Chapter 16, Problem 16A.1ST

Interpretation Introduction

Interpretation:

The time taken by the $200\text{mg}$ of $\text{Cs}$ atoms to effuse out of the oven is to be calculated.

Concept introduction:

Effusion is the motion of the gas particles inside a chamber through a hole. The rate of effusion according to Graham’s Law is inversely proportional to the mass of the gas particles raised to the power half or square root of the mass.

Answer to Problem 16A.1ST

The time taken by the $200\text{mg}$ of $\text{Cs}$ atoms to effuse out of the oven is $8.77\times {10}^3\text{s}$.

Explanation of Solution

The formula to calculate the rate of effusion is shown below.

    $\text{Rate}\text{of}\text{effusion}=\frac{p{A}_{°}{N}_{\text{A}}}{\sqrt{2\pi MRT}}$        (1)

Where,

• $p$ is the vapor pressure.
• $A_{°}$ is the area of the hole.
• $M$ is the molar mass of gas in kilograms.
• $N_{\text{A}}$ is the Avogadro’s number.
• $R$ is the gas constant.
• $T$ is the temperature.

The change in mass of atoms effuse out is given by the formula shown below.

    $\Delta m=\frac{p{A}_{°}{M}^{1/2}}{\sqrt{2\pi RT}}\times \Delta t$        (2)

Where,

• $\Delta t$ is the time taken by the $\Delta m$ mass of atoms to effuse out.

The given data and some known values are stated below.

The value of the diameter of the hole is $1.0\text{mm}$.

The value of $T$ is given $500°\text{C}$.

The value of $R$ is $8.314\text{J}{\text{K}}^{-\text{ 1}}{\text{mol}}^{-1}$.

The molar mass $\left(M\right)$ of cesium in kilograms is $\text{132}\text{.91}\times {\text{10}}^{-3}\text{kg}{\text{mol}}^{-1}$.

The value of $N_{\text{A}}$ is $6.022\times {10}^{23}{\text{mol}}^{-1}$.

The value of $p$ is given $16\text{Pa}$.

The value of $\Delta m$ is given $200\text{mg}$.

The area of the hole is given by the formula shown below.

    $A_{°}=\frac{\pi d^2}{4}$        (3)

Where,

• $d$ is the diameter of the hole.

Substitute the value of diameter in the equation (3) to find out area as shown below.

    $\begin{array}{c}{A}_{°}=\frac{\pi d^2}{4}\\ =\frac{3.14\times {\left(1.0\text{mm}\right)}^2}{4}\\ =\frac{3.14}{4}{\text{mm}}^2\\ =0.785\times {10}^{-6}{\text{m}}^2\left(\because 1\text{mm}={10}^{-3}\text{m}\right)\end{array}$

Substitute the value of $\Delta m,M,T,p,R$, and $A_{°}$ in equation (2) as shown below.

    $\begin{array}{l}200\text{mg}=\frac{16\text{Pa}\left(0.785\times {10}^{-6}{\text{m}}^2\right){\left(\text{132}\text{.91}\times {\text{10}}^{-3}\text{kg}{\text{mol}}^{-1}\right)}^{1/2}}{\sqrt{2\pi \left(8.314\text{J}{\text{K}}^{-\text{ 1}}{\text{mol}}^{-1}\right)\left(\left(500+273.15\right)\text{K}\right)}}\times \Delta t\\ 200\text{mg}=\frac{16\text{Pa}\left(0.785\times {10}^{-6}{\text{m}}^2\right)\left(\text{3}\text{.65}\times {\text{10}}^{-1}{\text{kg}}^{1/2}{\text{mol}}^{-1/2}\right)}{\sqrt{2\pi \left(8.314\text{J}{\text{K}}^{-\text{ 1}}{\text{mol}}^{-1}\right)\left(773.15\text{K}\right)}}\times \Delta t\end{array}$

Rearrange above equation to calculate $\Delta t$ as shown below.

    $\begin{array}{c}\Delta t=\frac{200\text{mg}\sqrt{2\pi \left(8.314\text{J}{\text{K}}^{-\text{ 1}}{\text{mol}}^{-1}\right)\left(773.15\text{K}\right)}}{16\text{Pa}\left(0.785\times {10}^{-6}{\text{m}}^2\right)\left(\text{3}\text{.65}\times {\text{10}}^{-1}{\text{kg}}^{1/2}{\text{mol}}^{-1/2}\right)}\\ =\frac{200\text{mg}\sqrt{40367.65\text{J}{\text{mol}}^{-1}}}{45.84\times {10}^{-7}\text{Pa}{\text{m}}^2{\text{kg}}^{1/2}{\text{mol}}^{-1/2}}\\ =\frac{200\text{mg}\left(200.92{\text{J}}^{1/2}{\text{mol}}^{-1/2}\right)}{45.84\times {10}^{-7}\text{Pa}{\text{m}}^2{\text{kg}}^{1/2}{\text{mol}}^{-1/2}}\\ =\frac{200\times {10}^{-6}\text{kg}\left(200.92{\text{J}}^{1/2}{\text{mol}}^{-1/2}\right)}{45.84\times {10}^{-7}\text{Pa}{\text{m}}^2{\text{kg}}^{1/2}{\text{mol}}^{-1/2}}\left(\because 1\text{mg}={10}^{-6}\text{kg}\right)\end{array}$

Simplify the above equation.

    $\begin{array}{c}\Delta t=\frac{200\times {10}^{-6}\text{kg}\left(200.92{\text{kg}}^{1/2}\text{m}{\text{s}}^{-1}{\text{mol}}^{-1/2}\right)}{45.84\times {10}^{-7}\text{kg}{\text{m}}^{-1}{\text{s}}^{-2}{\text{m}}^2{\text{kg}}^{1/2}{\text{mol}}^{-1/2}}\left(\because \text{J}=\text{kg}{\text{m}}^2{\text{s}}^{-2},\text{Pa}=\text{kg}{\text{m}}^{-1}{\text{s}}^{-2}\right)\\ =\frac{4.02\times {10}^{-2}}{45.84\times {10}^{-7}}\text{s}\\ =0.0877\times {10}^5\text{s}\\ =8.77\times {10}^3\text{s}\end{array}$

Therefore, the time taken by the $200\text{mg}$ of $\text{Cs}$ atoms to effuse out of the oven is $8.77\times {10}^3\text{s}$.