Question

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

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

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, \(v_{rms (He)} = \sqrt{\dfrac{3RT_1}{M_{He}}}\) \(v_{rms (UF6)} = \sqrt{\dfrac{3RT_2}{M_{UF6}}}\)

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.

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.

Key concepts

Molar Mass

Molar mass plays a crucial role in chemical calculations. It tells us the mass of a given substance divided by the amount of substance, measured in moles. The units typically used are grams per mole (g/mol).
It's straightforward to calculate if you know the atomic masses of the elements involved. For helium (\(\mathrm{He}\)), with an atomic mass of 4 g/mol, each mole weighs 4 grams. This is quite lightweight compared to uranium hexafluoride (\(\mathrm{UF_6}\)), which boasts a molar mass of 352 g/mol due to the heavy uranium and multiple fluorine atoms.

• He: Light gas, 4 g/mol.
• UF6: Heavier gas, 352 g/mol.

Understanding molar mass helps in determining how gases will behave under identical conditions of temperature and pressure. It’s integral in calculating the root mean square velocity and various other gas-related calculations.

Ideal Gas Law

The ideal gas law is a powerful equation in chemistry, relating the key aspects of gases—pressure, volume, temperature, and number of moles. This relationship is expressed as:\[ PV = nRT \]where

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant, and
• \( T \) is the temperature in Kelvin.

In the problem at hand, both the helium and uranium hexafluoride samples have the same volume and pressure, and contain the same number of moles. This sets a foundation for comparing their properties, such as root mean square velocity, under controlled and idealized conditions. By utilizing this law, we can explore how changing one variable, like temperature, impacts others in the equation while keeping certain aspects constant, which in this case, aids in finding the temperature ratio necessary for equal root mean square velocities.

Temperature Ratio

Determining the temperature ratio is critical when comparing different gases. In this case, we need the temperatures of the helium and uranium hexafluoride samples such that their root mean square velocities are identical. The formula used is:\[ \dfrac{T_1}{T_2} = \dfrac{M_{He}}{M_{UF6}} \]where \( T_1 \) and \( T_2 \) are the absolute temperatures of helium and UF6 respectively, and \( M_{He} \) and \( M_{UF6} \) are their respective molar masses.This formula arises from equating their root mean square velocity equations, which are both derived from kinetic molecular theory. Since the root mean square velocity of a gas depends inversely on its molar mass, a lighter gas like helium must be at a lower relative temperature than a heavier gas such as UF6 to match their velocities.The calculated temperature ratio of approximately 0.0114 means helium must be at a significantly lower temperature compared to UF6 to achieve the same kinetic behavior in terms of velocity.