From the ideal gas equation $PV = NkT$, the number density is $n = \frac{N}{V} = \frac{P}{kT}$. The mass density $\rho$ is mass per unit volume, so $\rho = m \times n = \frac{Pm}{kT}$.

We know $M = \frac{\rho RT}{P}$. Given $T_A = T_B$, $P_A = 2P_B$, and $\rho_A = 1.5\rho_B$. The ratio of molecular weights is $\frac{M_A}{M_B} = (\frac{\rho_A}{\rho_B}) \times (\frac{P_B}{P_A}) = 1.5 \times \frac{1}{2} = 0.75 = \frac{3}{4}$.

Since $\frac{\rho}{P} = \frac{M}{RT}$, the ratio is inversely proportional to the absolute temperature $T$. Thus, $x_2 = x_1 (\frac{T_1}{T_2}) = x \times \frac{10 + 273}{110 + 273} = x \times \frac{283}{383}$.

The number of moles $n = \frac{\text{given mass}}{\text{molar mass}}$. For $5 \text{ g}$ of $O_2$ gas, $n = \frac{5}{32}$. Substituting this into the ideal gas equation $PV = nRT$, we get $PV = (\frac{5}{32})RT$.

The critical temperature $T_c$ for a real gas obeying the Van der Waals equation is theoretically derived as $T_c = \frac{8a}{27Rb}$, where $a$ and $b$ are the Van der Waals constants.

Kinetic energy is directly proportional to the absolute temperature ($E_k = \frac{3}{2}kT$). At absolute zero ($T = 0 \text{ K}$), the translational kinetic energy becomes zero, implying that all molecular motion stops.

An increase in temperature raises the thermal agitation (root-mean-square speed) of gas molecules. This causes them to travel faster, increasing the frequency of their collisions with the container walls.

RMS velocity depends only on temperature: $v_{rms} \propto \sqrt{T}$. Let $v_1$ and $v_2$ be rms speeds at $T_1 = 27^{\circ}\text{C} = 300\text{ K}$ and $T_2 = 127^{\circ}\text{C} = 400\text{ K}$. $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}}$. So $v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}}\text{ ms}^{-1}$.

A monoatomic gas molecule has 3 degrees of freedom (translational). By the law of equipartition of energy, energy per degree of freedom is $\frac{1}{2}k_BT$. Total average thermal energy = $3 \times \frac{1}{2}k_BT = \frac{3}{2}k_BT$.