The pressure inside a tank of neon is $150atm$. The temperature is $25°C$. On average, how many atomic diameters does a neon atom move between collisions?

Given :

The pressure inside a tank of neon is : $150atm$.

The temperature is : $25°C$

Theory used :

The number density of a system is defined as the number of atoms or molecules per cubic meter, and it determines how closely the atoms are bound.

The ideal gas law is $$\begin{gathered} pV=N{k}_{B}T \\ \Rightarrow \frac{N}{V}=\frac{p}{k_{B}T} \end{gathered}$$

Boltzmann's constant,$k_B$, has a value of$1.38\times {10}^{-23}J/K$in SI units.

Because the pressure is expressed in atm, we must convert it to Pascal using the formula

$p=(150atm)\frac{1.013\times {10}^{5}Pa}{1atm}=152\times {10}^{5}Pa$

Conversion between the Celsius and Kelvin scales is given by :

$$\begin{gathered} T_K=T_c+273=25°C+273 \\ =298K \end{gathered}$$

To get number density number by number density, we plug the values for $p,T,\mathrm{and}{k}_B$into equation :

$$\begin{gathered} numberdensity=\frac{p}{k_{B}T} \\ =\frac{152x{10}^{5}Pa}{(1.38x{10}^{-23}J/K)(298K)} \\ =3.7x{10}^{27}{m}^{-3} \end{gathered}$$

When a molecule in a gas travels a long distance, it collides with a lot of other molecules. The molecules take a long time to disperse to a new place as a result of these collisions. The mean free path $\lambda$is defined as the average distance travelled by a molecule between collisions, and it is given by $\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}$.

The radius of the neon atom is denoted by$r$. Due to the fact that neon is a monatomic gas, its radius is$r=0.5x{10}^{-10}m$. We get :

$$\begin{gathered} \lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2} \\ =\frac{1}{4\sqrt{2}\pi (3.7x{10}^{27}{m}^{-3}){(0.5x{10}^{-10}m)}^2} \\ =61x{10}^{-10}m \end{gathered}$$

Because neon is a monatomic gas, it has a diameter of$d=1.0x{10}^{-10}m$, the number of atomic diameters it moves is :

$$\begin{gathered} N=\frac{61x{10}^{-10}m}{1.0x{10}^{-10}m} \\ =\boxed{\mathbf{61}} \end{gathered}$$