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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. \(\mathrm{O}_{2}\) 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 by step solution

01

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.
02

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.
03

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.
04

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

Which of the following statements best explains why a closed balloon filled with helium gas rises in air? (a) Helium is a monatomic gas, whereas nearly all the molecules that make up air, such as nitrogen and oxygen, are diatomic. (b) The average speed of helium atoms is greater than the average speed of air molecules, and the greater speed of collisions with the balloon walls propels the balloon upward. (c) Because the helium atoms are of lower mass than the average air molecule, the helium gas is less dense than air. The mass of the balloon is thus less than the mass of the air displaced by its volume. (d) Because helium has a lower molar mass than the average air molecule, the helium atoms are in faster motion. This means that the temperature of the helium is greater than the air temperature. Hot gases tend to rise.

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Consider the combustion reaction between \(1.00 \mathrm{~L}\) of liquid methanol (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ) and \(500 \mathrm{~L}\) of oxygen gas measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\), molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-},\) forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$\mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s)$$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of \(152.0 \mathrm{kPa}\) at \(21^{\circ} \mathrm{C},\) how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

(a) Calculate the density of dinitrogen tetroxide gas \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) at \(111.5 \mathrm{kPa}\) and \(0{ }^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a gas if 2.70 g occupies \(0.97 \mathrm{~L}\) at \(134.7 \mathrm{~Pa}\) and \(100^{\circ} \mathrm{C}\).

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