As discussed in the “Chemistry Put to Work” box in Section 10.8, enriched uranium can be produced by effusion of gaseous UF6 across a porous membrane. Suppose a process were developed to allow effusion of gaseous uranium atoms, U(g). Calculate the ratio of effusion rates for 235U and 238U, and compare it to the ratio for UF6 given in the essay.

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Identify the formula for the effusion rate ratio using Graham's law: \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r_1 \) and \( r_2 \) are the effusion rates of two gases, and \( M_1 \) and \( M_2 \) are their molar masses.

Determine the molar masses of the isotopes: \( M_{235U} = 235 \text{ g/mol} \) and \( M_{238U} = 238 \text{ g/mol} \).

Substitute the molar masses of \( 235U \) and \( 238U \) into Graham's law to find the effusion rate ratio: \( \frac{r_{235U}}{r_{238U}} = \sqrt{\frac{238}{235}} \).

Calculate the molar mass of \( UF_6 \) for both isotopes: \( M_{235UF_6} = 235 + 6(19) \) and \( M_{238UF_6} = 238 + 6(19) \).

Substitute the molar masses of \( UF_6 \) into Graham's law to find the effusion rate ratio for \( UF_6 \): \( \frac{r_{235UF_6}}{r_{238UF_6}} = \sqrt{\frac{M_{238UF_6}}{M_{235UF_6}}} \).

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 lighter gases effuse faster than heavier gases. For two gases, A and B, the relationship can be expressed as (Rate A / Rate B) = √(Molar Mass B / Molar Mass A), which is essential for comparing the effusion rates of isotopes like 235U and 238U.

Isotopes and Molar Mass

Isotopes are variants of a chemical element that have the same number of protons but different numbers of neutrons, resulting in different molar masses. In this case, 235U and 238U are isotopes of uranium with molar masses of approximately 235 g/mol and 238 g/mol, respectively. Understanding the differences in molar mass is crucial for applying Graham's Law to calculate their effusion rates.

Porous Membrane and Effusion

A porous membrane allows certain gases to pass through while restricting others based on size and kinetic energy. In the context of uranium effusion, the membrane's properties influence the separation of isotopes. The efficiency of this process can be analyzed by comparing the effusion rates of the gaseous forms of uranium, which is critical for understanding the enrichment process discussed in the question.

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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.

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.

Textbook Question

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.

Textbook Question

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?

Textbook Question

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.)

Textbook Question

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