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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.

Short Answer

Expert verified
Statements (a) and (d) are true.

Step by step solution

01

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.
02

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.
03

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.
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A person weighing \(75 \mathrm{~kg}\) is standing on a threelegged stool. The stool momentarily tilts so that the entire weight is on one foot. If the contact area of each foot is \(5.0 \mathrm{~cm}^{2},\) calculate the pressure exerted on the underlying surface in (a) bars, \((\mathbf{b})\) atmospheres, and (c) pounds per square inch.

The temperature of a 5.00-L container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\). If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; \((\mathbf{b})\) the rootmean- square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; d) the total number of collisions of molecules with walls per second.

When a large evacuated flask is filled with argon gas, its mass increases by \(3.224 \mathrm{~g}\). When the same flask is again evacuated and then filled with a gas of unknown molar mass, the mass increase is \(8.102 \mathrm{~g}\). (a) Based on the molar mass of argon, estimate the molar mass of the unknown gas. (b) What assumptions did you make in arriving at your answer?

Suppose you are given two 2 -L flasks and told that one contains a gas of molar mass 28 , the other a gas of molar mass 56 , both at the same temperature and pressure. The mass of gas in the flask \(A\) is \(1.0 \mathrm{~g}\) and the mass of gas in the flask \(\mathrm{B}\) is \(2.0 \mathrm{~g}\). Which flask contains the gas of molar mass \(28,\) and which contains the gas of molar mass \(56 ?\)

Perform the following conversions: (a) 0.912 atm to torr, (b) 0.685 bar to kilopascals, (c) \(655 \mathrm{~mm}\) Hg to atmospheres, (d) \(1.323 \times 10^{5}\) Pa to atmospheres, (e) 2.50 atm to psi.

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