/*! 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 29 A monatomic ideal gas that is in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 29

A monatomic ideal gas that is initially at a pressure of \(1.50 \times 10^{5}\) Pa and has a volume of 0.0800 \(\mathrm{m}^{3}\) is compressed adiabatically to a volume of 0.0400 \(\mathrm{m}^{3} .\) (a) What is the final pressure? (b) How much work is done by the gas? (c) What is the ratio of the final temperature of the gas to its initial temperature? Is the gas heated or cooled by this compression?

Short Answer

Expert verified
(a) 5.19 脳 10鈦� Pa, (b) -10.4 脳 10鲁 J, (c) \(T_2/T_1 = 1.59\); the gas is heated.

Step by step solution

01

Understanding the Adiabatic Process

In an adiabatic compression, no heat is exchanged with the environment. We apply the adiabatic condition for an ideal gas: \[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \]where \(P_1\) and \(P_2\) are initial and final pressures, \(V_1\) and \(V_2\) are initial and final volumes, and \(\gamma = \frac{C_p}{C_v}\) is the heat capacity ratio. For a monatomic gas, \(\gamma = \frac{5}{3}\).
02

Calculate the Final Pressure

We know \(P_1 = 1.50 \times 10^5\) Pa, \(V_1 = 0.0800\) m鲁, and \(V_2 = 0.0400\) m鲁. Using the adiabatic condition:\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{\gamma} \].Substitute the values and \(\gamma = \frac{5}{3}\):\[ P_2 = 1.50 \times 10^5 \times \left( \frac{0.0800}{0.0400} \right)^{\frac{5}{3}} \].Calculate to find \(P_2 = 5.19 \times 10^5\) Pa.
03

Calculate Work Done by the Gas

For an adiabatic process, work done by the gas can be calculated as:\[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \].Substitute \(P_1 = 1.50 \times 10^5\) Pa, \(V_1 = 0.0800\) m鲁, \(P_2 = 5.19 \times 10^5\) Pa, \(V_2 = 0.0400\) m鲁, and \(\gamma = \frac{5}{3}\):\[ W = \frac{1.50 \times 10^5 \times 0.0800 - 5.19 \times 10^5 \times 0.0400}{\frac{5}{3} - 1} \].Calculate to find \(W = -10.4 \times 10^3\) J (negative sign indicates work done on the gas).
04

Ratio of Final Temperature to Initial Temperature

For an adiabatic process, the temperature ratio is related by:\[ \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \].Substitute \(V_1 = 0.0800\) m鲁, \(V_2 = 0.0400\) m鲁, and \(\gamma = \frac{5}{3}\):\[ \frac{T_2}{T_1} = \left( \frac{0.0800}{0.0400} \right)^{\frac{5}{3} - 1} \].Calculate to find \(\frac{T_2}{T_1} = 1.59\).
05

Determine if the Gas is Heated or Cooled

Since the ratio \(\frac{T_2}{T_1} > 1\), it indicates that the final temperature is higher than the initial temperature. Thus, the gas is heated by the compression.

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
An ideal gas is a theoretical concept used in physics and chemistry to simplify the behavior of gases. It assumes a few key properties which make it easier to predict how gases respond under different conditions:
  • Gas particles are small and occupy no volume on their own.
  • There are no interactions between gas molecules, except for perfectly elastic collisions.
  • Gas behavior adheres to the ideal gas law: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the amount of substance (moles), \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
Understanding these assumptions helps predict real-world gases' behavior under certain conditions. This is crucial for solving problems involving temperature, volume, and pressure changes during processes like adiabatic compression.
Adiabatic Compression
Adiabatic compression refers to a process where a gas is compressed without exchanging heat with its surroundings. This means the total heat energy remains constant during compression:
  • No heat flow occurs into or out of the gas.
  • The gas's internal energy changes solely due to work done on it.
For an ideal gas undergoing adiabatic compression, the relation is given by:\[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \]where \( P_1 \) and \( P_2 \) are initial and final pressures, \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( \gamma \) is the heat capacity ratio.
This relation helps calculate the final pressure after compression. When a gas is compressed adiabatically, its temperature increases because the particles collide more frequently and energetically, which elevates their kinetic energy without losing energy to the surroundings.
Heat Capacity Ratio
The heat capacity ratio, also known as the adiabatic index or \( \gamma \), is crucial in determining how gases behave during processes like adiabatic compression. It is defined as the ratio of the molar heat capacities:
  • \( C_p \): Heat capacity at constant pressure.
  • \( C_v \): Heat capacity at constant volume.
The ratio is given by:\[ \gamma = \frac{C_p}{C_v} \].For a monatomic ideal gas, \( \gamma \) is often \( \frac{5}{3} \).This ratio indicates how much a gas will heat up when compressed without heat loss. Higher values indicate a more significant temperature change for a given volume change during an adiabatic process. This concept is essential in atmospheric physics, engineering, and understanding energy changes in gases under various 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

Bio Doughnuts: Breakfast of Champions! A typical doughnut contains 2.0 \(\mathrm{g}\) of protein, 17.0 \(\mathrm{g}\) of carbohydrates, and 7.0 \(\mathrm{g}\) of fat. The average food energy values of these substances are 4.0 \(\mathrm{kcal} / \mathrm{g}\) for protein and carbohydrates and 9.0 kcal/g for fat. (a) During heavy exercise, an average person uses energy at a rate of 510 kcal/h. How long would you have to exercise to "work off" one doughnut? (b) If the energy in the doughnut could somehow be converted into the kinetic energy of your body as a whole, how fast could you move after eating the doughnut? Take your mass to be \(60 \mathrm{kg},\) and express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h} .\)

A cylinder contains 0.250 mol of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\) . The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 Atm on the gas. The gas is heated until its temperature increases to \(127.0^{\circ} \mathrm{C}\) . Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) -diagram for this process. (b) How much work is done by the gas in this process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

A quantity of air is taken from state \(a\) to state \(b\) along a path that is a straight line in the \(p V\) -diagram (Fig. Pl9.39). (a) In this process, does the temperature of the gas increase, decrease, or stay the same? Explain. (b) If \(V_{a}=0.0700 \mathrm{m}^{3}\) , \(V_{b}=0.1100 \mathrm{m}^{3}, \quad p_{a}=1.00 \mathrm{x}\) \(10^{5} \mathrm{Pa},\) and \(p_{b}=1.40 \times 10^{5}\) \(\mathrm{Pa},\) what is the work \(W\) done by that gas in this process? Assume that the gas may be treated as ideal.

High-Altitude Research. A large research balloon containing \(2.00 \times 10^{3} \mathrm{m}^{3}\) of helium gas at 1.00 atm and a temperature of \(15.0^{\circ} \mathrm{C}\) rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm (Fig. Pl9.58. Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air. (a) Calculate the volume of the gas at the higher altitude. (b) Calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?

CALC The temperature of 0.150 mol of an ideal gas is held constant at \(77.0^{\circ} \mathrm{C}\) while its volume is reduced to 25.0\(\%\) of its initial volume. The initial pressure of the gas is 1.25 atm. (a) Determine the work done by the gas. (b) What is the change in its internal energy? (c) Does the gas exchange heat with its surroundings? If so, how much? Does the gas absorb or liberate heat?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Translational Dynamics

Read Explanation

Linear Momentum

Read Explanation

Energy Physics

Read Explanation

Medical Physics

Read Explanation

Quantum 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.