Show that, for an ideal gas, $$ P=\frac{1}{3} \rho v_{\mathrm{rms}}^{2} $$ where \(P\) is the pressure, \(\rho\) is the mass density, and \(v_{\mathrm{rms}}\) is the rms speed of the gas molecules.

In the kinetic theory of gases, we make a few simplifying assumptions: 1. Gas molecules move in straight lines with constant average speed and collide elastically. 2. The size of the gas molecules is much smaller than the mean free path between collisions. 3. The forces between gas molecules are negligible except during collisions. These assumptions allow us to use the concepts of momentum and kinetic energy to describe the behavior of an ideal gas.

Consider a cubic container of side length L and a gas molecule with mass m and velocity \(\vec{v}\). When the gas molecule collides with the container wall perpendicular to the x-axis, its momentum changes from \(m\vec{v}\) to \(m\vec{v'}\), where \(\vec{v'}\) is the velocity after the collision. As the collision is elastic, the x-component of momentum will change by \(2m v_x\), and other components (y and z) will remain the same. The time interval between two successive collisions of the same molecule with the same wall can be given as \(\frac{2L}{v_x}\) (molecule travels to opposite wall and comes back). Now, we can calculate the force exerted by the molecule on the container wall using the formula: $$ F = \frac{\Delta p}{\Delta t} = \frac{2mv_x}{\frac{2L}{v_x}} = \frac{mv_x^2}{L} $$

Now, consider N gas molecules in the container. The total force exerted by these molecules on the container wall can be obtained by summing the forces exerted by each molecule: $$ F_{tot} = \sum_{i=1}^{N} \frac{m v_{x,i}^{2}}{L} $$ We also know that the pressure in an ideal gas is given by: $$ P = \frac{F_{tot}}{A} = \frac{F_{tot}}{L^2} $$ where A is the area of the container wall. Thus, we have: $$ P = \frac{F_{tot}}{L^2} = \frac{1}{L^2} \sum_{i=1}^{N} \frac{m v_{x,i}^{2}}{L} = \frac{m}{L^3}\sum_{i=1}^{N} v_{x,i}^{2} $$

The mass density, \(\rho\), is given by: $$ \rho = \frac{m_total}{V} = \frac{mN}{L^3} $$ The RMS speed \(v_{rms}\) is defined as the square root of the average value of the squared speeds of molecules: $$ v_{rms}^2 = \frac{1}{N} \sum_{i=1}^{N} v_i^{2} = \frac{1}{N}\sum_{i=1}^{N} (v_{x,i}^2 + v_{y,i}^2 + v_{z,i}^2) $$ Since the motion of gas molecules is isotropic, we can write: $$ \langle v_{x}^2 \rangle = \langle v_{y}^2 \rangle = \langle v_{z}^2 \rangle = \frac{1}{3}\langle v^2 \rangle = \frac{1}{3} v_{rms}^2 $$ This equation gives us: $$ v_x^2 = \frac{1}{N}\sum_{i=1}^{N} v_{x,i}^2 = \frac{1}{3} v_{rms}^2 $$

Now, we use the equations for pressure, mass density, and RMS speed to show the given relationship: $$ P = \frac{m}{L^3}\sum_{i=1}^{N} v_{x,i}^{2} = \frac{mN}{L^3} \frac{1}{3}v_{rms}^{2} $$ Now, substitute the equation for mass density, $$ \rho = \frac{mN}{L^3} $$ into the equation above, and we get: $$ P =\rho \frac{1}{3}v_{rms}^{2} $$ This proves the desired relationship for an ideal gas.