Problem 10
Question
Which pair of the gases diffuses with the same rate at same temperature and pressure? (a) \(\mathrm{CO}\) and \(\mathrm{NO}\) (b) \(\mathrm{NO}_{2}\) and \(\mathrm{CO}_{2}\) (c) \(\mathrm{NH}_{3}\) and \(\mathrm{PH}_{3}\) (d) \(\mathrm{NO}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\)
Step-by-Step Solution
Verified Answer
The gases NO and C₂H₆ diffuse at the same rate.
1Step 1: Understand the Concept of Diffusion
The diffusion rate of a gas is inversely proportional to the square root of its molar mass, according to Graham's Law of Diffusion. For two gases to diffuse at the same rate under identical conditions, they must have the same molar mass.
2Step 2: Calculate Molar Masses
Calculate the molar mass for each of the gases provided in the pairs:- For CO, the molar mass is \(12 + 16 = 28\) g/mol.- For NO, the molar mass is \(14 + 16 = 30\) g/mol.- For NO₂, the molar mass is \(14 + 16 \times 2 = 46\) g/mol.- For CO₂, the molar mass is \(12 + 16 \times 2 = 44\) g/mol.- For NH₃, the molar mass is \(14 + 1 \times 3 = 17\) g/mol.- For PH₃, the molar mass is \(31 + 1 \times 3 = 34\) g/mol.- For C₂H₆, the molar mass is \(12 \times 2 + 1 \times 6 = 30\) g/mol.
3Step 3: Compare Molar Masses
Compare the molar masses calculated in the previous step for each pair:
- CO (28 g/mol) and NO (30 g/mol) have different molar masses.
- NO₂ (46 g/mol) and CO₂ (44 g/mol) have different molar masses.
- NH₃ (17 g/mol) and PH₃ (34 g/mol) have different molar masses.
- NO (30 g/mol) and C₂H₆ (30 g/mol) have the same molar mass.
4Step 4: Conclusion
Since NO and C₂H₆ have the same molar mass of 30 g/mol, they will diffuse at the same rate under the same temperature and pressure conditions.
Key Concepts
Molar MassDiffusion RateGas DiffusionChemistry Education
Molar Mass
Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance, expressed in grams per mole (g/mol).
The molar mass is calculated by summing the atomic masses of all the elements in a compound, as found on the periodic table.
For example, carbon monoxide (CO) is composed of one carbon atom and one oxygen atom. The molar mass of CO is calculated as the sum of the atomic masses of carbon (12 g/mol) and oxygen (16 g/mol), giving a total of 28 g/mol.
The molar mass is calculated by summing the atomic masses of all the elements in a compound, as found on the periodic table.
For example, carbon monoxide (CO) is composed of one carbon atom and one oxygen atom. The molar mass of CO is calculated as the sum of the atomic masses of carbon (12 g/mol) and oxygen (16 g/mol), giving a total of 28 g/mol.
- Molar mass helps to determine how much of a substance is needed for chemical reactions.
- It plays a crucial role in calculating the diffusion rates of gases using Graham's Law of Diffusion.
Diffusion Rate
The diffusion rate of a gas measures how fast it spreads out in an environment.
According to Graham's Law of Diffusion, the diffusion rate of a gas is inversely proportional to the square root of its molar mass.
This means that lighter gases diffuse more rapidly than heavier ones. In mathematical terms, for two gases A and B with molar masses \(M_A\) and \(M_B\), the relationship is given by:\[\frac{\text{Rate of diffusion of A}}{\text{Rate of diffusion of B}} = \sqrt{\frac{M_B}{M_A}} \]
According to Graham's Law of Diffusion, the diffusion rate of a gas is inversely proportional to the square root of its molar mass.
This means that lighter gases diffuse more rapidly than heavier ones. In mathematical terms, for two gases A and B with molar masses \(M_A\) and \(M_B\), the relationship is given by:\[\frac{\text{Rate of diffusion of A}}{\text{Rate of diffusion of B}} = \sqrt{\frac{M_B}{M_A}} \]
- Gases with the same molar mass will have the same diffusion rate under identical conditions of temperature and pressure.
- Knowing the diffusion rates is crucial for processes such as respiration and industrial gas separation.
Gas Diffusion
Gas diffusion is the process by which gas molecules spread from an area of high concentration to an area of lower concentration.
This process continues until equilibrium is reached, meaning the concentration of gas is uniform throughout.
Gas diffusion is influenced by factors such as temperature, pressure, and the molar mass of the gas.
This process continues until equilibrium is reached, meaning the concentration of gas is uniform throughout.
Gas diffusion is influenced by factors such as temperature, pressure, and the molar mass of the gas.
- At higher temperatures, gas molecules move faster, increasing the rate of diffusion.
- Lower pressure means less atmospheric resistance, promoting more rapid diffusion.
- Lighter gases diffuse faster than heavier ones due to their lower molar mass.
Chemistry Education
Chemistry education aims to provide students with a deep understanding of chemical principles and their real-world applications.
Topics such as molar mass, diffusion rates, and gas diffusion are not only academic concepts but have practical implications in everyday life. Modern chemistry education leverages various teaching methodologies to make learning engaging:
Topics such as molar mass, diffusion rates, and gas diffusion are not only academic concepts but have practical implications in everyday life. Modern chemistry education leverages various teaching methodologies to make learning engaging:
- Interactive simulations can visually demonstrate concepts like gas diffusion, making it easier for students to grasp.
- Problem-solving exercises, like the one about comparing gas diffusion rates, allow students to apply their knowledge practically.
- Collaborative projects promote teamwork and help students understand the interdisciplinary nature of chemistry.
Other exercises in this chapter
Problem 8
Which of the following law leads us to arrive at the conclusion that 1 g-molecule of each gas at STP occupies a volume of \(22.4 \mathrm{~L}\) ? (a) Dalton's la
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\(0.1\) mole of gas absorbs \(41.75 \mathrm{~J}\) of heat, the rise in temperature oecurs \(20^{\circ} \mathrm{C}\), the gas must be (a) triatomic (b) diatomic
View solution Problem 11
Van der Waals real gas, act as an ideal gas, at which condition? (a) high temperature, low pressure (b) low temperature, high pressure (c) high temperature, hig
View solution Problem 12
The compressibility factor of an ideal gas is (a) 1 (b) 2 (c) 4 (d) 0
View solution