.$$$$\Delta U_B=\tfrac32(.700)(8.314)(290-200)=785.673\,\mathrm J.$$The temperature average is weighted by molecular amount because the monatomic translational capacities per mole are equal.

Physical check

The final temperature is closer to the initial temperature of the larger molecular population.