Solution: Temperature of a Gas in a Moving Container
Step 1: Energy Conservation Principle
When the container moving with velocity $v$ stops suddenly, the ordered kinetic energy of the gas (due to the container’s motion) is converted into disordered internal energy, raising the temperature of the gas.
where $M$ is the molar mass (assuming 1 mole for calculation), and $f$ is the degrees of freedom.
Step 2: Analysis of Degrees of Freedom
The problem states that rotational degrees of freedom unfroze at temperature $T_0$.
- For $T < T_0$: Gas behaves as monoatomic, $f = 3$.
- For $T > T_0$: Gas behaves as diatomic (rigid rotor), $f = 5$.
Step 3: Solving for Different Velocity Ranges
Case I: Low Velocity ($T_f < T_0$)
The temperature rises but remains below the threshold $T_0$. The gas remains effectively monoatomic ($f=3$).
$$ \frac{3}{2} R T_f = \frac{3}{2} R T_i + \frac{1}{2} M v^2 $$
$$ T_f = T_i + \frac{M v^2}{3 R} $$
This is valid for $v \le \sqrt{\frac{3R(T_0 – T_i)}{M}}$.
Case II: High Velocity ($T_f > T_0$)
The final temperature exceeds $T_0$. In the final state, the gas has 5 degrees of freedom.
Applying energy conservation between the initial state ($f=3$) and final state ($f=5$):
$$ \frac{5}{2} R T_f = \frac{3}{2} R T_i + \frac{1}{2} M v^2 $$
$$ 5 R T_f = 3 R T_i + M v^2 $$
$$ T_f = \frac{3 T_i}{5} + \frac{M v^2}{5 R} $$
Case III: Intermediate Velocity
There is a transition range where the energy supplied is just enough to excite the rotational modes at $T_0$. In this regime, the temperature remains effectively at $T_0$.
$$ T_f = T_0 $$