The way in which the masses of hydrogen molecules and oxygen molecules are compared by the use of given information is to be determined. The way in which the masses of hydrogen molecules and oxygen molecules are compared by the use of information in the periodic table is to be determined. Concept introduction: Diffusion is a process in which particles move from higher concentration area to lower concentration area. The relation between rate of diffusion and effusion of gases with their molecular mass is given by Graham’s law of effusion. The Graham’s law of effusion is mathematically represented as, effusion rate A effusion rate B = Molecular mass of B Molecular mass of A

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Chapter 6, Problem 6.60E

Interpretation Introduction

Interpretation:

The way in which the masses of hydrogen molecules and oxygen molecules are compared by the use of given information is to be determined. The way in which the masses of hydrogen molecules and oxygen molecules are compared by the use of information in the periodic table is to be determined.

Concept introduction:

Diffusion is a process in which particles move from higher concentration area to lower concentration area. The relation between rate of diffusion and effusion of gases with their molecular mass is given by Graham’s law of effusion. The Graham’s law of effusion is mathematically represented as,

effusion rate Aeffusion rate B=Molecular mass of BMolecular mass of A

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At a given temperature and pressure, it takes 9.07 minutes for a 1.5-L sample of He (helium) to effuse through a membrane. How long does it take for 1.5 L of F2 (fluorine) to effuse under the same conditions? Express your answer in units of minutes. Express your answer to the appropriate number of significant figures.

This relationship is known as Graham’s Law of Effusion. Since both gases are at the same temperature, they must have the same average kinetic energy (½ mv2), where m is mass and v is velocity (like speed). Since both gases have the same average kinetic energy, you can state that ½ mLvL2 = ½ mHvH2. Multiplying both sides by 2 gives you mLvL2 = mHvH2. Rearranging the equation to get both masses on the same side of the equation will give you mL/mH = VH2/VL2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, VH2/VL2 = ¼. If the light gas was Ne, what would be a reasonable identity for the heavy gas? Explain.

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