/*! 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 53 In a certain process, \(2.15 \ti... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a certain process, \(2.15 \times 10^{5} \mathrm{J}\) of heat is liberated by a system, and at the same time the system contracts under a constant external pressure of \(9.50 \times 10^{5} \mathrm{Pa}\) . The internal energy of the system is the same at the beginning and end of the process. Find the change in volume of the system. (The system is not an ideal gas.)

Short Answer

Expert verified
The change in volume is -0.2263 m³.

Step by step solution

01

Understanding the First Law of Thermodynamics

The problem involves heat, work, and change in volume. Use the first law of thermodynamics: \[ \Delta U = Q - W \]where \( \Delta U \) is the change in internal energy, \( Q \) is the heat exchanged, and \( W \) is the work done by the system. Since the internal energy does not change in this process (\( \Delta U = 0 \)), we have:\[ 0 = Q - W \].
02

Relate Work to Pressure and Change in Volume

Work done by the system can also be expressed as the product of pressure and change in volume:\[ W = P \Delta V \]Substitute this into the modified first law equation:\[ Q = P \Delta V \]
03

Solve for Change in Volume

Rearrange the equation to solve for \( \Delta V \):\[ \Delta V = \frac{Q}{P} \]Substitute the given values:\[ Q = -2.15 \times 10^{5} \text{ J} \] (heat is liberated, hence negative) and \( P = 9.50 \times 10^{5} \text{ Pa} \).
04

Calculate the Change in Volume

Use the formula:\[ \Delta V = \frac{-2.15 \times 10^{5}}{9.50 \times 10^{5}} = -0.2263 \] Thus, the change in volume \( \Delta V \) is \(-0.2263 \text{ m}^3\). The negative sign indicates a decrease in volume.

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.

Change in Volume
In thermodynamics, the change in volume is an important concept, especially when analyzing systems under pressure. When a system undergoes a process, its volume can expand or contract depending on the pressure and the heat exchange involved. In this particular scenario, we are tasked with finding the change in volume of a system that contracts under constant external pressure.

Utilizing the First Law of Thermodynamics, we can connect the change in volume to the work done by the system. According to the formula, work is the product of pressure and change in volume: \[ W = P \Delta V \] In the exercise, we've established that the internal energy does not change, leading us to the equation: \[ Q = P \Delta V \] Here, work done is considered in terms of pressure and change in volume. By rearranging this equation, the change in volume can be expressed as: \[ \Delta V = \frac{Q}{P} \] This formula allows us to calculate how much a system's volume changes during a process.

The negative sign in the resulting \( \Delta V \) value indicates a reduction in volume, showing that the system contracts rather than expands. Understanding this concept is essential, especially in classical thermodynamics where such calculations frequently occur when analyzing gas laws, engine efficiency, and thermodynamic cycles.
Internal Energy
Internal energy is a fundamental concept in thermodynamics representing the total energy contained within a system. It encompasses all forms of kinetic and potential energy associated with the random motion and interactions of molecules. When analyzing any thermodynamic process, internal energy serves as a key variable.

The First Law of Thermodynamics, described by the equation: \[ \Delta U = Q - W \] relates changes in internal energy \( \Delta U \) to heat exchange \( Q \) and work \( W \). In this particular exercise, the internal energy remains constant throughout the process, meaning \( \Delta U = 0 \). This simplifies our calculations and tells us that any heat liberated or absorbed in the process must be equal and opposite to the work done by the system.

Since the system does not exhibit any change in its internal energy, it signifies that the entire amount of heat removed is converted into mechanical work against an external pressure. This is an illustration of energy conservation in isolated systems, and it highlights the power of the First Law of Thermodynamics in predicting system behavior during various physical processes.
Heat Exchange
Heat exchange refers to the transfer of thermal energy between a system and its surroundings. In thermodynamics, heat is a form of energy transfer that occurs due to a temperature difference between a system and its environment. It can be positive when absorbed by the system or negative when released.

In the stated problem, the system liberates heat, which means that heat is being transferred from the system to its surroundings. In thermodynamic terms, this is expressed as a negative heat value, \( Q = -2.15 \times 10^5 \text{ J} \). The negative sign signifies an exothermic process, where energy is emitted by the system.

Understanding heat exchange is crucial because it directly affects the internal energy and work performed by a system. As discussed in the First Law of Thermodynamics, heat exchange is one of the components that governs the energy balance within a system. In this case, since no change in internal energy is noted, the liberated heat is fully utilized for performing work on the system, specifically in changing its volume. Recognizing how heat and work interact allows us to comprehensively analyze energy exchanges in a wide range of physical and chemical processes.

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 monatomic ideal gas expands slowly to twice its original volume, doing 300 \(\mathrm{J}\) of work in the process. Find the heat added to the gas and the change in internal energy of the gas if the process is (a) isothermal; (b) adiabatic; (c) isobaric.

Five moles of an ideal monatomic gas with an initial temperature of \(127^{\circ} \mathrm{C}\) expand and, in the process, absorb 1200 \(\mathrm{J}\) of heat and do 2100 \(\mathrm{J}\) of work. What is the final temperature of the gas?

A gas in a cylinder expands from a volume of 0.110 \(\mathrm{m}^{3}\) to 0.320 \(\mathrm{m}^{3} .\) Heat flows into the gas just rapidly enough to keep the pressure constant at \(1.65 \times 10^{5}\) Pa during the expansion. The total heat added is \(1.15 \times 10^{5} \mathrm{J}\) (a) Find the work done by the gas. (b) Find the change in internal energy of the gas. (c) Does it matter whether the gas is ideal? Why or why not?

Acylinder contains 0.0100 mol of helium at \(T=27.0^{\circ} \mathrm{C}\) . (a) How much heat is needed to raise the temperature to \(67.0^{\circ} \mathrm{C}\) while keeping the volume constant? Draw a \(p V\) -diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from \(27.0^{\circ} \mathrm{C}\) to \(67.0^{\circ} \mathrm{C} ?\) Draw a \(p V\) -diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?

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.