Problem 83 - Chemistry: The Central Science in SI Units, Global Edition

Question

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.

Step-by-Step Solution

Answer

Statements (a) and (d) are true. \(\mathrm{O}_{2}\) will effuse faster than \(\mathrm{Cl}_{2}\) due to its lower molar mass, as described by Graham's law of effusion. The mean free path of a gas is shorter when the gas has a higher density, as there are more molecule collisions in a given volume.

Step 1: Statement (a) Evaluation

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.

Step 2: Statement (b) Evaluation

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.

Step 3: Statement (c) Evaluation

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.

Step 4: Statement (d) Evaluation

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.

Key Concepts

Effusion

Effusion is a process where gas molecules pass through a small opening from one compartment to another, typically moving from an area of higher pressure to an area of lower pressure. This concept is quantitatively described by Graham's Law of Effusion, which provides a way to compare how different gases effuse based on their molar masses. According to Graham's Law:\[ \frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \]This equation shows that the rate of effusion for gas 1 is inversely proportional to the square root of its molar mass compared to gas 2. This means that lighter gases, with a lower molar mass, will effuse faster than heavier gases. For example, oxygen gas (\(\mathrm{O}_{2}\)) with a molar mass of 32 g/mol effuses faster than chlorine gas (\(\mathrm{Cl}_{2}\)), which has a molar mass of 71 g/mol. Therefore, lighter molecules like oxygen migrate quicker through openings compared to heavier ones.

Diffusion

Diffusion is the process by which molecules spread from an area of higher concentration to an area of lower concentration, usually resulting in an even distribution of the molecules over time. Unlike effusion, diffusion occurs without the need for a small opening. Instead, it relies on the natural, random motion of particles in a gas or liquid, causing them to mix and eventually form a homogenous mixture. Diffusion is central to many everyday processes. For instance:

• Perfume molecules move and spread throughout a room by diffusion, allowing the scent to reach our noses.
• In our bodies, oxygen diffuses from the lungs into the bloodstream, and carbon dioxide follows an opposite path to be expelled from the body.

Diffusion is a slower process than effusion because it involves many molecular collisions and is not restricted to passage through a small hole. Understanding diffusion aids in explaining how substances like gases move through different mediums.

Mean Free Path

The mean free path refers to the average distance that a gas molecule travels before it collides with another molecule. This concept is important for understanding gas behavior and properties like diffusion and effusion. The mean free path can be calculated using the equation:\[ \lambda = \frac{kT}{\sqrt{2} \pi d^{2} P} \]where \(\lambda\) represents 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 exerted by the gas.Several factors affect the mean free path:

• **Temperature**: As temperature increases, molecules move faster and the mean free path gets longer.
• **Pressure and Density**: Higher pressure or density results in more frequent collisions, reducing the mean free path.

Understanding the mean free path helps explain why gases with high densities, where molecules are closer together, tend to have shorter distances between collisions. This concept helps paint a clearer picture of how gases interact at a molecular level, influencing their diffusion rate and the speed at which they effuse through a barrier.