Problem 83 - Chemistry The Central Science In Si Units

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.

Step 1: Understanding Effusion and Diffusion

Effusion is the process by which gas molecules escape through a small hole into a vacuum. Diffusion is the spreading of gas molecules throughout a container. They are distinct processes, so statement (b) is false.

Step 2: Comparing Effusion Rates Using Graham's Law

According to Graham's Law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Cl_2 has a molar mass of approximately 71 g/mol and O_2 has a molar mass of about 32 g/mol. Therefore, O_2 effuses faster than Cl_2, making statement (a) true.

Step 3: Perfume Molecules Travel by Diffusion

Perfume molecules spread out in a room and reach your nose through diffusion, not effusion, as diffusion involves spreading in a space already filled with gas. Statement (c) is false.

Step 4: Density and Mean Free Path Relationship

The mean free path of a gas molecule is the average distance it travels before colliding with another molecule. A higher density indicates more molecules are present, hence collisions are more frequent and the mean free path is shorter. Therefore, statement (d) is true.

Key Concepts

Graham's Law

Graham's Law is a crucial concept when discussing gas movement through effusion. It states that the rate at which a gas will effuse is inversely proportional to the square root of its molar mass. This means that lighter gases will effuse faster due to their smaller molar mass.
For instance, if we compare oxygen (\(\mathrm{O}_{2}\)) with chlorine (\(\mathrm{Cl}_{2}\)), we find that oxygen is lighter due to its smaller molar mass of about 32 g/mol, compared to chlorine's 71 g/mol.
Using Graham's Law, we know:\[\frac{\text{rate of effusion of O}_2}{\text{rate of effusion of Cl}_2} = \sqrt{\frac{M_2}{M_1}}\] where \(M_1\) and \(M_2\) are the molar masses of the respective gases.- This equation shows why oxygen effuses faster than chlorine.- It's an important tool for understanding the behavior of gases under different conditions.

Mean Free Path

The mean free path is another key concept in gas behavior, describing the average distance a gas molecule travels before colliding with another molecule.
When we talk about mean free path, we consider factors like temperature, pressure, and most importantly, density. - A higher density means more molecules are present per unit space. - More molecules result in more collisions and therefore shorter mean free paths. To visualize, imagine a crowded subway train. The more people (or molecules) in the train, the shorter the distance any single person can walk without bumping into someone. Similarly, in a dense gas, molecules don't travel far without colliding, indicating a shorter mean free path. This concept helps us understand why gases behave differently under varying conditions, such as high pressure or different temperatures.

Molar Mass Comparison

Comparing molar masses of different gases helps predict their relative behavior in processes like effusion and diffusion.
Molar mass, which is the mass of one mole of a substance, is a factor that highly influences how gases interact and spread.- \(\mathrm{O}_2\) has a molar mass of about 32 g/mol, while \(\mathrm{Cl}_2\) stands at roughly 71 g/mol.- The lighter oxygen molecules move faster than the heavier chlorine molecules.This comparison explains why oxygen (a lighter gas) will undergo processes like effusion at a faster rate than chlorine. Understanding these differences in molar mass is essential for applications that involve gas mixtures and reactions, as it impacts both the rate and the efficiency of these processes.