Explain the following seeming contradiction: You have two gases, \(A\) and \(B\), in two separate containers of equal volume and at equal pressure and temperature. Therefore, you must have the same number of moles of each gas. Because the two temperatures are equal, the average kinetic energies of the two samples are equal. Therefore, since the energy given such a system will be converted to translational motion (that is, move the molecules), the root mean square velocities of the two are equal, and thus the particles in each sample move, on average, with the same relative speed. Since \(A\) and \(B\) are different gases, they each must have a different molar mass. If \(A\) has a higher molar mass than \(B\), the particles of \(A\) must be hitting the sides of the container with more force. Thus the pressure in the container of gas \(A\) must be higher than that in the container with gas \(B .\) However, one of our initial assumptions was that the pressures were equal.

Step by step solution

01

Understand the ideal gas law and the mean molecular mass of a substance

The ideal gas law relates the volume, pressure, temperature, and the number of moles of a gas: \[PV = nRT\], where \(P\) is the pressure of the gas, \(V\) is the volume of the gas, \(n\) is the number of moles of the gas, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. Additionally, mean molecular mass is a measure of the average mass of one molecule in a mixture of molecules, usually measured in units of atomic mass units (amu) or grams per mole.

02

Establish a relationship between the average kinetic energy, the molar mass, and the root mean square velocity

From the kinetics theory of gases, we know that the average kinetic energy of a gas molecule (K.E) is directly proportional to the temperature (\(T\)): \[K.E = \dfrac{3}{2}kT\], where \(k\) is the Boltzmann constant. We also know that the average kinetic energy is given by the equation: \[K.E = \dfrac{1}{2}m\_avg * v^2_{rms}\], where \(m\_avg\) is the mean molecular mass, and \(v^2_{rms}\) is the root mean square velocity of the gas molecules. Combining these two relationships, we can write: \[\dfrac{3}{2}kT = \dfrac{1}{2}m\_avg * v^2_{rms}\]

03

Derive the expression for the root mean square velocity in terms of the molecular mass

From the relationship established in Step 2, we can derive an expression for the root mean square velocity with respect to molecular mass as follows: \[v_{rms} = \sqrt{ \dfrac{3kT}{m\_avg}}\] This equation shows that the root mean square velocity (\(v_{rms}\)) is inversely proportional to the square root of the mean molecular mass (\(m\_avg\)), and directly proportional to the square root of the temperature (\(T\)).

04

Use the expression for the root mean square velocity of gases A and B to demonstrate that the pressure is indeed the same

Now, we can use the equation derived in Step 3 and apply it to both gases A and B: \[v_{rms} = \sqrt{ \dfrac{3kT}{m\_avg}} \Rightarrow \begin{cases} v_{rmsA} = \sqrt{ \dfrac{3kT}{m_A}} \\ v_{rmsB} = \sqrt{ \dfrac{3kT}{m_B}} \end{cases}\] Since both gases have the same temperature (\(T\)), we can analyze the pressures by comparing their kinetic properties and applying the ideal gas law. Pressure is created by the gas molecules hitting the walls of the container; the force of these collisions is related to the momentum of the gas molecules (m*v). Therefore, the pressure created by gas A is proportional to \(m_A*v_{rmsA}\), and the pressure created by gas B is proportional to \(m_B*v_{rmsB}\). However, as m_A is inversely proportional to its \(v_{rmsA}\) and the same for B, these factors cancel each other out in the calculation. Thus, they produce equal pressure on the walls of their containers, and there is no contradiction.

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