Question

Calculate the ratio of rates of effusion of \({ }^{235} \mathrm{UF}_{6}\) and \({ }^{238} \mathrm{UF}_{6}\), where \({ }^{235} \mathrm{U}\) and \({ }^{238} \mathrm{U}\) are isotopes of uranium. The atomic weights are \({ }^{235} \mathrm{U}, 235.04 \mathrm{amu} ;{ }^{238} \mathrm{U}, 238.05 \mathrm{amu} ;{ }^{19} \mathrm{~F}\) (the only naturally occurring isotope), 18.998 amu. Carry five significant figures in the calculation.

Short answer

The ratio of effusion rates of \( {}^{235}\mathrm{UF}_6 \) to \( {}^{238}\mathrm{UF}_6 \) is approximately 1.0043.

Step-by-step solution

Understand the Concept of Effusion

According to Graham's law of effusion, the ratio of effusion rates for two gases is inversely proportional to the square root of their molar masses. We can write this mathematically as \( \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \), where \( R_1 \) and \( R_2 \) are the rates of effusion, and \( M_1 \) and \( M_2 \) are their respective molar masses.

Calculate the Molar Masses

For \( {}^{235}\mathrm{UF}_6 \), the molar mass is calculated as: \[ M_1 = 235.04 + 6 \times 18.998 = 235.04 + 113.988 = 349.028 \, \text{amu} \]For \( {}^{238}\mathrm{UF}_6 \), the molar mass is:\[ M_2 = 238.05 + 6 \times 18.998 = 238.05 + 113.988 = 352.038 \, \text{amu} \]

Apply Graham's Law

Substitute the molar masses calculated above into Graham's law:\[ \frac{R_1}{R_2} = \sqrt{\frac{352.038}{349.028}} \]Calculate the numerical value of the ratio of effusion rates.

Perform the Calculation

Compute the square root for the ratio derived in the previous step:\[ \frac{R_1}{R_2} = \sqrt{\frac{352.038}{349.028}} \approx \sqrt{1.008635416} \approx 1.0043 \]This gives the ratio of effusion rates as approximately 1.0043 when considering five significant figures.

Key concepts

Effusion Rate

Effusion is the process by which gas molecules pass through a tiny opening from an area of higher pressure to lower pressure. This phenomenon is fascinating because it highlights the behavior of gases at the molecular level. Graham's Law of Effusion provides us with a mathematical way to understand and compare the effusion rates of different gases. The law is defined by the formula:\[\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}}\]Here, \( R_1 \) and \( R_2 \) represent the rates of effusion of two gases, and \( M_1 \) and \( M_2 \) are their respective molar masses. The key insight from this law is that the lighter a gas is, the faster it will effuse. This makes sense because lighter molecules move faster and are more likely to pass through the opening. Understanding this concept is crucial when working with gases that need to be separated or purified based on their molecular weights.

Molar Mass Calculation

Calculating the molar mass of a compound involves simply adding the atomic masses of all the atoms present in the molecule. This step is essential when using Graham's Law, as it aids in determining how quickly a gas will effuse compared to another. For example, in the exercise given:- For \( {}^{235}\mathrm{UF}_6 \), the calculation includes one uranium atom with an atomic mass of 235.04 amu and six fluorine atoms each with an atomic mass of 18.998 amu. Thus, the molar mass is: - \( M_1 = 235.04 + 6 \times 18.998 = 349.028 \, \text{amu} \)- For \( {}^{238}\mathrm{UF}_6 \), the uranium atom has a slightly different atomic mass at 238.05 amu, but the calculation process remains similar: - \( M_2 = 238.05 + 6 \times 18.998 = 352.038 \, \text{amu} \)These molar mass calculations are crucial for accurately applying Graham's Law and determining effusion rates.

Isotopes of Uranium

An isotope is a variant of an element that has the same number of protons but a different number of neutrons. This causes them to have different atomic masses. Uranium, a naturally occurring element, has several isotopes, with \( {}^{235}\text{U} \) and \( {}^{238}\text{U} \) being the most significant. These isotopes vary slightly in mass but can have dramatic appearances in physical and chemical properties due to their different nuclear compositions.- **\( {}^{235}\text{U} \)** - This isotope is rare but critical, especially in nuclear reactors and atomic bombs. - It is fissile, meaning it can sustain a nuclear chain reaction.- **\( {}^{238}\text{U} \)** - This isotope is more abundant but not directly used for nuclear chain reactions. - It plays a role as a breeder material to create plutonium-239 in reactors.Understanding the nuances of these isotopes, including their molar masses, is vital in nuclear chemistry and helps in tasks like the separation of isotopes using methods like effusion.