Question

Photons of light scatter off molecules, and the distance you can see through a gas is proportional to the mean free path of photons through the gas. Photons are not gas molecules, so the mean free path of a photon is not given by Equation$20.3$, but its dependence on the number density of the gas and on the molecular radius is the same. Suppose you are in a smoggy city and can barely see buildingslocalid="1648634576764" role="math" $500$away.

a. How far would you be able to see if all the molecules around you suddenly doubled in volumelocalid="1648634590441" $?$

b. How far would you be able to see if the temperature suddenly rose from ${20}^{°}\mathrm{C}$to a blazing hot${500}^{°}\mathrm{C}$with the pressure unchanged?

Short answer

a.The distance for the molecules around you suddenly doubled in volume is$300\mathrm{m}$.

b.The distance for the temperature suddenly rose from${20}^{°}\mathrm{C}$to a blazing hot${1500}^{°}\mathrm{C}$with the pressure unchanged is$3000\mathrm{m}$.

Step-by-step solution

Calculation for distance of the molecules around you suddenly doubled in volume  (part a)

a.

volume of the space is $\mathrm{V}$.

where $\mathrm{N}$particles of dust are scattered in.

If we assume a spherical shape.We have ,

${\mathrm{V}}_2=2{\mathrm{V}}_1$

$\frac{4}{3}{\mathrm{\pi r}}_2^3=2\frac{4}{3}{\mathrm{\pi r}}_1^3$

The new radius is ,

${\mathrm{r}}_2={\mathrm{r}}_1\sqrt[3]{2}$

The radius's dependence on the mean free path, on the other hand, is inversely proportional to its square.

Then the new distance will be,

${\mathrm{d}}_2={\mathrm{d}}_1\frac{1}{\sqrt[3]{2}{)}^2}={\mathrm{d}}_1{2}^{-\frac{2}{3}}$

Substituting the first distance,

${\mathrm{d}}_2=500·2^{-\frac{2}{3}}=310\mathrm{m}$

Calculation for distance of temperature changed (part b)

b.

Volume in number density,

$\mathrm{pV}={\mathrm{Nk}}_{\mathrm{B}}\mathrm{T}\Rightarrow \frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{p}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}$

Substitute in the expression for the mean free path, we will have

$\mathrm{\lambda }\propto \frac{1}{\left(\mathrm{p}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\right){\mathrm{r}}^2}=\frac{{\mathrm{k}}_{\mathrm{B}}}{{\mathrm{pr}}^2}\mathrm{T}$

This means that the distance is proportional to the temperature.

$d_2=d_1\frac{T_2}{T_1}$localid="1648635077835" ${\mathrm{d}}_2={\mathrm{d}}_1\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}$

The temperature was increased fromlocalid="1648635092520" role="math" $293\mathrm{K}$tolocalid="1648635099956" $1773\mathrm{K}$.

The new line of sight will be,

localid="1648635112249" ${\mathrm{d}}_2=500·\frac{1773}{293}=3000\mathrm{m}$