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Chapter 10: Problem 80

Consider the following gases: \(\mathrm{He}, \mathrm{SO}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{Cl}_{2}\) (a) Which has the largest density (assuming that all gases are at the same T and P)? (b) Which gas will effuse fastest through a porous plate?

Short Answer

Expert verified
(a) \(\mathrm{Cl}_{2}\) has the largest density. (b) \(\mathrm{He}\) will effuse fastest.

Step by step solution

01

Understand Density of Gases

The density of a gas is dependent on its molar mass. Assuming all gases are under the same temperature and pressure, the gas with the highest molar mass will have the highest density.
02

Calculate Molar Mass of Each Gas

Calculate the molar mass of each gas:- \(\mathrm{He}\) has a molar mass of 4 g/mol.- \(\mathrm{SO}_{2}\) has a molar mass of 64 g/mol (\(\mathrm{S} = 32\ g/mol + 2 \times \mathrm{O} = 16\ g/mol\)).- \(\mathrm{CO}_{2}\) has a molar mass of 44 g/mol (\(\mathrm{C} = 12\ g/mol + 2 \times \mathrm{O} = 16\ g/mol\)).- \(\mathrm{Cl}_{2}\) has a molar mass of 71 g/mol (\(2 \times \mathrm{Cl} = 35.5\ g/mol\)).
03

Identify Gas with Largest Density

Compare the molar masses calculated: - \(\mathrm{Cl}_{2}\) has the highest molar mass of 71 g/mol, so it will have the largest density.
04

Understand Effusion of Gases

According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Therefore, the gas with the smallest molar mass will effuse fastest.
05

Identify Gas with Fastest Effusion

Compare the molar masses again and identify the gas with the smallest molar mass:- \(\mathrm{He}\) has the smallest molar mass of 4 g/mol, so it will effuse the fastest.

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Key Concepts

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

density of gases
Understanding the density of gases is crucial when comparing different gases under the same conditions of temperature and pressure. Density, which is mass per unit volume, is heavily influenced by a gas's molar mass. Since the conditions are constant for temperature and pressure, the gas with the highest molar mass will possess the greatest density.

For example, let's examine the gases: He, SO\( _2 \), CO\( _2 \), and Cl\(_2 \). Calculating their molar masses tells us:
  • Helium (He) has a molar mass of 4 g/mol.

  • Sulfur dioxide (SO\( _2 \)) has a molar mass of 64 g/mol.

  • Carbon dioxide (CO\(_2\)) has a molar mass of 44 g/mol.

  • Chlorine (Cl\(_2 \)) has a molar mass of 71 g/mol.

Since Cl\(_2 \) has the highest molar mass, it also has the largest density among these gases.
Graham's law of effusion
Graham's law of effusion provides insight into how quickly different gases pass through a small opening. The rate of effusion is inversely proportional to the square root of the molar mass of the gas. This means gases with smaller molar masses will effuse faster than those with larger molar masses.

For the gases in question, calculating their molar masses, we find that Helium (He) has the smallest molar mass, 4 g/mol. Hence, He will effuse the fastest through a porous plate compared to the other gases, such as SO\(_2\), CO\(_2\), and Cl\(_2\).
  • Small molar mass = faster effusion rate
  • Large molar mass = slower effusion rate

Utilizing this law helps in predicting behaviors of gases in various applications, from industrial processes to scientific experiments.
molar mass calculations
Molar mass calculations form the backbone of understanding many properties of gases, such as density and effusion rates. To compute the molar mass, you sum the atomic masses of the elements in a molecule.

For instance:
  • Helium (He) is a single element with a molar mass of 4 g/mol.

  • Sulfur dioxide (SO\(_2\)) includes sulfur and oxygen: 32 g/mol (Sulfur) + 2 x 16 g/mol (Oxygen) = 64 g/mol.

  • Carbon dioxide (CO\(_2\)) combines carbon and oxygen: 12 g/mol (Carbon) + 2 x 16 g/mol (Oxygen) = 44 g/mol.

  • Chlorine (Cl\(_2\)) requires multiplying the atomic mass of chlorine by two: 2 x 35.5 g/mol = 71 g/mol.

Knowing the molar mass aids in predicting and comparing the gas properties, making it an indispensable tool for chemists and students alike.

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Most popular questions from this chapter

\(.\) A sample of \(\mathrm{CO}_{2}\) gas has a pressure of \(56.5 \mathrm{mm}\) Hg in a 125 -mL flask. The sample is transferred to a new flask, where it has a pressure of \(62.3 \mathrm{mm}\) Hg at the same temperature. What is the volume of the new flask?

You have a \(550 .\) -mL tank of gas with a pressure of 1.56 atm at \(24^{\circ} \mathrm{C}\). You thought the gas was pure carbon monoxide gas, \(\mathrm{CO},\) but you later found it was contaminated by small quantities of gaseous \(\mathrm{CO}_{2}\) and \(\mathrm{O}_{2}\). Analysis shows that the tank pressure is 1.34 atm \(\left(\text { at } 24^{\circ} \mathrm{C}\right)\) if the \(\mathrm{CO}_{2}\) is removed. Another experiment shows that \(0.0870 \mathrm{g}\) of \(\mathrm{O}_{2}\) can be removed chemically. What are the masses of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) in the tank, and what is the partial pressure of each of the three gases at \(25^{\circ} \mathrm{C} ?\)

If you place 2.25 g of solid silicon in a \(6.56-\mathrm{L}\) flask that contains \(\mathrm{CH}_{3} \mathrm{Cl}\) with a pressure of \(585 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C},\) what mass of dimethyldichlorosilane, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2},\) can be formed? $$ \mathrm{Si}(\mathrm{s})+2 \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g}) \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \operatorname{SiCl}_{2}(\mathrm{g}) $$ What pressure of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(\mathrm{g})\) would you expect in this same flask at \(95^{\circ} \mathrm{C}\) on completion of the reaction? (Dimethyldichlorosilane is one starting material used to make silicones, polymeric substances used as lubricants, antistick agents, and water-proofing caulk.)

Silane, \(\operatorname{SiH}_{4}\), reacts with \(\mathrm{O}_{2}\) to give silicon dioxide and water: $$ \mathrm{SiH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SiO}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ A 5.20 -L sample of \(\mathrm{SiH}_{4}\) gas at \(356 \mathrm{mm}\) Hg pressure and \(25^{\circ} \mathrm{C}\) is allowed to react with \(\mathrm{O}_{2}\) gas. What volume of \(\mathrm{O}_{2}\) gas, in liters, is required for complete reaction if the oxygen has a pressure of \(425 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C} ?\)

One of the cylinders of an automobile engine has a volume of \(400 . \mathrm{cm}^{3} .\) The engine takes in air at a pressure of 1.00 atm and a temperature of \(15^{\circ} \mathrm{C}\) and compresses the air to a volume of \(50.0 \mathrm{cm}^{3}\) at \(77^{\circ} \mathrm{C} .\) What is the final pressure of the gas in the cylinder? (The ratio of before and after volumesin this case, 400: 50 or \(8: 1-\) is called the compression ratio.
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