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.

The ideal gas law describes the relationship between pressure (P), volume (V), the number of moles of gas (n), and temperature (T) in Kelvins. It is given by the equation: \[PV = nRT\] where R is the ideal gas constant. Given the scenario, we know that the gases A and B have the same pressure (P), volume (V), and temperature (T) in their respective containers. Therefore, we can write for both gases: \[P_A V_A = n_A R T_A\] and \[P_B V_B = n_B R T_B\] However, the pressures and volumes are the same, and the temperatures are given as equal: \[P_A = P_B\] \[V_A = V_B\] \[T_A = T_B\] So, we can substitute these equal values and write: \[n_A R T = n_B R T\] Where n_A represents the number of moles of gas A, and n_B represents the number of moles of gas B.

The equation for the average kinetic energy per molecule in a gas (denoted as E) is given by: \[E = \frac{3}{2} kT\] where k is Boltzmann's constant and T is the temperature in Kelvins. Since we are given that the average kinetic energies of the two gases are equal and they share the same temperature, we can write: \[E_A = E_B\]

The average kinetic energy of a gas can also be expressed in terms of the root mean square velocity (represented as \(v_{rms}\)) and the molar mass (M) of the gas molecules: \[E = \frac{1}{2} M {v_{rms}}^2\] Now using this relation for both gases A and B, and equating their average kinetic energies, we have: \[\frac{1}{2} M_A {v_{rms, A}}^2 = \frac{1}{2} M_B {v_{rms, B}}^2\] Here, our goal is to show, despite the different molar masses, the root mean square velocities of the two gases cannot be equal.

To prove that the root mean square velocities of the two gases can't be equal, we can proceed by dividing the equation for gas B by the equation for gas A: \[\frac{M_A {v_{rms, A}}^2}{M_B {v_{rms, B}}^2} = 1\] Next, we want to isolate the ratio of the root mean square velocities: \[\frac{{v_{rms, A}}^2}{{v_{rms, B}}^2} = \frac{M_B}{M_A}\] Now, take the square root of both sides: \[\frac{v_{rms, A}}{v_{rms, B}} = \sqrt{\frac{M_B}{M_A}}\]

From the equation above, we can see that the ratio of the root mean square velocities of the two gases is equal to the square root of the inverse ratio of their molar masses. This means that the root mean square velocities of the gases are not equal, contrary to the initial assumption in the problem statement. The seeming contradiction arises because we mistakenly assumed that the root mean square velocities of the gases must be the same for them to have equal average kinetic energies. We have shown that even though the average kinetic energies are the same, the root mean square velocities can be different due to the different molar masses of the gases. Thus, even with different molar masses, both gases can maintain equal pressures, volumes, and temperatures as described by the ideal gas law. In conclusion, there is no contradiction between the initial assumptions and the results obtained, as the root mean square velocities of the two gases are not equal, enabling the conditions of equal pressure, volume, and temperature to be maintained without violating the ideal gas law or the average kinetic energy equations.