/*! 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 6 You have several identical ballo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 6

You have several identical balloons. You experimentally determine that a balloon will break if its volume exceeds 0.900 \(\mathrm{L}\). The pressure of the gas inside the balloon equals air pressure \((1.00 \mathrm{atm} \). (a) If the air inside the balloon is at a constant temperature of \(22.0^{\circ} \mathrm{C}\) and behaves as an ideal gas, what mass of air can you blow into one of the balloons before it bursts? (b) Repeat part (a) if the gas is helium rather than air.

Short Answer

Expert verified
1.07 g of air and 0.148 g of helium can be blown into the balloon before it bursts.

Step by step solution

01

Convert Temperature to Kelvin

To use the ideal gas law, you need the temperature in Kelvin. Convert the temperature from Celsius to Kelvin using the formula: \[ T(K) = T(掳C) + 273.15 \] Given: \[ T = 22.0 + 273.15 = 295.15 \text{ K} \]
02

Use the Ideal Gas Law

The ideal gas law is given by: \[ PV = nRT \] Where:- \( P = 1.00 \text{ atm} \) (pressure)- \( V = 0.900 \text{ L} \) (volume)- \( R = 0.0821 \frac{L \, atm}{mol \, K} \) (ideal gas constant)- \( T = 295.15 \text{ K} \) (temperature)You need to find \( n \), the number of moles:\[ n = \frac{PV}{RT} \] Plug in the values:\[ n = \frac{(1.00) \times (0.900)}{(0.0821) \times (295.15)} \approx 0.037 \text{ mol} \]
03

Calculate Mass of Air Using Molar Mass

The molar mass of air (primarily nitrogen and oxygen) is approximately \( 28.97 \frac{g}{mol} \). Calculate the mass, \( m \), using:\[ m = n \times \, \text{Molar Mass} \]\[ m = 0.037 \times 28.97 \approx 1.07 \text{ g} \]
04

Calculate Mass of Helium Using Molar Mass

The molar mass of helium is \( 4.00 \frac{g}{mol} \). Use the same number of moles as calculated before:\[ m = n \times \, \text{Molar Mass} \]\[ m = 0.037 \times 4.00 \approx 0.148 \text{ g} \]

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.

Molar Mass
Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance. It's expressed in grams per mole (g/mol). Each element has a unique molar mass based on its atomic weight, and this can be found on the periodic table. For compounds or mixtures, like air, we use average molar masses which consider the proportions of different gases in the mixture.
Air, for instance, is primarily made up of nitrogen and oxygen, giving it an average molar mass of approximately 28.97 g/mol. This value is crucial when calculating the mass of air using chemical formulas. **Why is molar mass important?** - It helps us convert between moles and grams, which is essential for stoichiometry calculations. - For this problem, knowing the molar mass allows us to determine how much of a substance can be filled in a given volume before a balloon pops. Understanding molar mass helps in determining the mass of the gases in any reaction or scenario involving the ideal gas law.
Balloon Volume
Balloon volume is a measure of how much space the gas inside a balloon occupies. It's typically measured in liters (L). The volume of a balloon is crucial in calculations when using the ideal gas law. The ideal gas law formula is:\[ PV = nRT \]Where:- \( P \) is the pressure of the gas in atmospheres (atm).- \( V \) is the volume in liters.- \( n \) is the number of moles of the gas.- \( R \) is the ideal gas constant (0.0821 L atm/mol K).- \( T \) is the temperature in Kelvin.In the exercise, the balloon's volume is given as 0.900 L, which is important because it tells us the maximum volume before it bursts.
Knowing this, we can calculate the moles (n) of gas it can contain without exceeding this volume, ensuring we don't overfill the balloon with the air or helium.
Temperature Conversion
Temperature conversion is often necessary when dealing with gas laws, as they require temperatures in Kelvin. To convert Celsius to Kelvin, we use the formula:\[ T(K) = T(掳C) + 273.15 \]In our scenario, the air inside the balloon is at 22.0掳C. Conversion gives us:\[ T = 22.0 + 273.15 = 295.15 \text{ K} \]**Why use Kelvin?** - Kelvin is an absolute temperature scale with zero point at absolute zero, making it ideal for scientific calculations. - The ideal gas law uses Kelvin because it ensures direct proportionality with volume and pressure, integral for accurate calculations.Proper temperature conversion ensures accurate calculations and a correct understanding of how gases will behave under different conditions.

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

(a) Oxygen \(\left(\mathrm{O}_{2}\right)\) has a molar mass of 32.0 \(\mathrm{g} / \mathrm{mol}\) . What is the average translational kinetic energy of an oxygen molecule at a temperature of 300 \(\mathrm{K} ?\) (b) What is the average value of the square at a of its speed? (c) What is the root-mean-square speed? (d) What is the momentum of an oxygen molecule traveling at this speed? (e) Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.10 \(\mathrm{m}\) on a side. What is the average force the molecule exerts on one of the walls of the container? (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.) (f) What is the aver- age force per unit area? (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 \(\mathrm{atm} ?\) (h) Compute the number of oxygen molecules that are actually contained in a vessel of this size at 300 \(\mathrm{K}\) and atmospheric pressure. (i) Your answer for part (h) should be three times as large as the answer for part (g). Where does this discrepancy arise?

We have two equal-size boxes, \(A\) and \(B\) . Each box contains gas that behaves as an ideal gas. We insert a thermometer into each box and find that the gas in box \(A\) is at a temperature of \(50^{\circ} \mathrm{C}\) while the gas in box \(B\) is at \(10^{\circ} \mathrm{C}\) . This is all we know about the gas in the boxes. Which of the following statements must be true? Which could be true? (a) The pressure in \(A\) is higher than in \(B\) . There are more molecules in \(A\) than in \(B .(\mathrm{c}) A\) and \(B\) cannot contain the same type of gas. (d) The molecules in \(A\) have more average kinetic energy per molecule than those in \(B\) . (e) The molecules in \(A\) are moving faster than those in \(B\) . Explain the reasoning behind your answers.

Puffy cumulus clouds, which are made of water droplets, occur at lower altitudes in the atmosphere. Wispy cirrus clouds, which are made of ice crystals, occur only at higher altitudes. Find the altitude \(y\) (measured from sea level) above which only cirrus clouds can occur. On a typical day and at altitudes less than 11 \(\mathrm{km}\), the temperature at an altitude \(y\) is given by \(T=T_{0}-\alpha y,\) where \(T_{0}=15.0^{\circ} \mathrm{C}\) and \(\alpha=6.0 \mathrm{C}^{\circ} / 1000 \mathrm{m} .\)

A balloon whose volume is 750 \(\mathrm{m}^{3}\) is to be filled with hydrogen at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right) .\) (a) If the hydrogen is stored in cylinders with volumes of 1.90 \(\mathrm{m}^{3}\) at a gauge pressure of \(1.20 \times 10^{6} \mathrm{Pa},\) how many cylinders are required? Assume that the temperature of the hydrogen remains constant. (b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if the gas in the balloon and the surrounding air are both at \(15.0^{\circ} \mathrm{C}\) ? The molar mass of hydro\(\operatorname{gen}\left(\mathrm{H}_{2}\right)\) is 2.02 \(\mathrm{g} / \mathrm{mol}\) . The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.23 \(\mathrm{kg} / \mathrm{m}^{3}\) . See Chapter 14 for a discussion of buoyancy. (c) What weight could be supported if the balloon were filled with helium (molar mass 4.00 \(\mathrm{g} / \mathrm{mol}\) ) instead of hydrogen, again at \(15.0^{\circ} \mathrm{C} ?\)

A \(20.0-\mathrm{L}\) tank contains 0.225 \(\mathrm{kg}\) of helium at \(18.0^{\circ} \mathrm{C}\) . The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{inol}\) (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Energy Physics

Read Explanation

Modern Physics

Read Explanation

Thermodynamics

Read Explanation

Force

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.