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.

We are given that: 1. Gases A and B are in separate containers of equal volume. 2. Gases A and B are at equal pressure and temperature. 3. The number of moles of gas A and gas B are equal. 4. The molar mass of gas A is greater than the molar mass of gas B.

The root mean square velocity (RMS velocity) for a gas can be calculated using the formula: \[v_{rms} = \sqrt{\frac{3RT}{M}}\] where \(v_{rms}\) = root mean square velocity \(R\) = ideal gas constant \(T\) = temperature \(M\) = molar mass of the gas Since the temperatures are equal, and gas A has a higher molar mass than gas B, their RMS velocities calculated by the equation indeed shouldn't be equal. This refutes the assumption that the RMS velocities are the same.

The pressure exerted by the gas can be calculated using the ideal gas equation: \[PV = nRT\] where \(P\) = pressure \(V\) = volume \(n\) = number of moles \(R\) = ideal gas constant \(T\) = temperature Considering equal temperatures, volumes, and number of moles, as well as the equal pressure stated in the initial assumption, we can set the equations for both gases as follows: \[P_{A}V = nRT_A\] \[P_{B}V = nRT_B\] Since the temperature is equal for both gases \(T_A = T_B\). Therefore, we have: \[P_{A} = P_{B}\] This validates one of the initial assumptions and clarifies that the pressures exerted by the two gases remain equal. #Conclusion#: The main issue within the given problem was falsely assuming that the RMS velocities should equal due to equal temperatures, while in reality, the difference in molar mass should affect the RMS velocities. Despite the molar mass difference and unequal RMS velocities, the initial assumptions of equal pressure, volume and temperature hold true. Therefore, the contradiction is resolved.