Which of the following properties of a gas molecule is directly proportional to the absolute temperature? Also, calculate the final volume of a gas given initial and final temperature and pressure.

Initial volume $V_1 = 370 \text{ L}$.
Initial temperature $T_1 = 16^\circ \text{C} = 16 + 273 = 289 \text{ K}$.
Initial pressure $P_1 = 150 \text{ Pa}$.
Final temperature $T_2 = -21^\circ \text{C} = -21 + 273 = 252 \text{ K}$.
Final pressure $P_2 = 420 \text{ Pa}$.
We need to find the final volume $V_2$.

The Combined Gas Law states $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.
Rearrange the formula to solve for $V_2$:

The volume of the hydrogen gas at $-21^\circ \text{C}$ and $420 \text{ Pa}$ is $\boxed{115.23 \text{ L}}$.

Temperature $T = 0^\circ \text{C} = 0 + 273 = 273 \text{ K}$.
Boltzmann constant $k = 1.38 \times 10^{-23} \text{ J/K}$.
Avogadro's number $N_A = 6.022 \times 10^{23} \text{ molecule/mol}$.
Molar mass of nitrogen $M = 28 \text{ g/mol} = 0.028 \text{ kg/mol}$.

The Universal Gas Constant $R$ can be calculated as $R = k N_A$.

The formula for the root mean square speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.

The root mean square speed of a nitrogen molecule at $0^\circ \text{C}$ is $\boxed{493.06 \text{ m/s}}$.

Mass of nitrogen $m = 2.0 \text{ g}$.
Volume of tube $V = 30 \text{ cm}^3 = 30 \times (10^{-2} \text{ m})^3 = 30 \times 10^{-6} \text{ m}^3$.
Temperature $T = 20^\circ \text{C} = 20 + 273 = 293 \text{ K}$.
Molar mass of nitrogen $M = 28 \text{ g/mol} = 0.028 \text{ kg/mol}$.
Universal Gas Constant $R = 8.314 \text{ J/(mol}\cdot\text{K)}$.

The Ideal Gas Law is $PV = nRT$. Rearrange to solve for $P$:

The nitrogen pressure in the tube when it is warmed to $20^\circ \text{C}$ is $\boxed{5.81 \times 10^6 \text{ Pa}}$.