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Chapter 5: Problem 112

Consider separate \(1.0-\mathrm{L}\) samples of \(\mathrm{He}(g)\) and \(\mathrm{UF}_{6}(g),\) both at 1.00 atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?

Short Answer

Expert verified
The ratio of temperatures for He(g) and UF6(g) to have the same root mean square velocity is approximately 0.0114.

Step by step solution

01

Set up the root mean square velocity equation

First, set up the equation for the root mean square velocity: \(v_{rms} = \sqrt{\dfrac{3RT}{M}}\) We need to find the ratio of temperatures (\(T_{He}\) and \(T_{UF6}\)) for which the root mean square velocities of He(g) and UF6(g) are equal. So, we will set up the following equation: \(v_{rms (He)} = v_{rms (UF6)}\) Let \(T_{He} = T_1\) and \(T_{UF6} = T_2\). Then, \(v_{rms (He)} = \sqrt{\dfrac{3RT_1}{M_{He}}}\) \(v_{rms (UF6)} = \sqrt{\dfrac{3RT_2}{M_{UF6}}}\)
02

Find the molar masses of He and UF6

He has an atomic mass of 4 g/mol, so its molar mass, \(M_{He}\), is 4 g/mol or 0.004 kg/mol. UF6 has a molar mass of (1 × 238 U + 6 × 19 F) = 352 g/mol, so its molar mass, \(M_{UF6}\), is 352 g/mol or 0.352 kg/mol.
03

Set up the equation with molar masses and solve for the ratio of temperatures

Now, we have: \(\sqrt{\dfrac{3RT_1}{M_{He}}} = \sqrt{\dfrac{3RT_2}{M_{UF6}}}\) Squaring both sides, we get: \(\dfrac{3RT_1}{M_{He}} = \dfrac{3RT_2}{M_{UF6}}\) Next, we can write the ratio of temperatures as \(\dfrac{T_1}{T_2}\). Now, \(\dfrac{T_1}{T_2} = \dfrac{M_{He}}{M_{UF6}}\) Substitute the molar masses we found earlier: \(\dfrac{T_1}{T_2} = \dfrac{0.004\,\text{kg/mol}}{0.352\,\text{kg/mol}}\) Finally, solve for the ratio of temperatures: \(\dfrac{T_1}{T_2} = 0.0114\) So, the ratio of temperatures for He(g) and UF6(g) to have the same root mean square velocity is approximately 0.0114.

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