Chapter 25, Problem 25.12QP

Interpretation Introduction

Interpretation: The amount of water required in kilograms has to be calculated to get 2.0 L of ${\text{D}}_{\text{2}}$ at $\text{25}{}^{\text{o}}\text{C}$ and $\text{0}\text{.90}\text{atm}$

Concept introduction:

Ideal gas equation is given as follows,

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{P}\text{-}\text{pressure of gas; V}\text{-}\text{volume of gas; n}\text{-}\text{number of moles of gas}\\ \text{R}\text{-}\text{Gas constant; T - temperature in Kelvin}\end{array}$

Answer to Problem 25.12QP

The amount of water required to get 2.0 L of ${\text{D}}_{\text{2}}$ at $\text{25}{}^{\text{o}}\text{C}$ and $\text{0}\text{.90}\text{atm}$ is 11.04495Kg.

Explanation of Solution

Capture the given data

Pressure is given as $\text{0}\text{.90}\text{atm}$

Temperature is given as $\text{25}{}^{\text{o}}\text{C}$

Gas constant value is $\text{0}\text{.0821}\raisebox{1ex}{$\text{L}\cdot \text{atm}$}\!\left/ \!\raisebox{-1ex}{$\text{K}\cdot \text{mol}$}\right.$

The number of moles of deuterium is calculated by the use of modified ideal gas equation.

$\begin{array}{l}\text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{P}\text{-}\text{pressure of gas; V}\text{-}\text{volume of gas; n}\text{-}\text{number of moles of gas}\\ \text{R}\text{-}\text{Gas constant; T - temperature in Kelvin}\end{array}$

$\begin{array}{l}{\text{number of moles D}}_{\text{2}}\text{=}\frac{\left(\text{0}\text{.90}\text{atm}\right)\left(\text{2}\text{.0}\text{L}\right)}{\left(\text{0}\text{.0821}\raisebox{1ex}{$\text{L×atm}$}\!\left/ \!\raisebox{-1ex}{$\text{K×mol}$}\right.\right)\left(\text{298}\text{K}\right)}\\ \text{=}\left(\frac{\left(\text{0}\text{.90}\text{×}\text{2}\right)}{\left(\text{0}\text{.0821}\text{×}\text{298}\right)}\right)\text{moles}{\text{of D}}_{\text{2}}\\ \text{=}\left(\frac{\text{1}\text{.8}}{\text{24}\text{.4658}}\right)\text{moles}{\text{of D}}_{\text{2}}\\ \text{=}\text{0}\text{.07357}\text{moles}{\text{of D}}_{\text{2}}\end{array}$

The number of moles of deuterium is 0.07357 moles.

By using the modified ideal gas equation and given data the number of moles of deuterium is calculated as 0.07357 moles.

To determine: The number of moles of water

The abundance of deuterium is given as 0.015 percent.

$\begin{array}{l}\text{moles}\text{of}{\text{D}}_{\text{2}}\text{×}\frac{\text{percent}\text{of}{\text{H}}_{{}^{{}_{\text{2}}}}\text{O}}{\text{percent}\text{of}{\text{D}}_{{}^{{}_{\text{2}}}}}\text{=}\text{moles}\text{of}{\text{H}}_{{}^{{}_{\text{2}}}}\text{O}\\ \text{=}\text{0}\text{.07357}\text{mol}{\text{D}}_{\text{2}}\text{×}\frac{\text{100}\text{\%}{\text{H}}_{{}^{{}_{\text{2}}}}\text{O}}{\text{0}\text{.015}\text{\%}{\text{D}}_{{}^{{}_{\text{2}}}}}\\ \text{=}\left(\text{0}\text{.07357}\text{×}\text{6666}\text{.6666}\right)\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}\\ \text{=}\text{4}\text{.9047}\text{×}{\text{10}}^{{}_{{}^{\text{2}}}}\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}$

The number of moles of ${\text{H}}_{\text{2}}\text{O}$ is calculated as $\text{4}\text{.9047}\text{×}{\text{10}}^{\text{2}}$ .

By the use of calculated moles of deuterium and percent of deuterium the number of moles of water is calculated as $\text{4}\text{.9047}\text{×}{\text{10}}^{\text{2}}$

To determine: The amount of ${\text{H}}_{\text{2}}\text{O}$ needed to produce 2.0 L of ${\text{D}}_{\text{2}}$

Recovery is given as 80 percent. The amount of water needed is calculated as follows:

$\begin{array}{l}\text{amount}\text{of}\text{water}\text{in}\text{Kg}\text{=}\frac{\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{recovery}\text{of}{\text{D}}_{\text{2}}}\text{×}\frac{\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}\text{in}\text{Kg}}{\text{1}\text{mol}\text{of}{\text{H}}_{\text{2}}\text{O}}\\ \text{=}\frac{\text{4}\text{.9047}\text{×}{\text{10}}^{\text{2}}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{0}\text{.80}}\text{×}\frac{\text{0}\text{.0180153}\text{Kg}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \text{=}\left(\text{613}\text{.0875}\text{×}\text{0}\text{.0180153}\right){\text{H}}_{\text{2}}\text{O}\text{Kg}\\ \text{=}\text{11}\text{.04495}{\text{H}}_{\text{2}}\text{O}\text{Kg}\end{array}$

The amount of ${\text{H}}_{\text{2}}\text{O}$ needed to produce 2.0 L of ${\text{D}}_{\text{2}}$ is calculated as 11.04495Kg.

By the use of moles of water and recovery of deuterium the amount of water is needed to produce 2.0 L of deuterium is calculated as 11.04495 Kg.

Conclusion

The required amount of water to produce the given amount of deuterium was calculated.