Problem 104 - Chemistry An Atoms First Approach

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?

Step-by-Step Solution

Answer

The ratio of temperatures for the two samples of He(g) and UF6(g) that would produce the same root mean square velocity is 88.01.

Step 1: Write the rms velocity equation for He and UF6

For He: \(v_{rms, He} = \sqrt{\frac{3R(T_{He})}{M_{He}}}\) For UF6: \(v_{rms, UF6} = \sqrt{\frac{3R(T_{UF6})}{M_{UF6}}}\)

Step 2: Set the rms velocity equations equal to each other

\(\sqrt{\frac{3R(T_{He})}{M_{He}}} = \sqrt{\frac{3R(T_{UF6})}{M_{UF6}}}\)

Step 3: Square both sides of the equation to remove the square roots

\(\frac{3R(T_{He})}{M_{He}} = \frac{3R(T_{UF6})}{M_{UF6}}\)

Step 4: Solve for the ratio of temperatures

First, simplify by dividing both sides of the equation by 3R: \(\frac{T_{He}}{M_{He}} = \frac{T_{UF6}}{M_{UF6}}\) Now, solve for the ratio of temperatures (\(\frac{T_{He}}{T_{UF6}}\)): \(\frac{T_{He}}{T_{UF6}} = \frac{M_{UF6}}{M_{He}}\)

Step 5: Calculate the molar masses

Find the molar masses of He and UF6 in kg/mol. Molar mass of He = 4.00 g/mol (This is a well-known value, or you can find using the periodic table.) To convert to kg/mol, divide by 1000: Molar mass of He = \(\frac{4.00}{1000}\) kg/mol = 0.004 kg/mol Molar mass of UF6: 1 Uranium atom = 238.03 g/mol 6 Fluorine atoms = 6 × 19.00 g/mol = 114.00 g/mol Total molar mass of UF6 = 238.03 g/mol + 114.00 g/mol = 352.03 g/mol To convert to kg/mol, divide by 1000: Molar mass of UF6 = \(\frac{352.03}{1000}\) kg/mol = 0.35203 kg/mol

Step 6: Calculate the ratio of temperatures

Using the molar masses, we can find the ratio of temperatures: \(\frac{T_{He}}{T_{UF6}} = \frac{M_{UF6}}{M_{He}} = \frac{0.35203\, kg/mol}{0.004\, kg/mol} = 88.01\) The ratio of temperatures for the two samples that would produce the same root mean square velocity is 88.01.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that helps describe the behavior of gases. It combines several important gas laws into one equation: \(PV = nRT\), where:

• \(P\) is the pressure of the gas in atmospheres (atm) or pascals (Pa).
• \(V\) denotes the volume in liters (L) or cubic meters (m³).
• \(n\) is the number of moles of gas.
• \(R\) is the ideal gas constant, which can be expressed in various units such as \(8.314\, J/mol\cdot K\) or \(0.0821\, L\cdot atm/mol\cdot K\).
• \(T\) represents the temperature in Kelvin (K).

This law is key to understanding how gases will react under different conditions of pressure, volume, and temperature. It's important to first ensure that any given temperature is converted to Kelvin by adding 273.15 to the Celsius measurement.
In the context of the exercise, the Ideal Gas Law is fundamental in setting the conditions under which the root mean square velocity and temperatures are evaluated. Knowing the number of moles and volume allows students to solve related problems more efficiently.

Molar Mass Calculation

Calculating molar mass is a critical step in solving many chemistry problems. Molar mass tells us the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To determine the molar mass of a compound, you simply add up the atomic masses of all atoms present in a molecule. For instance:

• Helium (He) has an atomic mass of approximately \(4.00\, g/mol\).
• Uranium hexafluoride (UF₆) is more complex. It consists of one uranium (U) atom and six fluorine (F) atoms. The molar masses, derived from the periodic table, are:
• Uranium: \(238.03\, g/mol\)
• Fluorine: \(19.00\, g/mol\)
• Total for UF₆ = \((238.03 + 6 \times 19.00)\, g/mol\)

To use these values in calculations like the one in the exercise, they must be converted to kilograms per mole by dividing by \(1000\). This step aligns units when using the root mean square velocity equation, since it requires mass in terms of kilograms.

Kinetic Molecular Theory

The Kinetic Molecular Theory (KMT) provides insights into gas behavior based on the motion of molecules. It is a model that describes a gas as a large number of tiny particles (atoms or molecules) in constant, random motion. This motion results in properties such as pressure and temperature.
The theory explains several key points:

• Gas particles move rapidly in straight lines until they collide with other particles or the walls of their container.
• Collisions between gas particles or with the container walls are perfectly elastic. This means there is no loss of kinetic energy during collisions.
• The average kinetic energy of gas particles is directly proportional to the temperature of the gas. As the temperature increases, particles move faster.
• Gas pressure is a result of collisions between gas particles and the walls of their container.

The root mean square velocity discussed in the problem is derived from the KMT. It quantifies the average speed of gas molecules, directly linking the kinetic energy (and thus the temperature) to the molar mass of the gas. This helps explain why different gases at the same temperature may have different velocities based on their mass.