/*! 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 40 You fill a balloon with helium g... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 40

You fill a balloon with helium gas to a volume of \(2.68 \mathrm{~L}\) at \(23^{\circ} \mathrm{C}\) and \(789 \mathrm{mmHg}\). Now you release the balloon. What would be the volume of helium if its pressure changed to \(499 \mathrm{mmHg}\) but the temperature were unchanged?

Short Answer

Expert verified
The volume of helium is approximately 4.23 L.

Step by step solution

01

Understanding the Problem

We need to find the new volume of the helium balloon when the pressure changes from 789 mmHg to 499 mmHg. The temperature remains constant and we start with an initial volume of 2.68 L. This is a problem related to Boyle's Law, which states that the product of pressure and volume remains constant if the temperature is unchanged.
02

Applying Boyle's Law

Boyle's Law is expressed as \( P_1 V_1 = P_2 V_2 \), where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the new pressure and volume. Substituting the given values: \( 789 \, \text{mmHg} \times 2.68 \, \text{L} = 499 \, \text{mmHg} \times V_2 \).
03

Solving for the New Volume

To find \( V_2 \), we rearrange the equation: \( V_2 = \frac{789 \, \text{mmHg} \times 2.68 \, \text{L}}{499 \, \text{mmHg}} \). Calculate this to find the new volume of the helium.
04

Calculation

Perform the calculation: \( V_2 = \frac{789 imes 2.68}{499} \approx 4.23 \, \text{L} \). This value represents the new volume of the helium when the pressure changes to 499 mmHg, with the temperature constant.

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.

Gas Laws
Understanding gas laws is crucial in chemistry because they describe how gases behave under various conditions. These laws include Boyle's Law, Charles's Law, and Avogadro's Law, among others.
Boyle's Law, specifically, relates pressure and volume. Charles's Law connects volume with temperature, while Avogadro's Law shows the relationship between volume and the number of gas particles.
  • Boyle's Law: At constant temperature, the pressure of a gas is inversely proportional to its volume. This means if you increase the pressure, the volume decreases and vice versa.
  • Charles's Law: At constant pressure, the volume of a gas is directly proportional to its absolute temperature.
  • Avogadro's Law: At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas.
These laws together form the basis for comprehending more complex gas behaviors and are indispensable in the study of chemistry.
Pressure-Volume Relationship
Boyle's Law perfectly illustrates the pressure-volume relationship in gases. It is often used to predict the behavior of a gas when it undergoes changes in pressure or volume while keeping temperature constant.
According to Boyle's Law, expressed mathematically as \( P_1 V_1 = P_2 V_2 \), this principle highlights that the product of the initial pressure and volume of a gas equals the product of its new pressure and new volume for the same gas amount.
  • Inverse Relationship: The relationship is inverse, meaning as one variable increases, the other must decrease to maintain equality in the equation.
  • Practical Example: In the given exercise, the pressure of helium gas decreased from 789 mmHg to 499 mmHg. Hence, according to this relationship, the volume increased from 2.68 L to about 4.23 L.
  • Constant Temperature: This relationship holds true only when the temperature is maintained constant, as temperature can affect both pressure and volume.
The pressure-volume relationship is fundamental in calculations involving gases, useful in everyday applications, like inflating balloons and understanding how syringes work.
Helium Gas
Helium is a fascinating element, known for its lightness and as a noble gas, meaning it is quite unreactive. It's the second lightest element after hydrogen and is frequently used in applications where lighter-than-air gases are needed.
Helium is often used to fill balloons because it is non-flammable, unlike hydrogen, which can be dangerous in some circumstances. Additionally, helium's low density assists balloons in rising in the air without risk.
  • Noble Gas: Helium is non-reactive and belongs to group 18 on the periodic table, known as the noble gases.
  • Safe and Abundant: It is safe for a variety of uses and is in plentiful supply, obtained from natural gas extraction.
  • Unique Properties: Helium's properties, such as its low boiling point, make it valuable in cryogenics and as a coolant for superconducting magnets.
Understanding the role of helium in gas laws, especially in Boyle's Law scenarios, helps clarify its behavior under different pressures, making it a significant subject in scientific studies and real-world applications.

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

Consider the following setup, which shows identical containers connected by a tube with a valve that is presently closed. The container on the left has \(1.0 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) gas; the container on the right has \(1.0 \mathrm{~mol}\) of \(\mathrm{O}_{2}\). Which container has the greatest density of gas? Whieh container-has molecules that are moving at a faster average molecular speed? Which container has more molecules? If the valve is opened, will the pressure in each of the containers change? If it does, how will it change (increase, decrease, or no change)? \(2.0 \mathrm{~mol}\) of Ar is added to the system with the valve open. What fraction of the total pressure will be due to the \(\mathrm{H}_{2} ?\)

A mixture of Ne and Ar gases at \(350 \mathrm{~K}\) contains twice as many moles of Ne as of Ar and has a total mass of \(50.0 \mathrm{~g}\). If the density of the mixture is \(4.00 \mathrm{~g} / \mathrm{L},\) what is the partial pressure of \(\mathrm{Ne}\) ?

In a series of experiments, the U.S. Navy developed an undersea habitat. In one experiment, the mole percent composition of the atmosphere in the undersea habitat was \(79.0 \%\) He, \(17.0 \% \mathrm{~N}_{2}\), and \(4.0 \% \mathrm{O}_{2}\). What will the partial pressure of each gas be when the habitat is \(58.8 \mathrm{~m}\) below sea level, where the pressure is 6.91 atm?

A hydrocarbon gas has a density of \(1.22 \mathrm{~g} / \mathrm{L}\) at \(20^{\circ} \mathrm{C}\) and 1.00 atm. An analysis gives \(80.0 \% \mathrm{C}\) and \(20.0 \% \mathrm{H}\). What is the molecular formula?

Gas Laws and Kinetic Theory of Gases I Shown here are two identical containers labeled \(\mathrm{A}\) and \(\mathrm{B}\). Container A contains a molecule of an ideal gas, and container B contains two molecules of an ideal gas. Both containers are at the same temperature. (Note that small numbers of molecules and atoms are being represented in these examples in order that you can easily compare the amounts. Real containers with so few molecules and atoms would be unlikely.) How do the pressures in the two containers compare? Be sure to explain your answer. Shown below are four different containers \((\mathrm{C}, \mathrm{D}, \mathrm{E}\) and \(\mathrm{F}\) ), each with the same volume and at the same temperature. How do the pressures of the gases in the containers compare? Container \(\mathrm{H}\) below has twice the volume of container G. How will the pressure in the containers compare? Explain your reasoning. How will the pressure of containers \(\mathrm{G}\) and \(\mathrm{H}\) compare if you add two more gas molecules to container \(\mathrm{H}\) ? Consider containers I and J below. Container J has twice the volume of container \(\mathrm{I}\). Container \(\mathrm{I}\) is at a temperature of \(100 \mathrm{~K},\) and container \(\mathrm{J}\) is at \(200 \mathrm{~K}\). How does the pressure in container I compare with that in container \(\mathrm{J} ?\) Include an explanation as part of your answer.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

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.