Chapter 20, Problem 20.73QP

A 14-m by 10-m by 3.0-m basement had a high radon content. On the day the basement was sealed off from its surroundings so that no exchange of air could take place, the partial pressure of ²²²Rn was 1.2 × 10⁻⁶ mmHg. Calculate the number of ²²²Rn isotopes $(t\frac{1}{2}=3.8\text{d)}$ at the beginning and end of 31 days. Assume STP conditions.

Interpretation Introduction

Interpretation:

From the partial pressure of radon, the number of radon are there at beginning and end of $\text{31days}$ has to be calculated.

Concept introduction:

First-Order kinetic reaction:

In first-order reaction, the rate constant depends on concentration of reactant raised to first power.  The mathematical equation of first-order reaction is given as,

$\begin{array}{l}\text{ln}\frac{{\text{[A]}}_{\text{t}}}{{\text{[A]}}_{\text{0}}}\text{=}\text{-kt}\\ \\ {\text{ln[A]}}_{\text{t}}\text{=}\text{-kt}\text{+}{\text{ln[A]}}_{\text{0}}\end{array}$

${\text{[A]}}_{\text{t}}$ is the concentration of A at time t.

${\text{[A]}}_{\text{0}}$ is the concentration of A at time $\text{t}\text{=}\text{0}$

K is rate constant.

Answer to Problem 20.73QP

The number of radon is there at beginning is $\text{1}\text{.8}\text{×}{\text{10}}^{\text{19}}\text{atoms}$

The number of radon are there at end of $\text{31days}$ is $\text{6}{\text{.4×10}}^{\text{16}}\text{Rn}\text{atoms}$

Explanation of Solution

Given,

$\begin{array}{l}{\text{P}}_{\text{Rn}}\text{=}\text{1}\text{.2}\text{×}{\text{10}}^{\text{-6}}\text{mmHg}\\ \\ {\text{t}}_{\text{1/2}}\text{=}\text{3}\text{.8days}\end{array}$

From the half-life period of radon, the rate constant is calculated as

$\begin{array}{l}\text{k}\text{=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}\\ \\ \text{=}\frac{\text{0}\text{.693}}{\text{3}\text{.8days}}\text{=}\text{0}\text{.1824/day}\end{array}$

The volume of basement is calculated as

$\begin{array}{l}\text{volume}\text{=}\text{l}\text{×}\text{b}\text{×}\text{h}\\ \\ \text{=}\text{(14 m ×}\text{10m}\text{×}\text{3}\text{.0m)}\\ \\ \text{=}\text{4}\text{.2}\text{×}{\text{10}}^{\text{2}}{\text{m}}^{\text{3}}\text{(4}\text{.2}\text{×}{\text{10}}^{\text{5}}\text{L)}\end{array}$

The number of atoms of radon present in basement initially are calculated as

$\begin{array}{l}{\text{n}}_{\text{air}}\text{=}\frac{\text{PV}}{\text{RT}}\text{=}\frac{\text{(1}\text{.0atm)(4}\text{.2}\text{×}{\text{10}}^{\text{5}}\text{L)}}{\text{(0}\text{.0821L}\text{.atm/mol}\text{.K)(273K)}}\\ \\ \text{=}\text{1}\text{.9}\text{×}{\text{10}}^{\text{4}}\text{mol}\text{air}\\ \\ {\text{n}}_{\text{Rn}}\text{=}\frac{{\text{P}}_{\text{Rn}}}{{\text{P}}_{\text{air}}}\text{×}\text{(}\text{1}\text{.9}\text{×}{\text{10}}^{\text{4}}\text{mol)}\\ \\ \text{=}\frac{\text{(}\text{1}\text{.2}\text{×}{\text{10}}^{\text{-6}}\text{)}}{\text{(760mmHg)}}\text{×}\text{1}\text{.9}\text{×}{\text{10}}^{\text{4}}\text{mol}\\ \\ \text{=}\text{3}\text{.0}\text{×}{\text{10}}^{\text{-5}}\text{mol}\text{Rn}\\ \\ \text{3}\text{.0}\text{×}{\text{10}}^{\text{-5}}\text{mol}\text{Rn}\text{×}\frac{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}{\text{1mol}\text{Rn}}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{19}}\text{Rn}\text{atoms}\end{array}$

The first-order kinetics is obeyed by radioactive decays.  Thus, radon half-life is does not relies on the initial concentration.  The number of moles of radon present in basement at end is calculated as

$\begin{array}{l}\text{ln}\frac{{\text{[A]}}_{\text{t}}}{{\text{[A]}}_{\text{0}}}\text{=}\text{-kt}\\ \\ {\text{ln[A]}}_{\text{t}}\text{=}\text{-kt}\text{+}{\text{ln[A]}}_{\text{0}}\text{ln x = 2}\text{.303 log x}\\ \\ =\text{-0}\text{.1824/day}\text{×}\text{31days}\text{+}\ln (1.8\times {10}^{19})\\ \\ \text{=}\text{-0}\text{.1824/day}\text{×}\text{31days}\text{+}\text{44}\text{.34}\\ \\ \text{=}\text{38}\text{.74}\\ \\ {\text{ln[A]}}_{\text{t}}\text{=}\text{38}\text{.74}\\ \\ {\text{[A]}}_{\text{t}}\text{=}\text{6}{\text{.4×10}}^{\text{16}}\text{Rn}\text{atoms}\end{array}$