For a room$3.0m$tall, by roughly what percent does the probability of an air molecule being found at the ceiling differ from that of an equal speed molecule being found at the floor? Ignore any variation in temperature from floor to ceiling.

Boltzmann probability distribution:

$P\left(E_n\right)=A{e}^{-E_n/k_{B}T},$

where is a normalization constant, and T is temperature. This corresponds to the probability of finding a particle at an energy state$E_n$ for a system of N particles at temperature T .

To tackle this problem, we will be using the Boltzmann probability distribution:

$P\left(E_n\right)=A{e}^{-E_n/k_{B}T},$ ……. (1)

Under the influence of gravity, the total energy of an air molecule is the sum of its kinetic energy K and gravitational potential energy $U_g$:

$E=K+U_g=\frac{1}{2}m{v}^2+mgy$ ……. (2)

where m is the mass of the molecule, V is its speed, and Y is its position.

Using eq. (2), the total energy of an air molecule located at the floor must be:

$E_{y=0}=\frac{1}{2}m{v}^2.$

For an air molecule at a height $y=3\text{m}$, the total energy is:

$E_{y=3\text{m}}=\frac{1}{2}m{v}^2+3mg.$

The ratio of the probability of finding an air molecule at the height to that of another air molecule at the floor with the same speed is

$\begin{array}{rcl}\frac{P_{y=3m}}{P_{y=0}}& =& \frac{A{e}^{-\left(\frac{1}{2}m{v}^2+3mg\right)/k_{B}T}}{A{e}^{-\frac{1}{2}m{v}^2/kB}T}\\ & =& e^{-\left(\frac{1}{2}m{v}^2+3mg-\frac{1}{2}m{v}^2\right)/k_{B}T}\\ & =& e^{-3mg/k_{B}T}\\ & =& e^{-3\left(5.6\times {10}^{-26}\right)(9.8)/\left(1.38\times {10}^{-28}\right)\left(300\right)}\\ & =& 0.9996.\\ & & \end{array}$

To find the relative difference in percentage, we subtract the above value from 1 and multiply by ,$100\%$ we get:

$\begin{array}{rcl}\text{percent difference}& =& (1-0.9996)\times 100\%\\ & =& 0.04\%\\ & & \end{array}$

Hence, the percentage of the relative difference between the probability of finding an air molecule at height $y=3m$to that of another air molecule with the same speed at the floor is roughly $0.04\%$.