91Ó°ÊÓ

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

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

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

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

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 $0.04\%$.