Ch.10 - Gases

Brown - Chemistry: The Central Science 15th Edition

Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970당신이 사용하는 게 아니라요?교과서 변경

Brown 15th Edition

Ch.10 - Gases

문제 88

10장, 문제 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.)

검증된 단계별 안내

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 \( M_{\text{unknown}} \).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

관련 실천

교과서 질문

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.

1231

views

교과서 질문

(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.

1412

views

교과서 질문

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.

951

views

교과서 질문

Arsenic(III) sulfide sublimes readily, even below its melting point of 320 °C. The molecules of the vapor phase are found to effuse through a tiny hole at 0.52 times the rate of effusion of Xe atoms under the same conditions of temperature and pressure. What is the molecular formula of arsenic(III) sulfide in the gas phase?

1628

views

교과서 질문

Which statement concerning the van der Waals constants a and b is true? (a) The magnitude of a relates to molecular volume, whereas b relates to attractions between molecules. (b) The magnitude of a relates to attractions between molecules, whereas b relates to molecular volume. (c) The magnitudes of a and b depend on pressure. (d) The magnitudes of a and b depend on temperature.

1795

views