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Chapter 9: Q86E (page 512)

Question: During the discussion of gaseous diffusion for enriching uranium, it was claimed that \(^{{\rm{235}}}{{\rm{U}}_{\rm{6}}}\) diffuses 0.4% faster than\(^{{\rm{235}}}{{\rm{U}}_{\rm{6}}}\). Show the calculation that supports this value. The molar mass of \(^{{\rm{235}}}{{\rm{U}}_{\rm{6}}}{\rm{ = 235}}{\rm{.043930 + 6 \times 18}}{\rm{.998403 = 349}}{\rm{.034348 g/mol}}\), and the molar mass of \(^{{\rm{238}}}{{\rm{U}}_{\rm{6}}}{\rm{ = 238}}{\rm{.050788 + 6 \times 18}}{\rm{.998403 = 352}}{\rm{.041206 g/mol}}\).

Short Answer

Expert verified

We have to prove that \(^{{\rm{235}}}{{\rm{U}}_{\rm{6}}}\) diffuses \({\rm{0}}{\rm{.43\% }}\) faster than \(^{{\rm{238}}}{{\rm{U}}_{\rm{6}}}\).

Step by step solution

01

Concept Introduction

Graham's law states that a gas's rate of diffusion or effusion is inversely related to the square root of its molecular weight.

02

Relative Rates of Diffusion

Graham's finding relates the rate and the molar mass of a gas. If two gases \(A\) and \(B\) are at the same temperature and pressure, then the relative rates of diffusion for \(^{{\rm{235}}}{{\rm{U}}_{\rm{6}}}\) and \(^{{\rm{238}}}{{\rm{U}}_{\rm{6}}}\) will be:

\(\begin{array}{l}\frac{{{\rm{ rate of diffusion of}}{{\rm{ }}^{{\rm{235}}}}{{\rm{U}}_{\rm{6}}}}}{{{\rm{ rate of diffusion of}}{{\rm{ }}^{{\rm{238}}}}{{\rm{U}}_{\rm{6}}}}}{\rm{ = }}\frac{{\sqrt {{{\rm{M}}_{^{{\rm{238}}}{{\rm{U}}_{\rm{6}}}}}} }}{{\sqrt {{{\rm{M}}_{^{{\rm{235}}}{{\rm{U}}_{\rm{6}}}}}} }},where\;M\;\;is{\rm{ }}the{\rm{ }}molar{\rm{ }}mass.\;\\{\rm{ = }}\sqrt {\frac{{{\rm{352}}{\rm{.041206}}}}{{{\rm{349}}{\rm{.034348}}}}} \\{\rm{ = }}\sqrt {{\rm{1}}{\rm{.1117}}} \\{\rm{ = 1}}{\rm{.0043}}{\rm{.}}\end{array}\)

\(^{{\rm{235}}}{{\rm{U}}_{\rm{6}}}\)diffuses \({\rm{0}}{\rm{.0043}}\) times faster than \(^{{\rm{238}}}{{\rm{U}}_{\rm{6}}}\), which is:

\(\frac{{{\rm{0}}{\rm{.0043}}}}{{\rm{1}}}{\rm{ \times 100\% = 0}}{\rm{.43\% }}{\rm{.}}\)

Therefore, \(^{{\rm{235}}}{{\rm{U}}_{\rm{6}}}\) diffuses faster than \(^{{\rm{238}}}{{\rm{U}}_{\rm{6}}}\).

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Most popular questions from this chapter

Calculate the volume of oxygen required to burn \[12.00\;{\rm{L}}\] of ethane gas, \[{{\rm{C}}_2}{{\rm{H}}_6}\], to produce carbon dioxide and water, if the volumes of \[{{\rm{C}}_2}{{\rm{H}}_6}\]and \[{{\rm{O}}_2}\]are measured under the same conditions of temperature and pressure.

The pressure of a sample of gas is measured at sea level with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in: (a) torr (b) Pa (c) bar.

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Ethanol, C2H5OH,is produced industrially from ethylene, C2H4, by the following sequence of reactions:

\begin{aligned}{\rm{3}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{+2}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\to{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{HS}}{{\rm{O}}_{\rm{4}}}{\rm{+(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{{\rm{)}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\\{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{HS}}{{\rm{O}}_{\rm{4}}}{\rm{ +(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{{\rm{)}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{+3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\to{\rm{3}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH + 2}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \end{aligned}

What volume of ethylene at STP is required to produce 1000 metric tons (1000kg) of ethanol if the overall yield of ethanol is 90.1%?

A large scuba tank (see Figure \({\rm{9}}{\rm{.16}}\)) with a volume of \({\rm{18\;L}}\) is rated for a pressure of \({\rm{220 bar}}\). The tank is filled at \({\rm{2}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C}}\) and contains enough air to supply \({\rm{1860\;L}}\) of air to a diver at a pressure of \({\rm{2}}{\rm{.37\;atm}}\) (a depth of \({\rm{45}}\)feet). Was the tank filled to capacity at \({\rm{2}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C}}\)?
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