Chapter 3: Problem 125
Calculate relative rate of effusion of \(\mathrm{O}_{2}\) to \(\mathrm{CH}_{4}\) through a container containing \(\mathrm{O}_{2}\) and \(\mathrm{CH}_{4}\) in. \(3: 2\) mass ratio. (a) \(\frac{3 \sqrt{2}}{4}\) (b) \(\frac{3}{4 \sqrt{2}}\) (c) \(\frac{3}{2 \sqrt{2}}\) (d) none of these
Short Answer
Expert verified
The relative rate of effusion of \( \mathrm{O}_{2} \) to \( \mathrm{CH}_{4} \) is \( \frac{3}{4 \sqrt{2}} \) which corresponds to option (b).
Step by step solution
01
- Understand Graham's Law of Effusion
Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The rate of effusion can be compared between two gases with the equation \( R_1 / R_2 = \sqrt{M_2 / M_1} \) where \( R_1 \) and \( R_2 \) are the rates of effusion, and \( M_1 \) and \( M_2 \) are their molar masses, respectively.
02
- Calculate Molar Masses of \( \mathrm{O}_{2} \) and \( \mathrm{CH}_{4} \)
The molar mass of \( \mathrm{O}_{2} \) is 32 g/mol (16 g/mol for each oxygen atom) and that of \( \mathrm{CH}_{4} \) is 16 g/mol (12 g/mol for carbon and 1 g/mol for each of the four hydrogen atoms).
03
- Set up the Equation for Relative Rate of Effusion
Using Graham's Law of Effusion, set up the equation for the relative rate of effusion of \( \mathrm{O}_{2} \) to \( \mathrm{CH}_{4} \) as \( R_{O_2} / R_{CH_4} = \sqrt{M_{CH_4} / M_{O_2}} \). Substitute the molar masses into the equation to calculate the relative rate.
04
- Calculate the Relative Rate of Effusion
Substitute the molar masses into the equation from Step 3: \( R_{O_2} / R_{CH_4} = \sqrt{16 / 32} = \sqrt{1 / 2} = 1 / \sqrt{2} \).
05
- Incorporate the Mass Ratio into the Calculation
Given the mass ratio of \( \mathrm{O}_{2} \) to \( \mathrm{CH}_{4} \) is 3:2, we must multiply the relative rate of effusion by a factor that reflects this ratio. Since we are asked for the relative rate of effusion of \( \mathrm{O}_{2} \) to \( \mathrm{CH}_{4} \) and \( \mathrm{O}_{2} \) is heavier than \( \mathrm{CH}_{4} \), the mass ratio implies that there are more moles of \( \mathrm{CH}_{4} \) effusing for each mole of \( \mathrm{O}_{2} \). Therefore, the mass ratio needs to be converted to a mole ratio.
06
- Calculate the Mole Ratio Based on Mass Ratio
To find the mole ratio, divide the mass of each component by its molar mass. Since there is 3 grams of \( \mathrm{O}_{2} \) for every 2 grams of \( \mathrm{CH}_{4} \), divide 3 by the molar mass of \( \mathrm{O}_{2} \) (32 g/mol) and 2 by the molar mass of \( \mathrm{CH}_{4} \) (16 g/mol): \(\frac{3}{32}\) moles of \( \mathrm{O}_{2} \) and \(\frac{2}{16}\) moles of \( \mathrm{CH}_{4} \) which simplifies to \(\frac{3}{32} \) and \(\frac{1}{8}\) moles, respectively. This gives a ratio of \( (3/32) / (1/8) = 3 / 4 \) moles of \( \mathrm{O}_{2} \) for each mole of \( \mathrm{CH}_{4} \).
07
- Combine the Mole Ratio with the Relative Rate of Effusion
Putting it all together, the ratio of the rates of effusion of \( \mathrm{O}_{2} \) to \( \mathrm{CH}_{4} \) would be \( (1 / \sqrt{2}) \times (3 / 4) = 3 / (4 \sqrt{2}) \). This accounts for both the intrinsic rate of effusion due to molecular weight and the mass ratio of the gases in the container.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rate of Effusion
Understanding the rate of effusion is essential for grasping how gases escape through tiny openings. This concept is not just a theoretical study but has real-world implications in fields such as industrial gas production and environmental science. The rate of effusion is directly linked to the speed at which a gas particle can pass through an orifice. According to Graham's Law of Effusion, lighter gases effuse more quickly than heavier ones.
For students preparing for competitive exams like the IIT JEE in Physical Chemistry, mastering this concept is crucial. Knowing how to compute and compare the rates of different gases under variegated conditions could vastly improve problem-solving efficiency and accuracy, which is essential for such high-stakes exams.
For students preparing for competitive exams like the IIT JEE in Physical Chemistry, mastering this concept is crucial. Knowing how to compute and compare the rates of different gases under variegated conditions could vastly improve problem-solving efficiency and accuracy, which is essential for such high-stakes exams.
Molar Mass
One of the pivotal factors affecting the rate of effusion is the molar mass of a gas. It's the mass of one mole of a substance, typically expressed in grams per mole (g/mol). When comparing the effusion rates of two gases, we need to understand that a gas with a lower molar mass effuses faster than one with a higher molar mass.
Understanding the intricacies of molar mass is beneficial for students when attempting exercises from the IIT JEE Physical Chemistry syllabus, as it underpins multiple principles in gas laws and stoichiometry. Familiarity with the standard molar masses of common elements and compounds gives students the proficiency to tackle complex calculations with confidence.
Understanding the intricacies of molar mass is beneficial for students when attempting exercises from the IIT JEE Physical Chemistry syllabus, as it underpins multiple principles in gas laws and stoichiometry. Familiarity with the standard molar masses of common elements and compounds gives students the proficiency to tackle complex calculations with confidence.
Gas Effusion Calculations
To compute gas effusion rates accurately, students must follow a systematic approach. This involves knowing the molar masses of the involved gases, setting up the proportion as dictated by Graham's Law, and accounting for any given ratios or experimental data. The technique of these calculations is a frequent aspect of IIT JEE Physical Chemistry problems.
For improved comprehension, students should practice with varied exercises, including those with mass or volume ratios, mixed gas samples, and even real-life scenarios. Participating in such exercises promotes a deeper understanding and the ability to apply the theory to practical contexts.
For improved comprehension, students should practice with varied exercises, including those with mass or volume ratios, mixed gas samples, and even real-life scenarios. Participating in such exercises promotes a deeper understanding and the ability to apply the theory to practical contexts.
IIT JEE Physical Chemistry
The Indian Institutes of Technology Joint Entrance Examination (IIT JEE) is one of the most challenging engineering entrance exams, and Physical Chemistry forms a substantial part of its syllabus. Topics like Graham's Law of Effusion are integral in the IIT JEE, as they blend conceptual understanding with practical problem-solving skills.
In preparation for such topics, IIT JEE aspirants are encouraged to dissect problems into fundamental concepts, familiarize themselves with the mathematical relationships governing these principles, and apply them in diverse contexts. This rigorous preparation not only aids them in performing well in the exam but also lays a strong foundation for their future as engineers and scientists.
In preparation for such topics, IIT JEE aspirants are encouraged to dissect problems into fundamental concepts, familiarize themselves with the mathematical relationships governing these principles, and apply them in diverse contexts. This rigorous preparation not only aids them in performing well in the exam but also lays a strong foundation for their future as engineers and scientists.