91影视

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 ,

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

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

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,

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

b.

Volume in number density,

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

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

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

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" ${\mathrm{d}}_2=500路\frac{1773}{293}=3000\mathrm{m}$