Chapter 21, Problem 21.78QP

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_1/2 = 3.8 days) at the beginning and end of 31 days. Assume STP conditions.

Interpretation Introduction

Interpretation: The number of ${\text{Rn}}^{\text{222}}$ isotope at the beginning and end of 31 days should be determined. Half –life of  ${\text{Rn}}^{\text{222}}$ isotope  is 3.8 days.

Concept Introduction:

• Number of moles can be calculated in accordance with ideal gas equation,

  $\text{n}\text{=}\frac{\text{PV}}{\text{RT}}$    P, V, T and R are the pressure, volume, temperate and universal gas constant.

• Equation for Number of moles of a substance, from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

• Partial pressure is the pressure which the gas would have if it alone occupied the volume
• $\text{ Number of moles of Radon }\text{=}\frac{\text{Partial}\text{pressure}\text{of}\text{radon}}{\text{Total}\text{pressure}\text{of}\text{air}}\text{×}\text{number}\text{of}\text{moles}\text{of}\text{air}$
• $\text{Number}\text{of}\text{atom}\text{=}\text{Number}\text{of}\text{mole}\text{×}\frac{\text{Avogadro}\text{number}}{\text{1}\text{mole}}$
• Unstable nuclei emit radiation spontaneously to become stable nuclei by losing energy. This process of emission of radiation by unstable nuclei is known as radioactive decay.
• Radioactive decay is in the first order kinetics. Rate of radioactive decay at a time t is,

  $\begin{array}{l}Rateofdecayatatime\text{R = k N}\\ \text{k-}\text{First}\text{order}\text{rate}{\text{constant and its unit is t}}^{-1}\\ \text{N-}\text{number}\text{of}\text{radioactive}\text{nuclei}\text{present}\text{at}\text{time}\text{'t}\end{array}$

  Suppose the number of radioactive nuclei at time zero is ${\text{N}}_{\text{0}}$ and at a time t is ${\text{N}}_{\text{t}}$,

  $\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{=}\text{-kt}$

• Half-life of radioactive decay is the time required for a radioactive sample to decay to one half of the atomic nucleus.

  Half-life of the radiation,  $t_{1/2}=\frac{0.693}{k}$

To calculate: the number of moles of  ${\text{Rn}}^{\text{222}}$

Answer to Problem 21.78QP

The number of ${\text{Rn}}^{\text{222}}$ atoms at the beginning: $\text{1}{\text{.8×10}}^{\text{19}}\text{Rn}\text{atoms}$.

The number of ${\text{Rn}}^{\text{222}}$ atoms at end of 31 days: $\text{6}{\text{.4×10}}^{\text{16}}\text{Rn}\text{atoms}$.

Explanation of Solution

The number of ${\text{Rn}}^{\text{222}}$ atoms can be calculated by using ideal gas equation.

According to ideal gas equation,  $\text{n}\text{=}\frac{\text{PV}}{\text{RT}}$    P, V, T and R are the pressure, volume, temperate and universal gas constant.

Volume of the basement occupied by  ${\text{Rn}}^{\text{222}}$ atom is,

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

      The number of moles of air ${\text{n}}_{\text{air}}\text{=}\frac{\text{PV}}{\text{RT}}\text{=}\frac{\text{(1}\text{.0}\text{atm)(4}{\text{.2×10}}^{\text{5}}\text{L)}}{\text{(0}\text{.0821}\text{L}\text{.atm/mol)(273}\text{K)}}$

          $\text{=}\text{1}\text{.9}{\text{×10}}^{\text{4}}\text{mol}\text{air}$

From the number of moles of air, number of moles of  ${\text{Rn}}^{\text{222}}$ can be calculated as follows,

Partial pressure of radon is $\text{1}{\text{.2×10}}^{\text{-6}}\text{mmHg}$ and for air is 760 mmHg.

$\text{ Number of moles of Radon }\text{=}\frac{\text{Partial}\text{pressure}\text{of}\text{radon}}{\text{Total}\text{pressure}\text{of}\text{air}}\text{×}\text{number}\text{of}\text{moles}\text{of}\text{air}$

${\text{n}}_{\text{Rn}}\text{=}\frac{{\text{P}}_{\text{Rn}}}{{\text{P}}_{\text{air}}}\text{×(1}\text{.9}{\text{×10}}^{\text{4}}\text{mol)}$

        $\text{=}\frac{\text{1}{\text{.2×10}}^{\text{-6}}\text{mmHg}}{\text{760}\text{mmHg}}\text{×(1}{\text{.9×10}}^{\text{4}}\text{mol)}$

        $\text{=}\text{3}{\text{.0×10}}^{\text{-5}}\text{mol}\text{Rn}$

From the number of moles of Rn above calculated the number of atom at the beginning can be calculated as follows,

$\text{Number}\text{of}\text{atom}\text{=}\text{Number}\text{of}\text{mole}\text{×}\frac{\text{Avogadro}\text{number}}{\text{1}\text{mole}}$

$\text{Number}\text{of}\text{Rn}\text{atoms}\text{=}\text{Number}\text{of}\text{moles}\text{of}\text{Rn}\text{atom}\text{×}\frac{\text{6}\text{.022}{\text{×10}}^{\text{23}}\text{Rn}\text{atom}}{\text{1}\text{mole}\text{Rn}}$

So, $\text{Number}\text{of}{\text{Rn}}^{\text{222}}\text{atom}\text{at}\text{the}\text{beginning}\text{=}\text{(3}\text{.0}\text{×}{\text{10}}^{\text{-5}}\text{mol}\text{Rn)×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{Rn}\text{atoms}}{\text{1}\text{mol}\text{Rn}}$

     = $\text{1}{\text{.8×10}}^{\text{19}}\text{Rn}\text{atom}$

${\text{Rn}}^{\text{222}}$ is a radioactive element, its Unstable nuclei emit radiation spontaneously to become stable nuclei by losing energy. This process of emission of radiation by unstable nuclei is known as radioactive decay. All radioactive decays follows first order kinetics.

Half-life of radioactive decay is the time required for a radioactive sample to decay to one half of the atomic nucleus.

Half-life of the radiation that follows first order kinetics is  $t_{1/2}=\frac{0.693}{k}$

From this the rate constant $\text{k}\text{=}\frac{\text{0}\text{.693}}{{\text{t}}_{\frac{\text{1}}{\text{2}}}}$ ,

The rate constant k for the given reaction  $\text{k}\text{=}\frac{\text{0}\text{.693}}{\text{3}\text{.8}\text{day}}$ $\text{=}\text{0}\text{.182}\text{day}{}^{\text{-1}}$

Radioactive decay is in the first order kinetics. Rate of radioactive decay at a time t is,

$\begin{array}{l}Rateofdecayatatime\text{R = k N}\\ \text{k-}\text{First}\text{order}\text{rate}{\text{constant and its unit is t}}^{-1}\\ \text{N-}\text{number}\text{of}\text{radioactive}\text{nuclei}\text{present}\text{at}\text{time}\text{'t}\end{array}$

Suppose the number of radioactive nuclei at time zero is ${\text{N}}_{\text{0}}$ and at a time t is ${\text{N}}_{\text{t}}$,

$\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{=}\text{-kt}$

Radioactive nuclei at time zero and time t are the initial concentration and the final concentration (here considering 31 days).

So the equation can also be written as, $\text{ln}\frac{{\text{[A]}}_{\text{t}}}{{\text{[A]}}_{\text{0}}}\text{=}\text{-kt}$

Then the number of ${\text{Rn}}^{\text{222}}$ at the end of 31 days ${\text{[A]}}_{\text{t}}$ is taken as x, therefore,

              $\text{ln}\frac{\text{x}}{\text{1}{\text{.8×10}}^{\text{19}}}\text{=}\text{-}\text{(0}\text{.182}{\text{d}}^{\text{-1}}\text{)(31}\text{d)}$

$\text{x}\text{=}\text{6}\text{.4}{\text{×10}}^{\text{16}}\text{Rn}\text{atom}$

Conclusion

The number of ${\text{Rn}}^{\text{222}}$ isotope at the beginning and end of 31 days is determined.