Which one or more of the following statements are true? (a) \(\mathrm{O}_{2}\) will effuse faster than \(\mathrm{Cl}_{2}\). (b) Effusion and diffusion are different names for the same process. (c) Perfume molecules travel to your nose by the process of effusion. (d) The higher the density of a gas, the shorter the mean free path.

To determine if \(\mathrm{O}_{2}\) will effuse faster than \(\mathrm{Cl}_{2}\), we can use Graham's law of effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be represented as: \[\frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\] where \(Rate_{1}\) and \(Rate_{2}\) are the effusion rates of gases 1 and 2, and \(M_{1}\) and \(M_{2}\) are their respective molar masses. The molar mass of \(\mathrm{O}_{2}\) is 32 g/mol, and the molar mass of \(\mathrm{Cl}_{2}\) is 71 g/mol. Plug the values into the equation: \[\frac{Rate_{O_{2}}}{Rate_{Cl_{2}}} = \sqrt{\frac{71}{32}}\] Since the fraction inside the square root is greater than 1, the effusion rate of \(\mathrm{O}_{2}\) is greater than the effusion rate of \(\mathrm{Cl}_{2}\). Therefore, statement (a) is true.

Effusion refers to the process by which a gas passes through a small hole or opening, while diffusion is the process by which gas molecules move and spread out as a result of random motion. Although both processes involve the movement of gas molecules, they describe different physical phenomena and are not the same process. Therefore, statement (b) is false.

Perfume molecules travel through the air and reach our noses through the process of diffusion. As mentioned earlier, diffusion is the random movement of gas molecules that results in spreading out over time. In the case of perfume, the pleasant scent is the result of the perfume molecules moving and mixing with air molecules. This movement is due to diffusion and not effusion. Therefore, statement (c) is false.

The mean free path is the average distance a molecule travels between collisions with other molecules. It can be represented as: \[\lambda = \frac{kT}{\sqrt{2} \pi d^{2} P}\] where \(\lambda\) is the mean free path, \(k\) is the Boltzmann constant, \(T\) is the temperature, \(d\) is the diameter of the gas molecules, and \(P\) is the pressure of the gas. A higher density of gas implies that there are more gas molecules in a given volume. Therefore, the probability of collision between gas molecules will increase, which in turn will result in a shorter mean free path. This means statement (d) is true. In conclusion, statements (a) and (d) are true.