Chapter 21, Problem 21.73QP

Interpretation Introduction

Interpretation:

The pressure present inside the flask after $\text{78}\text{.8 min}$ has to be calculated using the given data.

Concept Introduction:

Radiocarbon dating: The dynamic equilibrium exists in all living organism by exhaling or inhaling, maintain the same ratio of ${}^{12}\text{C}$ and ${}^{14}\text{C}$ that takes in. When the living organism dies, the intake of $\text{C}$ stops and it’s the ratio no longer exhibits equilibrium since ${}^{14}\text{C}$ undergoes radioactive decay.

Half-life period: The time required to reduce to half of its initial value.

Formula used to calculate half-life:

$\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\\ \text{where,}\text{k}\text{is rate constant}\text{.}\\ \left(\text{or}\right)\\ {\text{ln[N]}}_{\text{t}}{\text{- ln[N]}}_0\text{= -kt}\end{array}$

Ideal gas equation:

An ideal gas equation is,

  $\begin{array}{l}\text{PV = n RT}\\ \text{where, P , pressure}\\ \text{ V is volume,}\\ \text{ n is mole of He}\\ \text{ R is gas constant}\\ \text{ T is temperature in Kelvin}\end{array}$

Answer to Problem 21.73QP

The value of pressure exist in the flask is $\text{22}\text{.42}\text{mmHg}$

Explanation of Solution

Given: Half-life of Bismuth $-214$ is ${\text{t}}_{1/2}\text{=}\text{19}\text{.7 min}$ by emitting alpha particle (Helium nuclei). Half-life of Bismuth $-214$ after $\text{78}\text{.8 min}$ is,

  $\begin{array}{c}{\text{t}}_{\text{1/2}}\text{= 19}\text{.7 min}\\ \\ \text{k = }\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}\text{=}\frac{\text{0}\text{.693}}{\text{19}\text{.7 min}}\\ \\ \text{=}\text{3}\text{.52}\times {10}^{-2}{\min }^{-1}\end{array}$

The decay constant is $\text{3}\text{.52}\times {10}^{-2}{\min }^{-1}$

Calculate the number of Bismuth atoms in the flask:

Mass of ${}^{\text{214}}\text{Bi}$ in the flask is $\text{5}\text{.26}\text{mg}\text{=}\text{5}\text{.26}\times {\text{10}}^{-3}g$

Number of Bismuth atoms in the flask is,

$\begin{array}{c}\text{No}\text{.of atoms}\text{=}\left(\text{Moles}\right)\left(\text{Avogadro's}\text{number}\right)\\ \\ \text{=}\left(\frac{\text{5}\text{.26}\times {\text{10}}^{-3}g}{\text{214}\text{g}\text{/mol}}\right)\left(\text{6}\text{.023×}{\text{10}}^{\text{23}}\text{atoms/mol}\right)\\ \\ \text{=}\text{1}\text{.48}\text{×}{\text{10}}^{19}\text{atoms}\end{array}$

The number of Bismuth atoms ${\text{[N]}}_{\text{0}}$ present initially, is $\text{1}\text{.48}\text{×}{\text{10}}^{19}\text{atoms}$

Predict the equation from first order decay process:

$\begin{array}{c}{\text{First order process equation: ln[N]}}_{\text{t}}-{\text{ ln[N]}}_{\text{0}}\text{= }-\text{ kt}\\ \\ \text{As known, k =}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}\text{substitute}\text{in}\text{above}\text{equation}\\ \\ {\text{ln[N]}}_{\text{t}}-{\text{ ln[N]}}_{\text{0}}\text{=}-\left(\text{3}\text{.52}\times {10}^{-2}{\min }^{-1}\right)\left(\text{78}\text{.8 min}\right)\\ \\ \text{ln}\left(\frac{{\text{[N]}}_{\text{t}}}{{\text{[N]}}_{\text{0}}}\right)\text{= }-\text{2}\text{.77}\\ \\ \left(\frac{{\text{[N]}}_{\text{t}}}{{\text{[N]}}_{\text{0}}}\right)=e^{-2.77}=6.27\times {10}^{-2}\end{array}$

By comparing the values,

${\text{[N]}}_{\text{t}}=\left(6.27\times {10}^{-2}\right)\left(\text{1}\text{.48}\text{×}{\text{10}}^{19}\text{atoms}\right)=9.28\text{×}{\text{10}}^{17}\text{atoms}$

The number of Bismuth atoms ${\text{[N]}}_t$ present after decay, is $9.28\text{×}{\text{10}}^{17}\text{atoms}$

Atoms of Helium is same as Atoms of Bismuth decayed

Atoms of He emitted is given as:

  $\begin{array}{c}{\text{[N]}}_{\text{0}}-{\text{[N]}}_t=\left(\text{1}\text{.48}\text{×}{\text{10}}^{19}\text{atoms}\right)-\left(9.28\text{×}{\text{10}}^{17}\text{atoms}\right)\\ \\ =1.3872\text{×}{\text{10}}^{19}\text{atoms}\end{array}$

Conversion of Helium atom into mole:

Mole of Helium $\text{= }\frac{1.3872\text{×}{\text{10}}^{19}\text{atoms}}{\text{6}\text{.023×}{\text{10}}^{\text{23}}\text{atoms/mol}}\text{=}\text{2}\text{.30}\text{×}{\text{10}}^{-5}\text{mol}\text{He}$

The mole of Helium emitted is $\text{2}\text{.30}\text{×}{\text{10}}^{-5}\text{mol}\text{He}$

Calculate the pressure in the flask:

  $\begin{array}{l}{\text{T= 40}}^{\text{o}}\text{C}\text{= 313}\text{K}\\ \\ \text{V}\text{=}\text{20}\text{.0}\text{mL}\text{=}\text{0}\text{.0200}\text{L}\end{array}$

An ideal gas equation is,

  $\begin{array}{l}\text{PV = n RT}\\ \text{where, P , pressure}\\ \text{ V is volume,}\\ \text{ n is mole of He}\\ \text{ R is gas constant}\\ \text{ T is temperature in Kelvin}\end{array}$

By substituting the known values into the above equation is,

  $\begin{array}{c}\text{P = }\frac{\text{n RT}}{\text{V}}\\ \\ \text{=}\frac{\left(\text{2}\text{.30}\text{×}{\text{10}}^{-5}\text{mol}\text{He}\right)\left(\text{0}\text{.08206}\frac{\text{L}\text{.atom}}{\text{K}\text{.mol}}\right)\left(\text{313K}\right)}{\text{0}\text{.0200}\text{L}}\\ \\ \text{=}\text{2}\text{.95}{\text{×10}}^{-2}\text{atm}\\ \\ \text{P = }\text{2}\text{.95}{\text{×10}}^{-2}\text{atm}\text{×}\frac{\text{760}\text{mmHg}}{\text{1atm}}\\ \\ \text{=}\text{22}\text{.42}\text{mmHg}\end{array}$

The value of pressure exist in the flask is $22.42mmHg$