Question

During a physics experiment, helium gas is cooled to a temperature of $10K$at a pressure of $0.10\text{atm.}$ What are (a) the mean free path in the gas, (b) the rms speed of the atoms, and (c) the average energy per atom?

Short answer

(a) The mean free path in gas is $\lambda =310\mathrm{nm}$

(b) The rms speed of atoms is $V_{\mathrm{rms}}=250\mathrm{m}/\mathrm{s}$

(c) The average energy per atom is ${\epsilon }_{\mathrm{avg}}=2.1\times {10}^{-22}\mathrm{J}$

Step-by-step solution

Step : 1 Introduction (part a)

(a) The thermodynamics depicts the link between the four state variables for an ideal gas, which are the container volume, the volume of the gas, the pressure$p$that the gas is exerted, temperature $T$of the gas and the number of moles $n$of the gas in the container. Ideal gas law is given by equation $18.13$in the form

$\begin{array}{r}pV=N{k}_{\mathrm{B}}T\\ \frac{N}{V}=\frac{p}{k_{\mathrm{B}}T}\end{array}$

Where $k_{\mathrm{B}}$is Boltzmann's constant and in SI unit its value is

$k_{\mathrm{B}}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

The pressure is given in atm, so we need to convert it into Pascal by

$p=\left(0.10\mathrm{atm}\right)\left(\frac{1.013\times {10}^5\mathrm{Pa}}{1\mathrm{atm}}\right)=1.013\times {10}^4\mathrm{Pa}$

We plug the values for $P,T$and $k_B$into equation $\left(1\right)$to get number density number by

Number density $=\frac{p}{k_{\mathrm{B}}T}$

$=\frac{1.013\times {10}^4\mathrm{Pa}}{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(10\mathrm{K}\right)}$

$=7.34\times {10}^{25}{\mathrm{m}}^{-3}$

Step :2 Explanation 

If a component in a vapour 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 average distance that the molecule moves between the collisions is called the mean free path $\lambda$and it is given by equation $20.3$in the form

$\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}$

Where $r$is the radius of the molecule, $N$is the number of molecules and $V$is the volume. Helium is a monatomic gas, so its radius is $r=0.5\times {10}^{-10}\mathrm{m}$. We plug the values for $(N/V)$and $r$into equation (2) to get $\lambda$by

$\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}$

$=\frac{1}{4\sqrt{2}\pi \left(7.34\times {10}^{25}{\mathrm{m}}^{-3}\right){\left(0.5\times {10}^{-10}\mathrm{m}\right)}^2}$

$=310\times {10}^{-9}\mathrm{m}$

$=310\mathrm{nm}$

Step :3  Kinetic energy (part b)

(b) The molecule with mass $m$and velocity $v$has an average translational kinetic energy and it is given by equation (20.19)in the form

${ϵ}_{\mathrm{avg}}=\frac{1}{2}m{v}_{\mathrm{rms}}^2$

The change in the temperature of the molecule affects its average translational kinetic energy, so it is related to the temperature $T$per molecule in the form

${ϵ}_{\mathrm{avg}}=\frac{3}{2}{k}_{\mathrm{B}}T$

As shown, both equations $\left(3\right)$and $\left(4\right)$have the same left side, so we can use these expressions to get an equation for root mean square velocity $v_{\mathrm{rms}}$

$\begin{array}{c}\frac{1}{2}m{v}_{\mathrm{rms}}^2=\frac{3}{2}{k}_{\mathrm{B}}T\\ v_{\mathrm{rms}}^2=\frac{3k_{\mathrm{B}}T}{m}\\ v_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm{B}}T}{m}}\end{array}$

The molecular mass of helium is $m=4U$. Converting this to $kg$, we get the mass of one atom of argon by

$m=4\mathrm{u}\times \frac{1.66\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}=6.64\times {10}^{-27}\mathrm{kg}$

Now, we plug the values for$k_{\mathrm{B}}$, $T$and $m$into equation $\left(3\right)$to get ${\mathcal{V}}_{\mathrm{rms}}$

$v_{\mathrm{rms}}=\sqrt{\frac{3k\mathrm{B}T}{m}}$

$=\sqrt{\frac{3\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(10\mathrm{K}\right)}{\left(6.64\times {10}^{-27}\mathrm{kg}\right)}}$

$=250\mathrm{m}/\mathrm{s}$

Step :4 Energy for helium (part c)

(c) Plug the values for $k_{\mathrm{B}}\text{and}T$into equation $\left(4\right)$to get the energy for helium

${ϵ}_{\mathrm{avg}}=\frac{3}{2}{k}_{\mathrm{B}}T$

$=\left(\frac{3}{2}\right)\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(10\mathrm{K}\right)$

$=2.1\times {10}^{-22}\mathrm{J}$