Question

At what temperature will a molecule of uranium hexafluoride, the densest gas known, have the same average speed as a molecule of the lightest gas, hydrogen, at 378C?

Short answer

Answer: Approximately 54652.6 K

Step-by-step solution

Convert given temperature to Kelvin

To convert the given temperature of hydrogen gas (37°C) to Kelvin, we add 273.15. This is because 0 K equals -273.15°C. \(T_\mathrm{H_2} = 37 + 273.15 = 310.15 \mathrm{K}\)

Find the molar mass of uranium hexafluoride and hydrogen

The molar mass can be calculated by multiplying the number of moles of each element with their respective atomic weight. For uranium hexafluoride, we have 1 uranium atom with an atomic weight of approximately 238 u and 6 fluorine atoms with atomic weights of approximately 19 u each. For hydrogen, there are 2 hydrogen atoms with atomic weights of approximately 1 u each. Molar mass of Uranium hexafluoride (UF6): \(M_\mathrm{UF_6} = 1 \cdot 238 + 6 \cdot 19 = 238 + 114 = 352 \mathrm{g/mol}\) Molar mass of Hydrogen (H2): \(M_\mathrm{H_2} = 2 \cdot 1 = 2 \mathrm{g/mol}\)

Calculate the average speeds of hydrogen and uranium hexafluoride

The average speed of a gas molecule can be derived from the kinetic theory of gases and is given by the formula: \(v = \sqrt{ \frac{8kT}{ \pi m} }\) where \(v\) is the average speed, \(k\) is the Boltzmann constant (\(1.38 \times 10^{-23} \mathrm{J/K}\)), \(T\) is the temperature in Kelvin, and \(m\) is the molecular mass in kg. For hydrogen gas at 310.15K: \(v_\mathrm{H_2} = \sqrt{ \frac{8 \cdot 1.38 \times 10^{-23} \cdot 310.15}{\pi \cdot 2 \cdot 10^{-3}} }\) For uranium hexafluoride at an unknown temperature \(T_\mathrm{UF_6}\): \(v_\mathrm{UF_6} = \sqrt{ \frac{8 \cdot 1.38 \times 10^{-23} \cdot T_\mathrm{UF_6}}{\pi \cdot 352 \cdot 10^{-3}} }\)

Set the average speeds equal and solve for the unknown temperature

We need to find the temperature at which the average speeds of hydrogen and uranium hexafluoride are equal, so we set the two equations equal and solve for \(T_\mathrm{UF_6}\): \(\sqrt{ \frac{8 \cdot 1.38 \times 10^{-23} \cdot 310.15}{\pi \cdot 2 \cdot 10^{-3}} } = \sqrt{ \frac{8 \cdot 1.38 \times 10^{-23} \cdot T_\mathrm{UF_6}}{\pi \cdot 352 \cdot 10^{-3}} }\) Squaring both sides of the equation and simplifying, we get: \(\frac{8 \cdot 1.38 \times 10^{-23} \cdot 310.15}{ 2 \cdot 10^{-3} } = \frac{8 \cdot 1.38 \times 10^{-23} \cdot T_\mathrm{UF_6}}{ 352 \cdot 10^{-3} }\) Now, solve for \(T_\mathrm{UF_6}\): \(T_\mathrm{UF_6} = \frac{310.15 \cdot 352}{2} = 54652.6 \mathrm{K}\) So, a molecule of uranium hexafluoride would have the same average speed as a molecule of hydrogen at 37°C when its temperature is approximately 54652.6 K.