Chapter 14, Problem 14.56PAE

(a)

Interpretation Introduction

To determine: What mass of octane $C_8{H}_{18}$ would have to undergo combustion to provide the same quantity of energy as that released by fission of one kilogram of ${}^{235}U$ to produce ${}^{90}Sr{\text{ and }}^{143}Xe$.

Explanation of Solution

Write the balanced nuclear reaction.

${}_{92}^{235}U{+}_0^{1}n{\to }_{38}^{90}Sr{+}_{54}^{143}NP+3_0^{1}n$

Mass of reactant $=m_u+m_n$ (putting the value from table)

$\begin{array}{l}=235.0439231u+1.0086649u\\ =236.0525880u\end{array}$

Mass of product $=m_{Sr}+m_{Xe}+3m_n$ (putting the value from table)

$\begin{array}{l}=89.90773764+142.9348900u+3\left(1.0086649u\right)\\ =235.8686223u\end{array}$

$\begin{array}{c}\Delta m=\text{mass of product}-\text{mass of eactant}\\ =\text{235}\text{.8686223u}-\text{236}\text{.0525880u}\\ \text{=}-0.1839657u\end{array}$

Negative sign indicated so mass convert into energy so energy is released.

Energy released on one atom fission of $u$

$=\left(0.1839657u\right)\times C^2$

$\begin{array}{l}1u=1.66053886\times {10}^{-27}\text{Kg}\\ \text{C}=3\times {10}^8\text{ m/s}\end{array}$

Putting these values we get,

$\begin{array}{l}=\left(0.1839657\times 1.66053886\times {10}^{-27}\right)\times {\left(3\times {10}^8\right)}^{2}J\\ =2.7455370\times {10}^{-11}J\end{array}$

Energy released per atom of ${}^{235}U=2.7455370\times {10}^{-11}J$

No. of atoms in $1{\text{ kg of }}^{235}U=\frac{1000g}{235.044\text{ g/mol}}\times 6.022\times {10}^{23}atoms/mol$

So energy released in fission of $1kg{}^{235}U=\left(2.7455370\times {10}^{-11}J\right)\times \left(\frac{1000}{235.044}\times 6.022\times {10}^{23}\right)$

$=7.0342675\times {10}^{13}J$

Combustion of octane

$\begin{array}{l}{C}_8{H}_{18}\left(l\right)+\frac{25}{2}{O}_2\to 8C{O}_2\left(g\right)+9H_{2}O\left(l\right)\\ \Delta H=-5471\text{ kJ/mol}\end{array}$

$1{\text{ mole of c}}_8{H}_{18}\left(l\right)\text{ released }5471\text{ kJ energy}\text{.}$

So no of moles released $7.0342675\times {10}^{13}J$ energy

$\begin{array}{l}=\frac{7.0342675\times {10}^{13}J}{5471\times {10}^{3}J}\\ =1.2857370\times {10}^7\text{ moles}\end{array}$

Mass of moles $=\text{moles}\times \text{molar mass}$

$\begin{array}{l}\text{=1}\text{.2857370}\times {10}^7\text{moles}\times 114\text{ g/mole}\\ \text{=1465740180 gram}\\ \text{=1465740}\text{.180 kg}\end{array}$

Mass of octant $\text{=1465740}\text{.180 kg}$

(b)

Interpretation Introduction

To determine: the amount of gallons of octane in this

Explanation of Solution

$\begin{array}{l}\frac{0.702\text{ g}}{1{\text{ cm}}^3}\end{array}$

$\begin{array}{l}(\begin{array}{l}1{\text{ cm}}^3=\frac{1}{1000}L\end{array}\end{array}$

$\begin{array}{l}1\text{ g}={10}^{-3}\text{ kg}\end{array})=\frac{0.702\times {10}^{-3}\text{ kg}}{\frac{1}{1000}L}\\ =0.702\text{}\text{ kg/L}$

Volume of octane $\left(\text{ in L}\right)=\frac{1465740.180\text{ kg}}{0.702\text{ kg/L}}$

$\begin{array}{c}1\text{ gallon }=3.78541\text{ lit}\end{array}$

$\begin{array}{c}\text{1lit}=\frac{1}{3.78541}\text{gallon}\end{array}$

$\begin{array}{c}\text{=}\frac{1465740.180\text{ kg}}{0.702\text{ kg/L}}\times \frac{1}{3.78541}\text{ gallon/LIt}\end{array}$

$\begin{array}{c}\text{=551578 gallons}\end{array}$

Hence gallons of octane required $\text{=551578 gallons}$

Conclusion

$1{\text{ kg }}^{235}U$ Gives energy as $1.46\times {10}^7\text{ kg}$ octane give.