Ch.10 - Gases

Brown - Chemistry: The Central Science 15th Edition

Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970Not the one you use?Change textbook

Brown 15th Edition

Ch.10 - Gases

Problem 88

Chapter 10, Problem 88

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 105 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 31 s for 1.0 L of O2 gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; in other words, rate is the amount that diffuses over the time it takes to diffuse.)

Name: Chemistry: The Central Science
Author: Brown
ISBN: 9780137542970

Verified step by step guidance

Identify the relationship between the rates of effusion and the molar masses of the gases using Graham's Law of Effusion: \( \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \), where Rate is inversely proportional to time.

Express the rates of effusion in terms of time: \( \text{Rate}_{\text{unknown}} = \frac{1.0 \text{ L}}{105 \text{ s}} \) and \( \text{Rate}_{\text{O}_2} = \frac{1.0 \text{ L}}{31 \text{ s}} \).

Substitute the rates into Graham's Law: \( \frac{\frac{1.0}{105}}{\frac{1.0}{31}} = \sqrt{\frac{32.00}{M_{\text{unknown}}}} \), where 32.00 g/mol is the molar mass of \( \text{O}_2 \).

Simplify the expression: \( \frac{31}{105} = \sqrt{\frac{32.00}{M_{\text{unknown}}}} \).

Square both sides to solve for the molar mass of the unknown gas: \( \left(\frac{31}{105}\right)^2 = \frac{32.00}{M_{\text{unknown}}} \), then rearrange to find \( M_{\text{unknown}} \).

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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. This means that lighter gases effuse faster than heavier gases. The relationship can be expressed mathematically as (Rate1/Rate2) = √(Molar Mass2/Molar Mass1), allowing for the comparison of effusion rates between two gases.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a critical property in stoichiometry and gas calculations, as it relates the mass of a substance to the number of particles it contains. In the context of effusion, knowing the molar mass allows for the determination of the rate at which a gas will effuse compared to another gas.

Rate of Effusion

The rate of effusion refers to the volume of gas that escapes through a small opening per unit of time. It is influenced by factors such as the size of the gas molecules and the temperature of the system. In this problem, the time taken for a specific volume of gas to effuse is used to calculate the rate, which is then compared to the rate of effusion of oxygen to find the molar mass of the unknown gas.

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Effusion Rate Example

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Textbook Question

Which one or more of the following statements are true? (a) O2 will effuse faster than Cl2. (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.

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Textbook Question

Calculate the pressure that CCl₄ will exert at 80 °C if 1.00 mol occupies 33.3 L, assuming that (a) CCl₄ obeys the ideal-gas equation (b) CCl₄ obeys the van der Waals equation. (Values for the van der Waals constants are given in Table 10.3.)

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Textbook Question

The planet Jupiter has a surface temperature of 140 K and a mass 318 times that of Earth. Mercury (the planet) has a surface temperature between 600 K and 700 K and a mass 0.05 times that of Earth. On which planet is the atmosphere more likely to obey the ideal-gas law? Explain.

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Textbook Question

(b) List two reasons why the gases deviate from ideal behavior.

Hydrogen has two naturally occurring isotopes, 1H and 2H. Chlorine also has two naturally occurring isotopes, 35Cl and 37Cl. Thus, hydrogen chloride gas consists of four distinct types of molecules: 1H35Cl, 1H37Cl, 2H35Cl, and 2H37Cl. Place these four molecules in order of increasing rate of effusion.

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Textbook Question

At constant pressure, the mean free path 1l2 of a gas molecule is directly proportional to temperature. At constant temperature, l is inversely proportional to pressure. If you compare two different gas molecules at the same temperature and pressure, l is inversely proportional to the square of the diameter of the gas molecules. Put these facts together to create a formula for the mean free path of a gas molecule with a proportionality constant (call it Rmfp, like the ideal-gas constant) and define units for Rmfp.

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