/*! 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 11 The gas inside a balloon will al... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 11

The gas inside a balloon will always have a pressure nearly equal to atmospheric pressure, since that is the pressure applied to the outside of the balloon. You fill a balloon with helium (a nearly ideal gas) to a volume of 0.600 L at a temperature of \(19.0^{\circ} \mathrm{C} .\) What is the volume of the balloon if you cool it to the boiling point of liquid nitrogen \((77.3 \mathrm{K}) ?\)

Short Answer

Expert verified
The volume of the balloon is approximately 0.159 liters when cooled to 77.3 K.

Step by step solution

01

Identify the Given Variables

We are given the initial volume of the balloon \( V_1 = 0.600 \) liters and initial temperature \( T_1 = 19.0^{\circ}C \). The final temperature \( T_2 \) is the boiling point of liquid nitrogen \( 77.3 \mathrm{K} \). First, convert \( T_1 \) from Celsius to Kelvin using the formula: \( T = \text{Celsius} + 273.15 \). Thus, \( T_1 = 19.0 + 273.15 = 292.15 \mathrm{K} \).
02

Apply Charles's Law

Charles's Law states that \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) for a constant pressure. Here, \( V_1 \) and \( T_1 \) are the initial conditions, and \( V_2 \) and \( T_2 \) are the final conditions.
03

Solve for the Final Volume \( V_2 \)

Re-arrange Charles's Law to solve for \( V_2 \): \[ V_2 = V_1 \times \frac{T_2}{T_1} \]Substitute the known values: \[ V_2 = 0.600 \times \frac{77.3}{292.15} \approx 0.159 \text{ liters} \]
04

Verify the Calculation

Double-check the calculations for any arithmetic errors. The calculation is correct, confirming that the final volume \( V_2 \) of the balloon is approximately \( 0.159 \) liters.

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.

Ideal Gas Behavior
Gases that behave ideally follow a specific set of rules that make predicting their behavior simpler. An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. It's an approximation used in many gas-related calculations, including Charles's Law.
  • An ideal gas follows the ideal gas law equation, which is a combination of various simpler gas laws including Boyle's and Charles's Laws.
  • Real gases can often be approximated as ideal gases when under conditions of high temperature and low pressure.
  • Helium, the gas used in balloons, approximates ideal gas behavior closely under many conditions, making it suitable for use in theoretical and practical exercises.
By assuming ideal gas behavior, calculations become more straightforward, as deviations due to molecular interactions are not considered. It's important for students to understand where approximations work and where they might lead to errors in real-world applications.
Temperature Conversion
Temperature conversion is essential when working with gas laws as they often require temperatures to be in Kelvin. The Kelvin scale is used because it begins at absolute zero, making it ideal for calculations in physics and chemistry. To convert from Celsius to Kelvin, you use this simple formula:
  • Add 273.15 to the Celsius temperature to obtain the temperature in Kelvin.
For the given problem:
  • The initial temperature was 19.0掳C. By converting to Kelvin, it becomes 292.15 K.
  • This ensures compatibility in calculations using gas laws, which require absolute temperature measurements.
Always remember that Kelvin is the appropriate unit for temperatures in many scientific formulas, ensuring calculations reflect real physical behavior.
Gas Laws
Gas laws are fundamental principles that describe how gases behave under various conditions of pressure, temperature, and volume.Charles鈥檚 Law is one of these laws, focusing specifically on the relationship between temperature and volume:
  • For a given mass of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature (measured in Kelvin).
  • This can be expressed mathematically as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).
  • This means if you increase the temperature, the volume will increase, provided the pressure remains constant.
Characteristic of gas laws, such as Charles's Law, is that they assume ideal gas behavior.This assumption works well under typical conditions but may vary under extreme pressures or temperatures.
Volume Change
Volume change in gases is directly linked to changes in temperature when the pressure is held constant, as explained by Charles鈥檚 Law.In our example, as the temperature of the helium-filled balloon is lowered to the boiling point of liquid nitrogen, a significant decrease in volume occurs.The formula from Charles's Law allows us to predict how much the volume will change:
  • Using the relationship \( V_2 = V_1 \times \frac{T_2}{T_1} \), you can calculate the final volume.
  • For the balloon, when cooled from 292.15 K to 77.3 K, the volume reduces from 0.600 L to approximately 0.159 L.
  • This illustrates the impact temperature changes have on the volume of a gas, crucial for applications and experiments involving gas expansion or compression.
Understanding the connection between temperature and volume aids in manipulating physical conditions in scientific experiments and practical 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

Smoke particles in the air typically have masses of the order of \(10^{-16} \mathrm{kg} .\) The Brownian motion (rapid, irregular movement) of these particles, resulting from collisions with air molecules, can be observed with a microscope. (a) Find the root-mean-square speed of Brownian motion for a particle with a mass of \(3.00 \times 10^{-16} \mathrm{kg}\) in air at 300 \(\mathrm{K}\) . (b) Would the root-mean-square speed be different if the particle were in hydrogen gas at the same temperature? Explain.

(a) A deuteron, \(_{1}^{2} \mathrm{H},\) is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a nuclear fusion reactor must be heated to about 300 million \(\mathrm{K} .\) What is the rms speed of the deuterons? Is this a significant fraction of the speed of light \(\left(c=3.0 \times 10^{8} \mathrm{m} / \mathrm{s}\right) ?\) (b) What would the temperature of the plasma be if the deuterons had an rms speed equal to 0.10\(c ?\)

Modern vacuum pumps make it easy to attain pressures of the order of \(10^{-13}\) atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of \(9.00 \times 10^{-14}\) atm and an ordinary temperature of \(300.0 \mathrm{K},\) how many molecules are present in a volume of 1.00 \(\mathrm{cm}^{3} ?\) (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

Helium gas with a volume of \(2.60 \mathrm{L},\) under a pressure of 0.180 atm and at a temperature of \(41.0^{\circ} \mathrm{C},\) is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol} .\)

(a) For what mass of molecule or particle is \(v_{\mathrm{rms}}\) equal to 1.00 \(\mathrm{mm} / \mathrm{s}\) at 300 \(\mathrm{K} ?\) (b) If the particle is an ice crystal, how many molecules does it contain? The molar mass of water is 18.0 \(\mathrm{g} / \mathrm{mol}\) . (c) Calculate the diameter of the particle if it is a spherical piece of ice. Would it be visible to the naked eye?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Famous Physicists

Read Explanation

Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Geometrical and Physical Optics

Read Explanation

Turning Points in Physics

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.