/*! 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 14 A mole of a chemical compound is... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 14

A mole of a chemical compound is a fixed number, \({ }^{16}\) like a dozen, of molecules (or atoms in the case of an element) of that compound. A mole of water, for example, is about 18 grams, or just over a half an ounce in your kitchen. Chemists often use the mole as the measure of the amount of a chemical compound. A mole of carbon dioxide has a fixed mass, but the volume \(V\) that it occupies depends on pressure \(p\) and temperature \(T\); greater pressure tends to compress the gas into a smaller volume, whereas increasing temperature tends to make the gas expand into a larger volume. If we measure the pressure in atmospheres ( 1 atm is the pressure exerted by the atmosphere at sea level), the temperature in kelvins, and the volume in liters, then the relationship is given by the ideal gas law: $$ p V=0.082 T \text {. } $$ a. Solve the ideal gas law for the volume \(V\). b. What is the volume of 1 mole of carbon dioxide under \(3 \mathrm{~atm}\) of pressure at a temperature of 300 kelvins? c. Solve the ideal gas law for pressure. d. What is the pressure on 1 mole of carbon dioxide if it occupies a volume of \(0.4\) liter at a temperature of 350 kelvins? e. Solve the ideal gas law for temperature. f. At what temperature will 1 mole of carbon dioxide occupy a volume of 2 liters under a pressure of \(0.3 \mathrm{~atm}\) ?

Short Answer

Expert verified
a) \( V = \frac{0.082T}{p} \) b) 8.2 liters c) \( p = \frac{0.082T}{V} \) d) 71.75 atm e) \( T = \frac{pV}{0.082} \) f) 7.32 K

Step by step solution

01

Solve for Volume V

The given ideal gas law is \( pV = 0.082T \). To solve for volume \( V \), we need to isolate \( V \) on one side of the equation. Divide both sides of the equation by \( p \) to get:\[ V = \frac{0.082T}{p} \]
02

Calculate Volume at Given Conditions

We are asked to find the volume \( V \) of 1 mole of carbon dioxide under \( 3 \text{ atm} \) of pressure at a temperature of \( 300 \text{ K} \). Using the formula derived:\[ V = \frac{0.082 \times 300}{3} = \frac{24.6}{3} = 8.2 \text{ liters} \]
03

Solve for Pressure p

To solve the ideal gas law for pressure \( p \), rearrange the equation to isolate \( p \):\[ p = \frac{0.082T}{V} \]
04

Calculate Pressure at Given Conditions

We need to find the pressure \( p \) when 1 mole of carbon dioxide occupies a volume of \( 0.4 \text{ liters} \) at a temperature of \( 350 \text{ K} \). Using the formula:\[ p = \frac{0.082 \times 350}{0.4} = \frac{28.7}{0.4} = 71.75 \text{ atm} \]
05

Solve for Temperature T

Rearrange the ideal gas law to solve for temperature \( T \):\[ T = \frac{pV}{0.082} \]
06

Calculate Temperature at Given Conditions

We are to find the temperature \( T \) at which 1 mole of carbon dioxide occupies a volume of 2 liters under a pressure of \( 0.3 \text{ atm} \). Using the formula:\[ T = \frac{0.3 \times 2}{0.082} = \frac{0.6}{0.082} \approx 7.32 \text{ K} \]

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.

Mole
The concept of the mole is fundamental in chemistry and describes a specific quantity of particles, such as atoms or molecules. One mole is equal to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles.
This large number helps chemists to count particles in a given mass of a substance easily because individual molecules or atoms are too small and numerous to count directly.
  • In practical terms, a mole allows for a bridge between the microscopic world of atoms and the macroscopic world that humans can measure and observe.
  • For instance, one mole of water weighs about 18 grams, equivalent to its molar mass.
Using moles, chemists can efficiently convert between mass and the number of particles, aiding in understanding and applying chemical reactions.
Pressure
Pressure is a measure of the force exerted by gas particles when they collide with the walls of their container. It's measured in units like atmospheres (atm), kilopascals (kPa), or millimeters of mercury (mmHg). In chemistry, pressure is an important variable when studying gases.
The Ideal Gas Law illustrates how pressure interacts with both temperature and volume. In the equation \( pV = nRT \), where \(p\) denotes pressure:
  • Pressure inversely relates to volume at a constant temperature, as Boyle's Law describes.
  • The pressure increases with the amount and temperature of the gas, describing how particles move more vigorously, leading to more frequent and forceful impacts against the container walls.
Understanding pressure is crucial for predicting how gases behave under different conditions.
Temperature
Temperature is a measure of the average kinetic energy of the particles in a substance. When discussing gases, temperature is typically measured in Celsius or Kelvins. Kelvins are used in the Ideal Gas Law to ensure calculations remain proportional to the energy of particles.
  • As the temperature of a gas increases, the kinetic energy of its particles increases, leading to louder collisions and increased pressure if volume remains constant, as stated in Gay-Lussac's Law.
  • Temperature changes can dramatically affect the behavior of gases. At higher temperatures, gases expand, while they contract at lower temperatures.
This expansion and contraction of gases as temperatures change makes understanding temperature fundamental when applying the Ideal Gas Law.
Volume
Volume is the amount of space a substance occupies and, in the context of gases, it can change with alterations in pressure and temperature. Measured in liters for gases in the Ideal Gas Law, volume reflects how much space the gas molecules are dispersed across.
  • According to Charles's Law, volume is directly proportional to temperature when pressure is constant, meaning a gas expands as it heats up.
  • In Boyle鈥檚 Law, volume is inversely proportional to pressure at a constant temperature, demonstrating how gases compress when pressure increases.
These relationships help predict how gases will behave in different scenarios, which is essential when working with the Ideal Gas Law\(\: pV = nRT \). Understanding these interactions allows chemists to manipulate gas conditions to desired states efficiently.

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

If you roll \(N\) dice, then the probability \(p=p(N)\) that you will get exactly 4 sixes is given by $$ p=\frac{N(N-1)(N-2)(N-3)}{24} \times\left(\frac{1}{6}\right)^{4}\left(\frac{5}{6}\right)^{N-4} $$ a. What is the probability, rounded to three decimal places, of getting exactly 4 sixes if 10 dice are rolled? How many times out of 1000 rolls would you expect this to happen? b. How many dice should be rolled so that the probability of getting exactly 4 sixes is the greatest?

Competition between populations: In this exercise we consider the question of competition between two populations that vie for resources but do not prey on each other. Let \(m\) be the size of the first population and \(n\) the size of the second (both measured in thousands of animals), and assume that the populations coexist eventually. Here is an example of one common model for the interaction: Per capita growth rate for \(m=5(1-m-n)\), Per capita growth rate for \(n=6(1-0.7 m-1.2 n)\). a. An isocline is formed by the points at which the per capita growth rate for \(m\) is zero. These are the solutions of the equation \(5(1-m-n)=0\). Find a formula for \(n\) in terms of \(m\) that describes this isocline. b. The points at which the per capita growth rate for \(n\) is zero form another isocline. Find a formula for \(n\) in terms of \(m\) that describes this isocline. c. At an equilibrium point the per capita growth rates for \(m\) and for \(n\) are both zero. If the populations reach such a point, they will remain there indefinitely. Use your answers to parts \(a\) and \(b\) to find the equilibrium point.

Motion toward or away from us distorts the pitch of sound, and it also distorts the wavelength of light. This phenomenon is known as the Doppler effect. In the case of light, the distortion is measurable only for objects moving at extremely high velocities. Motion of objects toward us produces a blue shift in the spectrum, whereas motion of objects away from us produces a red shift. Quantitatively, the red shift \(S\) is the change in wavelength divided by the unshifted wavelength, and thus red shift is a pure number that has no units associated with it. Cosmologists believe that the universe is expanding and that many stellar objects are moving away from Earth at radial velocities sufficient to produce a measurable red shift. Particularly notable among these are quasars, which have a number of important properties (some of which remain poorly understood). \({ }^{11}\) Quasars are moving rapidly away from us and thus produce a large red shift. The radial velocity \(V\) can be calculated from the red shift \(S\) using $$ V=c \times\left(\frac{(S+1)^{2}-1}{(S+1)^{2}+1}\right) $$ where \(c\) is the speed of light. a. Most known quasars have a red shift greater than 1. What would be the radial velocity of a quasar showing a red shift of 2 ? Report your answer as a multiple of the speed of light. b. Make a graph of the radial velocity (as a multiple of the speed of light) versus the red shift. Include values of the red shift from 0 to 5 . c. The quasar \(3 C 48\) shows a red shift of \(0.37\). How fast is \(3 C 48\) moving away from us? d. Find approximately the red shift that would indicate a radial velocity of half the speed of light. e. What is the maximum theoretical radial velocity that a quasar could achieve?

If you borrow \(\$ 120,000\) at an APR of \(6 \%\) in order to buy a home, and if the lending institution compounds interest continuously, then your monthly payment \(M=M(Y)\), in dollars, depends on the number of years \(Y\) you take to pay off the loan. The relationship is given by $$ M=\frac{120000\left(e^{0.005}-1\right)}{1-e^{-0.06 Y}}. $$ a. Make a graph of \(M\) versus \(Y\). In choosing a graphing window, you should note that a home mortgage rarely extends beyond 30 years. b. Express in functional notation your monthly payment if you pay off the loan in 20 years, and then use the graph to find that value. c. Use the graph to find your monthly payment if you pay off the loan in 30 years. d. From part b to part \(\mathrm{c}\) of this problem, you increased the debt period by \(50 \%\). Did this decrease your monthly payment by \(50 \%\) ? e. Is the graph concave up or concave down? Explain your answer in practical terms. f. Calculate the average decrease per year in your monthly payment from a loan period of 25 to a loan period of 30 years.

The per capita growth rate \(r\) (on an annual basis) of a population of grazing animals is a function of \(V\), the amount of vegetation available. A positive value of \(r\) means that the population is growing, whereas a negative value of \(r\) means that the population is declining. For the red kangaroo of Australia, the relationship has been given \({ }^{22}\) as $$ r=0.4-2 e^{-0.008 v} $$ Here \(V\) is the vegetation biomass, measured in pounds per acre. a. Draw a graph of \(r\) versus \(V\). Include vegetation biomass levels up to 1000 pounds per acre. b. The population size will be stable if the per capita growth rate is zero. At what vegetation level will the population size be stable?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

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.