Solution 26: Sudden Stop of Container
When the container stops, the kinetic energy of the gas ($v$) is converted to internal energy.
For monoatomic gas ($C_v = \frac{3}{2}R$):
$$ \frac{1}{2} (n M_{\text{molar}}) v^2 = n \frac{3}{2} R \Delta T $$
$$ \Delta T = \frac{M_{\text{molar}} v^2}{3R} $$
From Ideal Gas Law, density $\rho = \frac{PM_{\text{molar}}}{RT} \implies \frac{M_{\text{molar}}}{R} = \frac{\rho T}{P}$.
Substituting into $\Delta T$:
$$ \Delta T = \frac{v^2}{3} \left( \frac{\rho T}{P} \right) $$
Final Temperature $T_f = T + \Delta T$:
$$ T_f = T \left[ 1 + \frac{\rho v^2}{3P} \right] $$