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Chapter 11: Problem 48

Argon gas is 10 times denser than helium gas at the same temperature and pressure. Which gas is predicted to effuse faster? How much faster?

Short Answer

Expert verified
Helium effuses faster; it is 3.16 times faster than argon.

Step by step solution

01

Identify the Relationship

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. The formula can be expressed as: \( \frac{\text{Rate of effusion of Gas 1}}{\text{Rate of effusion of Gas 2}} = \sqrt{\frac{M_2}{M_1}} \), where \( M_1 \) and \( M_2 \) are the molar masses of the gases.
02

Density and Molar Mass Relation

Given that argon gas is 10 times denser than helium gas under the same conditions, we can relate density \( \rho \) and molar mass \( M \). Since \( \rho \propto M \) at constant temperature and pressure, if \( \rho_{\text{Ar}} = 10\rho_{\text{He}} \), then \( M_{\text{Ar}} = 10M_{\text{He}} \). We know \( M_{\text{He}} = 4 \text{ g/mol} \), therefore \( M_{\text{Ar}} = 40 \text{ g/mol} \), close approximation to known data.
03

Applying Graham's Law

Plug the molar masses into Graham's Law. Let Gas 1 be helium (\( M_1 = 4 \)), and Gas 2 be argon (\( M_2 = 40 \)). We get: \( \frac{\text{Rate of effusion of He}}{\text{Rate of effusion of Ar}} = \sqrt{\frac{40}{4}} = \sqrt{10} \).
04

Calculate the Rate

Calculate \( \sqrt{10} \), which is approximately 3.16. This value indicates that helium effuses 3.16 times faster than argon.

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Key Concepts

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

Rate of Effusion
The rate of effusion refers to how fast a gas can pass through a small opening from one container to another. Understanding this concept is excellent for students studying gas behavior in chemistry. One essential idea here involves Graham's Law of Effusion. According to this law, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. This means that lighter gases will effuse faster than heavier ones. For instance, if you have two gases, the one with the lower molar mass will pass through the opening more quickly. This relationship is expressed mathematically as:
  • \(\frac{\text{Rate of effusion of Gas 1}}{\text{Rate of effusion of Gas 2}} = \sqrt{\frac{M_2}{M_1}}\)
Molar Mass
Molar mass is a measure of the mass of a given amount (usually a mole) of a substance. It is typically expressed in grams per mole (g/mol). Understanding molar mass is fundamental when applying Graham's Law of Effusion. If you can identify the molar masses of different gases, you can predict how quickly they will effuse under the same conditions. For example:
  • Helium (\(M_{\text{He}}\)) has a molar mass of 4 g/mol.
  • Argon (\(M_{\text{Ar}}\)) has a molar mass of 40 g/mol.
In the framework of this exercise, it becomes easier to compare and predict the effusion rates of these gases by substituting their molar masses into Graham's Law equation.
Density and Molar Mass Relation
In this context, density and molar mass have an intrinsic relationship at constant temperature and pressure. When you know that one gas is denser than another, you can infer details about their respective molar masses. The relationship is given by:
  • \(\rho \propto M\)
Here, \(\rho\) is the density, and \(M\) is the molar mass. Because argon is said to be ten times denser than helium under the same conditions, we can immediately relate their molar masses:
  • \(\rho_{\text{Ar}} = 10 \rho_{\text{He}}\)
  • \(M_{\text{Ar}} = 10 \times M_{\text{He}}\)
Through this proportionality, understanding gas properties in different chemical and physical reactions becomes more accessible.
Argon and Helium Effusion Comparison
When comparing the effusion rates of argon and helium, Graham's Law comes into play. Helium, being lighter, effuses faster than argon. By calculating how much faster, we plug in their respective molar masses into the formula and find that:
  • \(\frac{\text{Rate of effusion of He}}{\text{Rate of effusion of Ar}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16\)
This calculation reveals that helium effuses approximately 3.16 times faster than argon. This type of comparison is crucial for fields that require precise handling of gases, such as chemical engineering and respiratory therapy, helping ensure the correct application of gases in both industrial and clinical settings.

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Most popular questions from this chapter

Analysis of a gaseous chlorofluorocarbon, \(\mathrm{CCl}_{x} \mathrm{F}_{y}\), shows that it contains \(11.79 \%\) C and \(69.57 \%\) Cl. In another experiment, you find that \(0.107 \mathrm{g}\) of the compound fills a 458 -mL. flask at \(25^{\circ} \mathrm{C}\) with a pressure of \(21.3 \mathrm{mm}\) Hg. What is the molecular formula of the compound?

Ethane burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) $$2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ (a) Four gases are involved in this reaction. Place them in order of increasing rms speed. (Assume all are at the same temperature.) (b) A \(3.26-\) - flask contains \(C_{2} H_{6}\) at a pressure of \(256 \mathrm{mm} \mathrm{Hg}\) and a temperature of \(25^{\circ} \mathrm{C} .\) Suppose \(\mathrm{O}_{2}\) gas is added to the flask until \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{O}_{2}\) are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of \(\mathrm{O}_{2}\) and what is the total pressure in the flask?

Equal masses of gaseous \(\mathrm{N}_{2}\) and \(\mathrm{Ar}\) are placed in separate flasks of equal volume at the same temperature. Tell whether each of the following statements is true or false. Briefly explain your answer in each case. (a) There are more molecules of \(\mathrm{N}_{2}\) present than atoms of Ar. (b) The pressure is greater in the Ar flask. (c) The Ar atoms have a greater rms speed than the \(\mathrm{N}_{2}\) molecules. (d) The \(\mathrm{N}_{2}\) molecules collide more frequently with the walls of the flask than do the Ar atoms.

A collapsed balloon is filled with He to a volume of 12.5 L. at a pressure of 1.00 atm. Oxygen, \(\mathrm{O}_{2}\), is then added so that the final volume of the balloon is \(26 \mathrm{L}\) with a total pressure of 1.00 atm. The temperature, which remains constant throughout, is \(21.5^{\circ} \mathrm{C}\) (a) What mass of He does the balloon contain? (b) What is the final partial pressure of He in the balloon? (c) What is the partial pressure of \(\mathrm{O}_{2}\) in the balloon? (d) What is the mole fraction of each gas?

The sodium azide required for automobile air bags is made by the reaction of sodium metal with dinitrogen monoxide in liquid ammonia: $$\begin{aligned} 3 \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+4 \mathrm{Na}(\mathrm{s})+\mathrm{NH}_{3}(\ell) & \rightarrow \\ & \mathrm{NaN}_{3}(\mathrm{s})+3 \mathrm{NaOH}(\mathrm{s})+2 \mathrm{N}_{2}(\mathrm{g}) \end{aligned}$$ (a) You have \(65.0 \mathrm{g}\) of sodium, a \(35.0-\mathrm{L}\). flask containing \(\mathrm{N}_{2} \mathrm{O}\) gas with a pressure of 2.12 atm at \(23^{\circ} \mathrm{C}\) and excess ammonia. What is the theoretical yield (in grams) of NaNg? (b) Draw a Lewis structure for the azide ion. Include all possible resonance structures. Which resonance structure is most likely? (c) What is the shape of the azide ion?
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