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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