Question

The number density in a container of neon gas is $5.00\times {10}^{25}{m}^{-3}$. The atoms are moving with an rms speed of $660m/s$. What are

(a) the temperature and

(b) the pressure inside the container?

Short answer

(a) The temperature is$\mathbf{351}\mathbf{}\mathit{K}$

(b) The pressure inside the container is$\mathbf{241}\mathbf{}\mathit{P}\mathit{a}$

Step-by-step solution

Given information and formula used

Given :

The number density of neon gas : $5.00\times {10}^{25}{m}^{-3}$

rms speed : $660m/s$

Theory used :

The pressure is calculated using equation :

$p=\frac{1}{3}\frac{N}{V}m{v^2}_{rms}$

and is dependent on the quantity of molecules in the container and their velocity.

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 is $k_B$, and its SI unit value is$k_B=1.38\times {10}^{-23}J/K$

Calculating the temperature 

(a)The pressure on the container's wall is caused by all of the molecular collisions.

Solving the equation $p=\frac{1}{3}\frac{N}{V}m{v^2}_{rms}$for $T$we have :

$$\begin{gathered} p=\frac{1}{3}\left(\frac{p}{k_{B}T}\right)m{v^2}_{rms} \\ \Rightarrow T=\frac{1}{3}\left(\frac{m{v^2}_{rms}}{k_B}\right) \end{gathered}$$

Neon has a molecular mass of $m=20u$. Converting this to kg, we get the mass of one atom of neon:

$$\begin{gathered} m=20u\times \frac{1.66x{10}^{-27}kg}{1u} \\ =33.2x{10}^{-2}kg \end{gathered}$$

We now input the numbers for $m,k_B,\mathrm{and}{v}_{rms}$into this equation to get $T$

$$\begin{gathered} T=\frac{1}{3}\left(\frac{m{v^2}_{rms}}{k_B}\right) \\ =\frac{1}{3}\left(\frac{(33.2x{10}^{-27}kg){(660m/s)}^2}{1.38x{10}^{-23}J/K}\right) \\ =\mathbf{}\mathbf{351}\mathbf{}\mathit{K} \end{gathered}$$

Calculating the pressure inside the container 

(b) In part (a), we calculated the pressure within the container. By plugging in the values of $\frac{N}{V},m,\mathrm{and}{v}_{rms}$into the equation, we can get:

$$\begin{gathered} p=\frac{1}{3}\frac{N}{V}m{v^2}_{rms} \\ =\frac{1}{3}(5\times {10}^{25}{m}^{-3})(33.2\times {10}^{-27}kg){(660m/s)}^2=241\times {10}^{3}kPa \\ =\mathbf{241}\mathbf{}\mathit{P}\mathit{a} \end{gathered}$$