Chapter 24, Problem 16QP

Interpretation Introduction

Interpretation:

The amount of water is to be calculated with given volume of deuterium, pressure, and temperature.

Concept introduction:

The ideal gas equation is as follows:

$PV=n\text{R}T$

Here, $P$ is the pressure and its unit is $\text{atm}$; $V$ represents volume and its unit is $\text{L}$; $n$ represents the number of moles; $\text{R}$ is the gas constant and it has value $\text{0}{\text{.0821 L atm mol}}^{-\text{1}}{\text{K}}^{-\text{1}}$; and $T$ represents the temperature and its unit is $\text{K (Kelvin)}$.

The number of moles of water is calculated as

$n_{H_{2}O}\text{}=n_{D_2}\times \frac{100\%\text{}{\text{H}}_2\text{O}}{\text{abundance}\text{of}{\text{D}}_{\text{2}}}$

Here $n_{H_{2}O}$ is number of moles of water and $n_{D_2}$ is number of moles of deuterium

The amount of water needed is calculated as represented below:

$\text{Amount}\text{of}\text{water}\text{=}\frac{\text{Mole}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{Recovery}\text{of}{\text{D}}_{\text{2}}}\text{×}\frac{\text{Mass}\text{of}\text{}{\text{H}}_{\text{2}}\text{O(kg)}}{\text{1}\text{mole}{\text{H}}_{\text{2}}\text{O}}$

Answer to Problem 16QP

Solution: $11\text{ kg}{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

Given information: Volume of ${\text{D}}_{\text{2}}=\text{2 L}$

Pressure, $P=\text{0}\text{.90 atm}$

Temperature, $T=\text{298 K}$

Abundance of ${\text{D}}_{\text{2}}=0.015\%$

Recovery ${\text{D}}_{\text{2}}=80\%$

The ideal expression is rearranged to give the number of moles and is as follows:

$n=\frac{PV}{\text{R}T}$

The number of moles is calculated as

$\begin{array}{c}{n}_{D_2}=\left(\frac{\text{0}\text{.90 atm}\times \text{2 L}}{\text{0}\text{.0821}{\text{L atm mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\times 298\text{ K}}\right)\\ =0.074\text{ moles}\end{array}$

As the abundance of ${\text{D}}_{\text{2}}\text{ is }0.015\%$, the number of moles of water is calculated as

$n_{H_{2}O}\text{}=n_{D_2}\times \frac{100\%\text{}{\text{H}}_2\text{O}}{\text{abundance}\text{of}{\text{D}}_{\text{2}}}$

$\left(0.074{\text{ mol of D}}_{\text{2}}\times \frac{100\%{\text{ H}}_{\text{2}}\text{O}}{0.015\%{\text{ D}}_{\text{2}}}\right)=\text{4}\text{.9 }\times {\text{ 10}}^{\text{2}}{\text{ mol H}}_{\text{2}}\text{O}$

For $80\%$ recovery of ${\text{D}}_{\text{2}}$, the amount of water needed is calculated as represented below.

$\text{Amount}\text{of}\text{water}\text{=}\frac{\text{Mole}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{Recovery}\text{of}{\text{D}}_{\text{2}}}\text{×}\frac{\text{Mass}\text{of}\text{}{\text{H}}_{\text{2}}\text{O(kg)}}{\text{1}\text{mole}{\text{H}}_{\text{2}}\text{O}}$

$\left(\frac{4.9\times {10}^2\text{ mol}{\text{H}}_{\text{2}}\text{O}}{0.80}\right)\left(\frac{0.01802\text{ kg }{\text{H}}_{\text{2}}\text{O}}{1\text{ mol}{\text{H}}_{\text{2}}\text{O}}\right)=11\text{ kg}{\text{H}}_{\text{2}}\text{O}$

Conclusion

The amount of water is $11\text{ kg}$.