Question

Two isotopes of uranium, \({ }^{235} \mathrm{U}\) and \({ }^{238} \mathrm{U}\), are separated by a gas diffusion process that involves combining them with fluorine to make the compound \(\mathrm{UF}_{6}\). Determine the ratio of the root-mean-square speeds of \(\mathrm{UF}_{6}\) molecules for the two isotopes. The masses of \({ }^{235} \mathrm{UF}_{6}\) and \({ }^{238} \mathrm{UF}_{6}\) are 349.03 amu and 352.04 amu, respectively.

Short answer

Answer: The ratio of the root-mean-square speeds of uranium isotopes ${}^{235}\mathrm{UF}_{6}$ and ${}^{238}\mathrm{UF}_{6}$ is approximately 1.0043.

Step-by-step solution

Root-mean-square speed formula and constants

The root-mean-square speed can be calculated using the following formula: \( v_{rms} = \sqrt{\frac{3RT}{M}} \) Where: \(v_{rms}\) = root-mean-square speed \(R\) = universal gas constant, which is \(8.314 \ \text{J} \text{(mol K)}^{-1}\) when expressed in SI units \(T\) = absolute temperature in Kelvin (K) \(M\) = molar mass of the gas in kg/mol In this exercise, we are not given the temperature, but since it is the ratio of the rms speeds of uranium isotopes we are interested in, the temperature factor will cancel out when we compare their speeds. Step 2: Convert atomic mass unit (amu) to molar mass in kg/mol

Conversion of amu to kg/mol

The atomic masses for \({ }^{235}\mathrm{UF}_{6}\) and \({ }^{238}\mathrm{UF}_{6}\) are given in atomic mass units (amu). To obtain the molar masses in kg/mol, we can use the conversion factor: \(1 \ \text{amu} = 1.66054 \times 10^{-27} \ \text{kg}\) So, we have: Molar mass of \({ }^{235}\mathrm{UF}_{6} = 349.03 \times 1.66054 \times 10^{-27} \ \text{kg}\) Molar mass of \({ }^{238}\mathrm{UF}_{6} = 352.04 \times 1.66054 \times 10^{-27} \ \text{kg}\) Now that we have converted the atomic masses to their molar masses, we can find the ratio of their root-mean-square speeds. Step 3: Find the ratio of rms speeds

Ratio of rms speeds for isotopes

Using the formula for rms speed, and recalling that the temperature factor will cancel out when comparing speeds, we can find the ratio of rms speeds for \({ }^{235}\mathrm{UF}_{6}\) and \({ }^{238}\mathrm{UF}_{6}\) as follows: \(\frac{v_{rms}({}^{235}\mathrm{UF}_{6})}{v_{rms}({}^{238}\mathrm{UF}_{6})} = \frac{\sqrt{\frac{3R\cancel{T}}{M({}^{235}\mathrm{UF}_{6})}}}{\sqrt{\frac{3R\cancel{T}}{M({}^{238}\mathrm{UF}_{6})}}} = \sqrt{\frac{M({}^{238}\mathrm{UF}_{6})}{M({}^{235}\mathrm{UF}_{6})}}\) Now, substitute the calculated molar masses from Step 2: \(\frac{v_{rms}({}^{235}\mathrm{UF}_{6})}{v_{rms}({}^{238}\mathrm{UF}_{6})} = \sqrt{\frac{352.04 \times 1.66054 \times 10^{-27}}{349.03 \times 1.66054 \times 10^{-27}}} = \sqrt{\frac{352.04}{349.03}} \approx 1.0043 \) The ratio of root-mean-square speeds of \({ }^{235}\mathrm{UF}_{6}\) and \({ }^{238}\mathrm{UF}_{6}\) is approximately 1.0043.