/*! 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 23 The temperature of a monatomic i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The temperature of a monatomic ideal gas remains constant during a process in which 4700 \(\mathrm{J}\) of heat flows out of the gas. How much work (including the proper \(+\) or \(-\) sign ) is done?

Short Answer

Expert verified
The work done is +4700 J.

Step by step solution

01

Understand the problem

We need to find the work done on or by a monatomic ideal gas. The problem states that the temperature remains constant and 4700 J of heat flows out of the gas. This is an isothermal process because the temperature does not change.
02

Use the First Law of Thermodynamics

For thermodynamic processes, the First Law of Thermodynamics is key, which states \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.
03

Analyze changes in internal energy

Since the temperature remains constant, the internal energy \( \Delta U \) also remains constant. Therefore, \( \Delta U = 0 \).
04

Solve for work done

Since \( \Delta U = 0 \), the equation \( Q = W \) holds for an isothermal process. We are given that 4700 J of heat flows out, meaning \( Q = -4700 \text{ J} \). Therefore, \( W = -(-4700 \text{ J}) = 4700 \text{ J} \).
05

Determine the sign of the work done

The work done on the gas is positive, as energy left the gas and work was done by the gas to release such energy. Hence, the work done is +4700 J.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isothermal Process
An isothermal process, in thermodynamics, is one in which the temperature of a system remains constant. This is a key characteristic that defines isothermal processes. To maintain constant temperature, any heat added to the system must be converted into work, or vice versa.

In an isothermal process involving an ideal gas, the internal energy of the system does not change, because the internal energy of an ideal gas is solely dependent on its temperature. Since the temperature remains unchanged, the internal energy remains unaffected. Thus, heat exchange in an isothermal process is fully converted into work done by or on the system.
  • The pressure and volume of the gas must adjust accordingly to maintain constant temperature.
  • The process can be represented mathematically for an ideal gas using the equation: \[PV = nRT\]
  • An isothermal process typically requires slower changes, allowing the system to exchange heat with its surroundings effectively.
Monatomic Ideal Gas
A monatomic ideal gas is a simple form of gas comprised of single-atom molecules. This characteristic makes it easier to analyze in the study of thermodynamics. The term "ideal gas" refers to a hypothetical gas that perfectly follows the ideal gas laws, such as Boyle's Law, Charles's Law, and Avogadro's Law. It is composed of tiny, non-interacting particles that do not occupy space.

Key features of a monatomic ideal gas include:
  • The kinetic energy of the gas particles is the main contributor to the internal energy.
  • The particles are in constant, random motion, but do not attract or repel each other.
  • The specific heat capacity at constant volume, denoted as \(C_v\), for a monatomic ideal gas can be shown as \(\frac{3}{2}R\), where \(R\) is the ideal gas constant.
Understanding these properties helps explain why the internal energy remains constant during an isothermal process. Since the energy depends only on temperature, a constant temperature means no change in internal energy.
First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental concept that revolves around the conservation of energy. It states that energy cannot be created or destroyed, only transferred or converted. The law is mathematically represented as: \[\Delta U = Q - W\]where:
  • \(\Delta U\) is the change in internal energy of the system,
  • \(Q\) is the heat exchanged (added to the system is positive, extracted is negative), and
  • \(W\) is the work done by the system (on the environment is positive; on the system itself is negative).
In an isothermal process, like the one described in the exercise, the internal energy \(\Delta U\) is zero due to constant temperature. Therefore, the equation simplifies to \(Q = W\). This means any heat entering or leaving the system is directly converted into work done by the system or on the system.

It illustrates the energy exchange perfectly: if a system loses a certain amount of energy as heat, it must have done an equivalent amount of work. In the provided exercise, 4700 J of heat exits the gas, transforming into 4700 J of work done by the gas. Thus, thermodynamics showcases how energy continues to move and transform efficiently within an isolated system.

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 52 -kg mountain climber, starting from rest, climbs a vertical distance of 730 \(\mathrm{m}\) . At the top, she is again at rest. In the process, her body generates \(4.1 \times 10^{6} \mathrm{J}\) of energy via metabolic processes. In fact, her body acts like a heat engine, the efficiency of which is given by Equation 15.11 as \(e=|W| / Q_{\mathrm{H}} |,\) where \(|W|\) is the magnitude of the work she does and \(\left|Q_{\mathrm{H}}\right|\) is the magnitude of the input heat. Find her efficiency as a heat engine.

A Carnot engine uses hot and cold reservoirs that have temperatures of 1684 and 842 \(\mathrm{K}\) , respectively. The input heat for this engine is \(\left|Q_{\mathrm{H}}\right|\) The work delivered by the engine is used to operate a Carnot heat pump. The pump removes heat from the \(842-\mathrm{K}\) reservoir and puts it into a hot reservoir at a temperature \(T^{\prime}\) . The amount of heat removed from the \(842-\mathrm{K}\) reservoir is also \(\left|Q_{\mathrm{H}}\right| .\) Find the temperature \(T^{\prime}\)

In exercising, a weight lifter loses 0.150 \(\mathrm{kg}\) of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is \(1.40 \times 10^{5} \mathrm{J}\) (a) Assuming that the latent heat of vaporization of perspiration is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\) find the change in the internal energy of the weight lifter. \((\mathbf{b})\) Determine the minimum number of nutritional Calories of food \((1 \text { nutritional Calorie }=4186 \mathrm{J})\) that must be consumed to replace the loss of internal energy.

Argon is a monatomic gas whose atomic mass is 39.9 u. The temperature of eight grams of argon is raised by 75 \(\mathrm{K}\) under conditions of constant pressure. Assuming that argon behaves as an ideal gas, how much heat is required?

Suppose that 31.4 \(\mathrm{J}\) of heat is added to an ideal gas. The gas expands at a constant pressure of \(1.40 \times 10^{4}\) Pa while changing its volume from \(3.00 \times 10^{-4}\) to \(8.00 \times 10^{-4} \mathrm{m}^{3}\) . The gas is not monatomic, so the relation \(C_{P}=\frac{5}{2} R\) does not apply. (a) Determine the change in the internal energy of the gas. (b) Calculate its molar specific heat capacity \(C_{p} .\)

See all solutions

Recommended explanations on Physics Textbooks

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.