Calculate the relative rates of diffusion for \({ }^{235} \mathrm{UF}_{6}\) and \({ }^{238} \mathrm{UF}_{6}\).

Short Answer

Expert verified
The relative rate of diffusion of 235 UF6 to 238 UF6 is sqrt(352 / 349), which is approximately 1.0043.

Step by step solution

01

Identify the Molecular Masses

Determine the molecular masses of both isotopes of UF6. The atomic mass of uranium-235 is 235 amu and for uranium-238 is 238 amu. The molecular mass of UF6 includes the mass of the uranium isotope plus the mass of 6 fluorine atoms (6 x 19 amu = 114 amu). Therefore, the molecular mass of - 235 UF6 is 235 amu + 114 amu = 349 amu, - 238 UF6 is 238 amu + 114 amu = 352 amu.
02

Use Graham's Law of Effusion

Apply Graham's Law of Effusion, which states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass. Mathematically, rate1 / rate2 = sqrt(M2 / M1) where M1 and M2 are the molar masses of the gases.
03

Compute the Square Roots of the Molecular Masses

Calculate the square roots of the molecular masses:- sqrt(349 amu) for 235 UF6, and - sqrt(352 amu) for 238 UF6.
04

Calculate the Relative Rates of Diffusion

Plug the square roots of the molecular masses into Graham's Law to find the relative rates:rate235 / rate238 = sqrt(352 / 349)Solve for the ratio of the rates to obtain the relative rates of diffusion.

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Key Concepts

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

Relative Rates of Diffusion
Understanding the relative rates of diffusion between two gases is insightful when examining how quickly one substance will spread out or effuse compared to another. Diffusion is fundamentally the movement of particles from a region of higher concentration to a region of lower concentration.

When considering gases, such as the isotopes of uranium hexafluoride ( ^{235}UF_{6} and ^{238}UF_{6} ), Graham's Law gives us the ability to determine which gas will effuse faster based on the inversely proportional relationship between the rate of effusion (or diffusion) and the square root of the gas's molar mass. Therefore, a lighter gas diffuses more rapidly than a heavier one.

This concept is not only practical in laboratory settings where diffusion plays a part, but also in industrial applications, such as gas purification or even environmental studies where diffusion rates affect the dispersion of pollutants.
Molecular Mass Calculation
Calculating the molecular mass of a compound is a foundational skill in chemistry. The molecular mass, often expressed in atomic mass units (amu), is the sum of the mass numbers of all the atoms in a single molecule of the substance.

For example, to calculate the molecular mass of uranium hexafluoride isotopes, we consider both the mass of the uranium isotope and the mass of six fluorine atoms. Given the masses provided in our exercise, we can readily calculate this for ^{235}UF_{6} and ^{238}UF_{6} .

  • ^{235}UF_{6} : 235 amu (Uranium-235) + 6 × 19 amu (Fluorine) = 349 amu
  • ^{238}UF_{6} : 238 amu (Uranium-238) + 6 × 19 amu (Fluorine) = 352 amu

Understanding this calculation is crucial when using Graham's Law to compare the effusion rates of different gases.
Isotopes in Physical Chemistry
Isotopes play a significant role in physical chemistry due to their applications in various chemical processes and radiometric dating techniques. An isotope of an element contains the same number of protons but a different number of neutrons in its nucleus, which results in distinct atomic masses.

In our exercise dealing with ^{235}UF_{6} and ^{238}UF_{6} , these uranium hexafluoride compounds differ only in the uranium isotope present. The physical properties of isotopes, including their mass, can influence the rate at which physical processes occur, such as effusion.

It's fascinating to note that even isotopes of the same element can behave differently in a chemical context. This distinction is especially important when considering physical phenomena that are influenced by the mass of a substance, such as effusion, diffusion, and sedimentation.
Gas Effusion
Gas effusion is a term that refers to the escape of gas molecules through a tiny opening, such as a pinhole, into a vacuum. Since this process depends on the speed of the gas molecules, lighter molecules effuse at a faster rate than heavier ones due to their higher average kinetic energy at the same temperature.

Applying Graham's Law to the effusion of gases allows us to compare effusion rates quantitatively. For instance, by calculating the molecular mass of uranium hexafluoride isotopes, we can determine how quickly one isotope will effuse relative to the other.

This process has practical applications in various scientific fields, including industrial separation technologies where gases need to be separated based on their molar masses. Understanding gas effusion is also crucial in areas like atmospheric science and astrophysics, where the behavior of gases in different environments is studied.

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Most popular questions from this chapter

A bulb of three litre capacity filled with air is heated from \(27^{\circ} \mathrm{C}\) to \(t^{\circ} \mathrm{C}\). The air thus, expelled measured \(1.45\) litre at \(17^{\circ} \mathrm{C}\). Considering the pressure to be 1 atm throughout the experiment and ignoring the expansion of bulb, calculate \(t\).

Equal volumes of two gases \(A\) and \(B\) diffuse through a porous pot in 20 and 10 seconds respectively. If the molar mass of \(A\) be 80, find the molar mass of \(B\)

One litre of a gas at 300 atm and \(473 \mathrm{~K}\) is compressed to a pressure of 600 atm and \(273 \mathrm{~K}\). The compressibility factors found to be \(1.072\) and \(1.375\) respectively at the initial and finally conditions. Calculate the final volume.

An evacuated bulb of unknown volume is filled with a sample of \(\mathrm{H}_{2}\) gas at a temperature \(T\). The pressure of the gas in the bulb is \(756 \mathrm{~mm} \mathrm{Hg}\). A portion of the \(\mathrm{H}_{2}\) gas is transferred to a different flask and found to occupy a volume of \(40.0 \mathrm{~mL}\) at \(1.00\) atm and the same temperature \(T\). The pressure of the \(\mathrm{H}_{2}\) gas remaining in the original bulb drops to 625 \(\mathrm{mm} \mathrm{Hg}\) at the same temperature \(T\). Assuming \(\mathrm{H}_{2}\) is an ideal gas, what is the volume of the bulb?

An open flask contains air at \(27^{\circ} \mathrm{C}\). Calculate the temperature at which it should be heated so that, (a) \(\frac{1}{3} \mathrm{rd}\) of air measured at \(27^{\circ} \mathrm{C}\) escapes out. (b) \(\frac{1}{3} \mathrm{rd}\) of air measured at final temperature escapes out.

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