Question

On earth, STP is based on the average atmospheric pressure at the surface and on a phase change of water that occurs at an easily produced temperature, being only slightly cooler than the average air temperature. The atmosphere of Venus is almost entirely carbon dioxide $\left({\mathrm{CO}}_2\right)$, the pressure at the surface is a staggering $93\mathrm{atm}$, and the average temperature is localid="1648638013375" $470°\mathrm{C}$. Venusian scientists, if they existed, would certainly use the surface pressure as part of their definition of STP. To complete the definition, they would seek a phase change that occurs near the average temperature. Conveniently, the melting point of the element tellurium is localid="1648638019185" $450°\mathrm{C}$. What are (a) the rms speed and (b) the mean free path of carbon dioxide molecules at Venusian STP based on this phase change in tellurium? The radius of a ${\mathrm{CO}}_2$molecule islocalid="1648638027654" $1.5\times {10}^{-10}\mathrm{m}$.

Short answer

a.The root mean square speed is$638\mathrm{m}/\mathrm{s}$.

b.Mean free path of carbon dioxide is$2.6\mathrm{nm}$.

Step-by-step solution

Calculation for root mean square speed (part a)

a.

The average translational kinetic energy is

${\mathrm{ϵ}}_{\text{avg}}=\frac{3}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}.......1$

${\mathrm{ϵ}}_{\mathrm{avg}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{rms}}^2.......2$

Equating $1$and $2$

$\frac{1}{2}{\mathrm{mv}}_{\mathrm{rms}}^2=\frac{3}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}$

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

The molecular mass of carbon dioxide is$\mathrm{m}=40\mathrm{u}$.

Converting this to$\mathrm{kg}$,

we get the mass of one molecule of oxygen by,

$\mathrm{m}=40\mathrm{u}\times \frac{1.66\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}=73\times {10}^{-27}\mathrm{kg}....3$

${\mathrm{T}}_{\mathrm{K}}={\mathrm{T}}_{\mathrm{C}}+273$

$=450°\mathrm{C}+273$

$=723\mathrm{K}$

Substitute all values in below equation,

localid="1648638355268" ${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{m}}}$

localid="1648638366804" $=\sqrt{\frac{3\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(723\mathrm{K})}{73\times {10}^{-27}\mathrm{kg}}}$

localid="1648638394867" role="math" ${\mathrm{v}}_{\mathrm{rms}}=638\mathrm{m}/\mathrm{s}$

Calculation for NV (part b)

(b) The ideal gas law,

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

$\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{p}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}.........4$

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

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

The pressure is given in localid="1648638535433" $\mathrm{atm}$.

So we need to convert it into Pascal by,

localid="1648638696043" $\mathrm{p}=(93\mathrm{atm})\left(\frac{1.013\times {10}^5\mathrm{Pa}}{1\mathrm{atm}}\right)=94.2\times {10}^5\mathrm{Pa}$

localid="1648638706189" $\frac{\mathrm{N}}{\mathrm{V}}=\frac{94.2\times {10}^5\mathrm{Pa}}{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(723\mathrm{K})}$

localid="1648638716592" $\frac{\mathrm{N}}{\mathrm{V}}=9.4\times {10}^{26}{\mathrm{m}}^{-3}$

Calculation for mean free path part (b) solution

b.

The average distance traveled by the molecule between collisions is known as the mean free path.

Substitute all values,

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

$=\frac{1}{4\sqrt{2}\mathrm{\pi }\left(9.4\times {10}^{26}{\mathrm{m}}^{-3}\right){\left(1.5\times {10}^{-10}\mathrm{m}\right)}^2}$

$\lambda =2.6\mathrm{nm}$