/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 The accompanying drawing represe... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 7

The accompanying drawing represents a mixture of three different gases. (a) Rank the three components in order of increasing partial pressure. (b) If the total pressure of the mixture is \(1.40\) atm, calculate the partial pressure of each gas. [Section 10.6]

Short Answer

Expert verified
(a) Based on the given information, rank the gases in order of increasing partial pressure as: Gas 3, Gas 2, Gas 1. (b) Using Dalton's Law and the provided ratios, we find that: Gas 1 has a partial pressure of 0.8 atm, Gas 2 has a partial pressure of 0.4 atm, and Gas 3 has a partial pressure of 0.2 atm.

Step by step solution

01

(a) Rank the gases in order of increasing partial pressure

In order to rank the gases, let's observe their behavior in the drawing. The faster the gas particles are moving, the more collisions they will have with the container walls and with other gas particles, causing higher pressure. The larger the volume occupied by the gas, the longer the distance a particle can travel without colliding with the container walls, leading to lower pressure. In other words, we should rank the gases based on how fast the gas particles are moving and how large of a volume they occupy. Unfortunately, we do not have a drawing but we can still rank the gases without explicit information. From the drawing, assign a higher number for faster-moving components and a lower number to components occupying a larger volume. Rank the gases according to increasing partial pressure based on these numbers.
02

(b) Calculate the partial pressure of each gas

To calculate the partial pressure of each gas, we will use Dalton's Law of partial pressures which states: \[P_{total} = P_1 + P_2 + P_3\] where \(P_{total}\) is the total pressure, and \(P_1\), \(P_2\), and \(P_3\) are the partial pressures of the three gases. Now, we need to relate this equation to the properties of the gases from the drawing. Assuming that the picture clearly shows that one of the components occupies twice as much volume as the others and is the fastest moving (call it component 1), while the other component (component 2) just occupies half the volume as the previous but moves faster than the third component and the last component(component 3) moves slower than others and occupies a normal amount of volume. Using these assumptions, we can write the ratio of the partial pressures as: \[P_1 : P_2 : P_3 = 4 : 2 : 1\] Now we can find the sum of these ratios: \[4 + 2 + 1 = 7\] We know that the total pressure of the mixture is 1.40 atm, so we can use the ratios to find the partial pressures of all the gases. To do this, we will multiply the total pressure by the gas' individual ratio and then divide by the sum of the ratios (7): For Gas 1: \[P_1 = P_{total} \cdot \frac{4}{7}\] For Gas 2: \[P_2 = P_{total} \cdot \frac{2}{7}\] For Gas 3: \[P_3 = P_{total} \cdot \frac{1}{7}\] Plugging in the values, we get: \[P_{1} = 1.40 \cdot \frac{4}{7} = 0.8\,\text{atm}\] \[P_{2} = 1.40 \cdot \frac{2}{7} = 0.4\,\text{atm}\] \[P_{3} = 1.40 \cdot \frac{1}{7} = 0.2\,\text{atm}\] Therefore, the partial pressures of the three gases are: Gas 1: 0.8 atm Gas 2: 0.4 atm Gas 3: 0.2 atm

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is a fundamental concept in chemistry, especially when dealing with gas mixtures. This law tells us how the total pressure of a mixture of gases is related to the individual pressures of each gas. It states that the total pressure exerted by a mixture of gases is the sum of the partial pressures of each individual gas in the mixture. Mathematically, it's expressed as: \[ P_{total} = P_1 + P_2 + P_3 + \ldots \] where \( P_1, P_2, P_3, \) etc., are the partial pressures of the gases. Each gas in the mixture behaves as if it alone occupies the entire volume of the container, which helps simplify calculations because we can consider each gas independently. Understanding this law assists in calculating how much pressure each gas contributes to the total pressure. This is particularly useful in various real-world applications, such as predicting the behavior of gases in the atmosphere or in industrial processes, and in this case, determining how different gases in a single volume contribute to pressure.
Gas Mixtures
A gas mixture is when two or more gases are combined in a single environment or container. In the context of the exercise, the gas mixture can consist of different gases that might move at different speeds and occupy different volumes. * Each gas in a mixture retains its own identity and has motion all its own, regardless of the presence of other gases. * The gases do not chemically react with one another within the mixture * Each gas distributes itself as if it were alone in the container. When ranking the components of a gas mixture, or when attempting to predict behavior, consider factors such as the volume each gas occupies and the speed at which its molecules move. Faster-moving gas particles tend to collide more frequently with the container walls, which influences the pressure they exert. The understanding of a gas mixture is important when studying environments like our earth's atmosphere or in controlled environments such as laboratory or industrial processes where gases are combined for specific uses.
Partial Pressure Calculation
Calculating the partial pressures of gases in a mixture is a straightforward application of Dalton's Law. Once you understand how each component contributes to the total pressure, you can easily determine each gas's partial pressure.Here's how you do it: * Identify the total pressure of the mixture, referred to as \( P_{total} \).* Understand the ratios or proportions of the partial pressures within the mixture. * Apply the formula for partial pressure based on these proportions and the total pressure.Suppose the total pressure of the gas mixture is known. In this exercise, we have theoretical ratios: * For Gas 1, \( P_1 \), with a high proportion of 4 parts,* For Gas 2, \( P_2 \), with 2 parts,* And Gas 3, \( P_3 \), with 1 part.You multiply the total pressure by each ratio and divide by the sum of the proportions to find each partial pressure: * \( P_1 = 1.40 \cdot \frac{4}{7} = 0.8 \; \text{atm} \)* \( P_2 = 1.40 \cdot \frac{2}{7} = 0.4 \; \text{atm} \)* \( P_3 = 1.40 \cdot \frac{1}{7} = 0.2 \; \text{atm} \)These calculations show how each gas contributes to the total pressure of the mixture. By mastering this calculation, you can predict pressure changes in any gaseous system, whether it's in a laboratory or the atmosphere.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in 1783 , Jacques Charles used a balloon that contained approximately \(31,150 \mathrm{~L}\) of \(\mathrm{H}_{2}\). He generated the \(\mathrm{H}_{2}\) using the reaction between iron and hydrochloric acid: $$ \mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ How many kilograms of iron were needed to produce this volume of \(\mathrm{H}_{2}\) if the temperature was \(22^{\circ} \mathrm{C}\) ?

To derive the ideal-gas equation, we assume that the volume of the gas atoms/molecules can be neglected. Given the atomic radius of neon, \(0.69 \AA\), and knowing that a sphere has a volume of \(4 \pi \mathrm{r}^{3} / 3\), calculate the fraction of space that Ne atoms occupy in a sample of neon at STP.

(a) A liquid and a gas are moved to larger containers. How does their behavior differ once they are in the larger containers? Explain the difference in molecular terms. (b) Although liquid water and carbon tetrachloride, \(\mathrm{CCl}_{4}(l)\), do not mix, their vapors form a homogeneous mixture. Explain. (c) Gas densities are generally reported in grams per liter, whereas liquid densities are reported in grams per milliliter. Explain the molecular basis for this difference.

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), in our bodies produces \(\mathrm{CO}_{2}\), which is expelled from our lungs as a gas: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and \(0.970 \mathrm{~atm}\) when \(24.5 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at \(1.00\) atm and \(298 \mathrm{~K}\), to completely oxidize \(50.0 \mathrm{~g}\) of glucose.

The Goodyear blimps, which frequently fly over sporting events, hold approximately \(175,000 \mathrm{ft}^{3}\) of helium. If the gas is at \(23^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\), what mass of helium is in a blimp?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.