91Ó°ÊÓ

a.

The number density is,

$\text{number density}=\frac{\mathrm{N}}{\mathrm{V}}.............1$

Ideal gas law is given by ,

$\mathrm{pV}={\mathrm{Nk}}_{\mathrm{B}}\mathrm{T}$

$\mathrm{p}=\left(\frac{\mathrm{N}}{\mathrm{V}}\right){\mathrm{k}}_{\mathrm{B}}\mathrm{T}$

Where ${\mathrm{k}}_{\mathrm{B}}$is Boltzmann's constant .

In $\mathrm{Sl}$unit its value is,

${\mathrm{k}}_{\mathrm{B}}=1.38×{10}^{-23}\mathrm{J}/\mathrm{K}$

The number of density is ,

$\mathrm{N}/\mathrm{V}=1\mathrm{atom}/{\mathrm{cm}}^3={10}^6\mathrm{atom}/{\mathrm{m}}^3$.

The values of$(\mathrm{N}/\mathrm{V}),$$\mathrm{T}$and$k_B$into equation$1$to get$\mathrm{p}$in$\mathrm{Pascal}$

$\mathrm{P}=\left(\frac{\mathrm{N}}{\mathrm{V}}\right){\mathrm{k}}_{\mathrm{B}}\mathrm{T}...........2$

$=\left({10}^6\mathrm{atom}/{\mathrm{m}}^3\right)\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)(3\mathrm{K})$

$=4.14×{10}^{-17}\mathrm{Pa}$

Pressure of atom in localid="1648636270485" $\mathrm{atm}$,

localid="1648636286605" $\mathrm{p}=\left(4.14×{10}^{-17}\mathrm{Pa}\right)\left(\frac{1\mathrm{atm}}{1.013×{10}^5\mathrm{Pa}}\right)=4.1×{10}^{-22}\mathrm{atm}$

b.

The average translational kinetic energy is,

${\mathrm{Ï\mu }}_{\mathrm{avg}}=\frac{3}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}.....3$

${\mathrm{Ï\mu }}_{\mathrm{avg}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{rms}}^2....4$

As shown, both equations$3$and$4$have the same left side,

Use these expressions to get an equation for root mean square velocity, $\frac{1}{2}{\mathrm{mv}}_{\mathrm{rms}}^2=\frac{3}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}$

${\mathrm{v}}_{\mathrm{rms}}^2=\frac{3\mathrm{k}\mathrm{BT}}{\mathrm{m}}$

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{m}}}............5$

The mass of the hydrogen atom is the same for the proton is,${\mathrm{m}}_{\mathrm{p}}=1.67×{10}^{-27}\mathrm{kg}$.

Substitute the values for${\mathrm{k}}_{\mathrm{B}}$,$\mathrm{T}$and ${\mathrm{m}}_{\mathrm{p}}$into equation $5$to get ${\mathrm{v}}_{\text{rms}}$,

$v_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm{B}}T}{m_{\mathrm{p}}}}$

localid="1648637031505" $=\sqrt{\frac{3\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)(3\mathrm{K})}{1.67×{10}^{-27}\mathrm{kg}}}$

localid="1648637058825" $v_{rms}=270\mathrm{m}/\mathrm{s}$

c.

Translational kinetic energy is the kinetic energy of a monatomic gas.

${\mathrm{E}}_{\mathrm{th}}=\frac{3}{2}{\mathrm{Nk}}_{\mathrm{B}}\mathrm{T}...........6$

Where $\mathrm{N}$is the number of particles

$\mathrm{R}$is the universal gas constant.

$\frac{\mathrm{N}}{\mathrm{V}}={10}^{6}atom/{\mathrm{m}}^3$

$\mathrm{N}={10}^6\mathrm{atom}/{\mathrm{m}}^3\left(\mathrm{V}\right)$

The volume is given by $\mathrm{V}=\mathrm{L}×\mathrm{L}×\mathrm{L}={\mathrm{L}}^3$.

So, the number of molecules is,

$\mathrm{N}={10}^6{\mathrm{L}}^3$

${\mathrm{E}}_{\mathrm{th}}=\frac{3}{2}{\mathrm{Nk}}_{\mathrm{B}}\mathrm{T}$

${\mathrm{E}}_{\text{th}}=\frac{3}{2}\left({10}^6{\mathrm{L}}^3\right){\mathrm{k}}_{\mathrm{B}}\mathrm{T}$

$\mathrm{L}=\sqrt[3]{\frac{2{\mathrm{E}}_{\text{th}}}{3×{10}^6{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}$

$=\sqrt[3]{\frac{2(1.0\mathrm{J})}{3×{10}^6\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)(3\mathrm{K})}}$

$\mathrm{L}=2.5×{10}^5\mathrm{m}$