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 the two samples of helium gas (He) and uranium hexafluoride gas (UF6) that would produce the same root mean square velocity is \(1:88\).

Step-by-step solution

Use rms velocity formula

The root mean square velocity is given by the equation: \(v_{rms} = \sqrt{\frac{3RT}{M}}\) where \(v_{rms}\) is the rms velocity, \(R\) is the ideal gas constant, \(T\) is the temperature in kelvin, and \(M\) is the molar mass of the gas.

Set equal rms velocities for both gases

Since we need to find the temperature ratio for both gases to have the same rms velocity, we can set their rms velocities equal to each other: \(\sqrt{\frac{3RT_{He}}{M_{He}}} = \sqrt{\frac{3RT_{UF6}}{M_{UF6}}}\)

Simplify the expression

First, we can square the equation to get rid of the square root sign on both sides: \(\frac{3RT_{He}}{M_{He}} = \frac{3RT_{UF6}}{M_{UF6}}\) Since 3 and R are constant, we can further simplify the equation by dividing both sides by 3R: \(\frac{T_{He}}{M_{He}} = \frac{T_{UF6}}{M_{UF6}}\)

Find the molar masses of He and UF6

We need to find the molar masses of He and UF6 in order to plug them into our equation. For helium (He): The atomic mass of helium is 4 g/mol. For uranium hexafluoride (UF6): Uranium (U): atomic mass = 238 g/mol Fluorine (F): atomic mass = 19 g/mol Since UF6 has 6 fluorine atoms and 1 uranium atom, its molar mass is: \(M_{UF6} = 1 \times 238 + 6 \times 19 = 238 + 114 = 352 \, g/mol\)

Plug in molar masses and solve for temperature ratio

Now that we have the molar masses, we can plug them into our equation from Step 3, and solve for the ratio of temperatures: \(\frac{T_{He}}{4} = \frac{T_{UF6}}{352}\) To find the ratio of temperatures, we need to isolate one of the temperatures and then divide it by the other temperature. We'll isolate \(T_{He}\) and find the ratio \(\frac{T_{He}}{T_{UF6}}\): \(T_{He} = \frac{4T_{UF6}}{352}\) Therefore, \(\frac{T_{He}}{T_{UF6}} = \frac{4T_{UF6}}{352T_{UF6}} = \frac{4}{352}\)

Simplify the temperature ratio

Finally, let's simplify the fraction to find the required temperature ratio: \(\frac{T_{He}}{T_{UF6}} = \frac{4}{352} = \frac{1}{88}\) So, the ratio of temperatures for the two samples that would produce the same root mean square velocity is \(1:88\).

Key concepts

Understanding the Ideal Gas Law

The ideal gas law is fundamental in understanding how gases behave under various conditions of pressure, volume, temperature, and amount. This law relates four state variables: pressure (P), volume (V), number of moles of gas (n), and temperature (T), and it's encapsulated in the equation PV = nRT, where R is the ideal gas constant.

When dealing with problems related to the root mean square velocity, we often rearrange this equation to solve for a specific variable, like temperature. In an ideal gas, the particles are considered as small points with no volume and no interactions with each other, allowing us to use the ideal gas law to deduce properties, like the root mean square velocity, that depend on these state variables.

Understanding the ideal gas law is not just about memorizing the equation, but also grasping the proportional relationships between the variables. For instance, if you increase the temperature while holding the pressure constant, the volume must also increase to maintain the balance in the PV = nRT equation. Thus, a deep understanding of the ideal gas law is valuable for solving various gas-related problems in chemistry and physics.

Significance of Molar Mass

Molar mass is a key concept that bridges the gap between the atomic or molecular scale and the macroscopic world we interact with. It represents the mass of one mole (approximately 6.022 x 10^23 particles) of a substance and is typically expressed in grams per mole (g/mol).

When calculating the root mean square velocity, the molar mass becomes crucial because it is inversely proportional to the velocity of gas particles. Heavier particles (higher molar mass) move more slowly than lighter ones at the same temperature. This relationship allows us to compare different gases, like helium and uranium hexafluoride, under the lens of their molar masses to deduce how their velocities will vary.

In problems involving gas behavior, it's important to accurately determine the molar mass by summing up the atomic masses of the constituent atoms. As shown in the exercise, the molar mass directly influences calculations related to the gas properties and is used to solve for variables such as temperature ratios for maintaining a constant root mean square velocity.

Interpreting Temperature Ratio of Gases

Temperature plays a central role in defining the kinetic energy and velocity of gas particles. Since the root mean square velocity is a measure of the average velocity of the gas particles, the temperature directly influences this value.

The concept of temperature ratio becomes significant when comparing two different gases, as it allows us to find a proportion where certain properties of these gases, like their velocities, are equal. In the exercise, the aim is to determine the temperature ratio that would equate the root mean square velocity of two gases with differing molar masses.

Here's a critical connection: higher temperature means greater thermal energy imparted to the particles, hence a higher velocity. By setting up a ratio, we link these temperatures to their corresponding gases' molar masses and solve for the conditions under which they achieve the same kinetic energy. It's like balancing two sides of a scale to achieve equilibrium, and it beautifully demonstrates how properties at the microscopic level relate to the measurable conditions we control.