/*! 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 27 (a) Amonton's law expresses the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 27

(a) Amonton's law expresses the relationship between pressure and temperature. Use Charles's law and Boyle's law to derive the proportionality relationship between \(P\) and \(T\). (b) If a car tire is filled to a pressure of \(32.0 \mathrm{lbs} / \mathrm{in}^{2}\) (psi) measured at \(75^{\circ} \mathrm{F}\), what will be the tire pressure if the tires heat up to \(120^{\circ} \mathrm{F}\) during driving?

Short Answer

Expert verified
Using Charles's Law and Boyle's Law, the derived proportionality relationship between pressure and temperature (Amonton's Law) is given by \(P_2T_1 = P_1T_2\). When initial tire pressure is \(32.0\, \mathrm{lbs} / \mathrm{in}^2\) at \(75^{\circ} \mathrm{F}\) and the temperature increases to \(120^{\circ} \mathrm{F}\), the tire pressure becomes approximately \(34.709\, \mathrm{lbs} / \mathrm{in}^2\).

Step by step solution

01

(a) Deriving Amonton's Law using Charles's Law and Boyle's Law

First, let's write down the expressions for Charles's Law and Boyle's Law: Charles's Law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) Boyle's Law: \(P_1V_1 = P_2V_2\) Our goal is to express \(P\) as a function of \(T\). Step 1: Replace \(V_2\) in Charles's Law with an expression from Boyle's Law From Boyle's Law, we know that: \(V_2 = \frac{P_1V_1}{P_2}\) Now we substitute \(V_2\) in Charles's Law: \(\frac{V_1}{T_1} = \frac{P_1V_1}{P_2T_2}\) Step 2: Simplify the equation Now let's simplify: \(P_2T_1 = P_1T_2\) This is the derived proportionality relationship between pressure and temperature, also known as Amontons's Law or Gay-Lussac's Law.
02

(b) Calculate the tire pressure at a higher temperature

Given: Initial Pressure, \(P_1 = 32.0 \, \mathrm{lbs}\,/\, \mathrm{in}^2\) Initial Temperature, \(T_1 = 75^{\circ}\mathrm{F}\) Final Temperature, \(T_2 = 120^{\circ}\mathrm{F}\) Step 1: Convert temperatures to Kelvin To work with the equation derived in part (a), we must convert the given temperatures to Kelvin. [掳F] to [K] conversion formula: \[K = \frac{9}{5}(掳 \mathrm{F} - 32) + 273.15\] Converting initial temperature: \(T_1 = \frac{9}{5}(75-32) + 273.15 = 297.039\, \mathrm{K}\) Converting final temperature: \(T_2 = \frac{9}{5}(120-32) + 273.15 = 322.039\, \mathrm{K}\) Step 2: Apply Amonton's Law From part (a), we derived that: \(P_2T_1 = P_1T_2\) Now, we can solve for the final tire pressure, \(P_2\): \(P_2 = \frac{P_1T_2}{T_1}\) Substitute the given values: \(P_2 = \frac{32.0 \,\mathrm{lbs/\,in}^2 \times 322.039\, \mathrm{K}}{297.039\, \mathrm{K}}\) Step 3: Calculate the final tire pressure \(P_2 = 34.709\, \mathrm{lbs/\,in}^2\) The tire pressure after heating up to \(120^{\circ} \mathrm{F}\) will be approximately \(34.709 \mathrm{\,lbs} /\mathrm{in}^2\).

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.

Charles's Law
Charles's Law is a fundamental principle in thermodynamics that describes how gases expand when heated. Specifically, it states that the volume of a gas is directly proportional to its temperature when the pressure is held constant. This can be mathematically expressed as:
  • \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)

Here, \( V_1 \) and \( V_2 \) are the initial and final volumes, while \( T_1 \) and \( T_2 \) are the initial and final temperatures, respectively. It implies that if the temperature of a gas increases, the volume will also increase provided that the pressure does not change.
The law is crucial when considering experiments involving ideal gases, where it helps predict how a change in temperature can affect volume. In deriving other gas laws, like Amonton's Law, it's often used in combination with other principles such as Boyle's Law.
Boyle's Law
Boyle's Law is another essential gas law that addresses the relationship between pressure and volume of a gas at a constant temperature. According to this law, the pressure of a gas is inversely proportional to its volume when the temperature remains unchanged. Mathematically, it is represented as:
  • \( P_1V_1 = P_2V_2 \)

In this equation, \( P_1 \) and \( P_2 \) represent the initial and final pressures, while \( V_1 \) and \( V_2 \) symbolize the initial and final volumes. This means that as the volume of a gas decreases, the pressure increases, provided the temperature remains constant.
Boyle's Law is particularly useful in calculating pressure changes in enclosed systems, such as a tire or a piston, where temperature can be controlled. Understanding this law helps to appreciate how variables in gas systems are interconnected, which is a key to deriving more complex relationships like those seen in determining Amonton's Law.
Pressure-Temperature Relationship
The pressure-temperature relationship, also referred to as Amonton's Law, provides insight into how gas pressure changes with temperature when volume is held constant. This relationship is derived by combining the concepts from Charles's and Boyle's Laws. The derived equation is:
  • \( P_2T_1 = P_1T_2 \)

Here, \( P_1 \) and \( P_2 \) are the initial and final pressures, with \( T_1 \) and \( T_2 \) being the initial and final temperatures. The key takeaway from this law is that if the temperature of a gas increases, so does the pressure, assuming the volume stays the same.
Amonton's Law is especially useful in real-world situations like predicting how the pressure in a car tire might change when it heats up from friction during driving. By using this relationship, one can understand and anticipate the behavior of contained gases under different thermal conditions, making it crucial for various scientific and engineering 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

Arsenic(III) sulfide sublimes readily, even below its melting point of \(320^{\circ} \mathrm{C}\). The molecules of the vapor phase are found to effuse through a tiny hole at \(0.28\) times the rate of effusion of Ar atoms under the same conditions of temperature and pressure. What is the molecular formula of arsenic(III) sulfide in the gas phase?

Suppose you are given two 1-L flasks and told that one contains a gas of molar mass 30 , the other a gas of molar mass 60 , both at the same temperature. The pressure in flask A is \(\mathrm{X}\) atm, and the mass of gas in the flask is \(1.2 \mathrm{~g}\). The pressure Which flask contains the gas of molar mass 30 , and which contains the gas of molar mass 60 ?

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? (a) Because Xe is a noble gas, there is less tendency for the Xe atoms to repel one another, so they pack more densely in the gaseous state. (b) Xe atoms have a higher mass than \(\mathrm{N}_{2}\) molecules. Because both gases at STP have the same number of molecules per unit volume, the Xe gas must be denser. (c) The Xe atoms are larger than \(\mathrm{N}_{2}\) molecules and thus take up a larger fraction of the space occupied by the gas. (d) Because the Xe atoms are much more massive than the \(\mathrm{N}_{2}\) molecules, they move more slowly and thus exert less upward force on the gas container and make the gas appear denser.

Indicate which of the following statements regarding the kinetic-molecular theory of gases are correct. (a) The average kinetic energy of a collection of gas molecules at a given temperature is proportional to \(\mathrm{m}^{1 / 2}\). (b) The gas molecules are assumed to exert no forces on each other. (c) All the molecules of a gas at a given temperature have the same kinetic energy. (d) The volume of the gas molecules is negligible in comparison to the total volume in which the gas is contained. (e) All gas molecules move with the same speed if they are at the same temperature.

Consider the combustion reaction between \(25.0 \mathrm{~mL}\) of liquid methanol (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ) and \(12.5 \mathrm{~L}\) of oxygen gas measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

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.