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

In order to make the root mean square velocities equal for separate samples of He(g) and UF6(g) containing the same number of moles, the ratio of temperatures between the helium and uranium hexafluoride samples should be 88:1.

Step-by-step solution

Recall the formula for root mean square velocity

We know that the root mean square (RMS) velocity for a gas molecule can be given by the formula: \[v_{rms} = \sqrt{\frac{3RT}{M}}\] where \(v_{rms}\) is the root mean square velocity, R is the universal gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas molecule in kg/mol. So, in order to find out the ratio of temperatures at which the two gases have the same root mean square velocity, we should set the root mean square velocity for He equal to the root mean square velocity for UF6, and solve for the temperature ratio.

Write the equation for the ratio of temperatures

Let \(T_{He}\) be the temperature of helium and \(T_{UF6}\) be the temperature of uranium hexafluoride. According to the given condition, the root mean square velocities of both gases should be equal. We can write this condition as follows: \[\sqrt{\frac{3RT_{He}}{M_{He}}} = \sqrt{\frac{3RT_{UF6}}{M_{UF6}}}\] Now, we should find the ratio of temperatures, \(T_{He}\) and \(T_{UF6}\), in order to make the root mean square velocities of both gases equal.

Solve for the temperature ratio

By squaring both sides of the equation, we get: \[\frac{3RT_{He}}{M_{He}} = \frac{3RT_{UF6}}{M_{UF6}}\] Since both samples contain the same number of moles, we can cancel out the 3 and R terms from both sides: \[\frac{T_{He}}{M_{He}} = \frac{T_{UF6}}{M_{UF6}}\] Now, we can divide both sides of the equation by the molar mass of He and then by the molar mass of UF6 to find the ratio of temperatures: \[\frac{T_{He}}{T_{UF6}} = \frac{M_{UF6}}{M_{He}}\] To find the ratio of temperatures, we need to know the molar masses of He and UF6. For helium (He), the molar mass is 4.00 g/mol, and for uranium hexafluoride (UF6), the molar mass is 238 g/mol (U) + 6 * 19 g/mol (F) = 352 g/mol.

Plug in the molar masses and find the ratio of temperatures

Now, we can plug in the molar masses of He and UF6 into the equation to find the ratio of temperatures: \[\frac{T_{He}}{T_{UF6}} = \frac{352}{4.00} = 88\] Therefore, in order to make the root mean square velocities equal, the ratio of temperatures between the helium and uranium hexafluoride samples should be 88:1.

Key concepts

Kinetic Molecular Theory

The Kinetic Molecular Theory (KMT) explains the behavior of gases, including how they diffuse, how pressure is formed, and their reaction to temperature changes. It describes gases as large numbers of submicroscopic particles (atoms or molecules), all of which are in constant, random motion.

The fundamental premises of KMT include the following:

• The particles in a gas are small compared to the distances between them and, as a result, the volume of the individual particles can be assumed to be negligible (zero).
• The particles are in constant random motion, colliding with each other and the walls of their container. These collisions are perfectly elastic, meaning there's no net loss of kinetic energy.
• The average kinetic energy of gas particles is directly proportional to the temperature of the gas in Kelvins. This implies that as the temperature of the gas increases, the average speed of the molecules—also reflected in the RMS velocity—will also increase.

By relying on these principles, we can understand how the root mean square velocity (\(v_{rms}\)) relates to the temperature and molecular mass of a gas. This relationship is crucial when we are solving problems such as the textbook exercise, comparing \(v_{rms}\) of different gases at varying temperatures.

Gas Laws

Gas laws are the scientific laws that describe the behavior of gases under different conditions of temperature, pressure, and volume. These laws are summarized by various equations that express these relationships, and at times they are interconnected by the ideal gas law, which is an equation of state for a hypothetical ideal gas.

The ideal gas law is represented as \(PV = nRT\), where \(P\) stands for pressure, \(V\) for volume, \(n\) for moles of the gas, \(R\) for the universal gas constant, and \(T\) for temperature. When dealing with RMS velocity, we extend this further to include the relationship between the kinetic energy of molecules and the mass of the molecules, as illustrated in the RMS velocity formula \(v_{rms} = \sqrt{\frac{3RT}{M}}\).Understanding gas laws allows us to manipulate and solve for the variables in question, such as finding the temperature ratio that would equate the RMS velocities of two different gases, as demonstrated in the given exercise. The analysis becomes easier as we realize we can cancel out the constant terms and focus solely on the variables that differ: the molar masses and the temperatures of the two gases.

Molecular Speed

Molecular speed refers to the speed of individual gas molecules, which depends on the temperature and molar mass of the gas. According to the kinetic molecular theory, all gas molecules have the same average kinetic energy at a given temperature. However, lighter molecules move faster than heavier molecules because they have less mass. This is why different gases at the same temperature will have different RMS velocities.

The root mean square velocity (\(v_{rms}\)) is a way to define the average speed of gas molecules, which is given by the equation \(v_{rms} = \sqrt{\frac{3RT}{M}}\). The equation takes into account the gas constant (\(R\)), the temperature (\(T\)), and the molar mass (\(M\)) to calculate this average speed. In the context of our exercise, to achieve the same \(v_{rms}\) for both helium (\(He\)) and uranium hexafluoride (\(UF_{6}\)), we must adjust the temperature for each gas. Since \(He\) molecules are much lighter than \(UF_{6}\) molecules, the temperature of \(He\) must be significantly higher to compensate for its lower mass and to match the molecular speeds of the two gases. As the exercise illustrates, this relationship can be used to find the required temperature ratio for both gases to have the same \(v_{rms}\).