/*! 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 9 A gas in a cylinder expands from... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 9

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 \(1.80 \times 10^{5}\) 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. it matter whether the gas is ideal? Why or why not?

Short Answer

Expert verified
(a) The work done by the gas is \(3.78 \times 10^{4}\) J. (b) The change in internal energy is \(7.72 \times 10^{4}\) J. The gas type does not affect these calculations.

Step by step solution

01

Understand the Scenario

The gas in a cylinder expands under constant pressure. You're given the initial and final volumes, the constant pressure, and the total heat added to the system. You need to find the work done by the gas during expansion and the change in internal energy of the gas.
02

Calculate Work Done by the Gas

Work done by the gas during expansion at constant pressure is calculated using the formula \( W = P \Delta V \), where \( P \) is the pressure and \( \Delta V \) is the change in volume.\[ \Delta V = V_{final} - V_{initial} = 0.320 \ \mathrm{m}^{3} - 0.110 \ \mathrm{m}^{3} = 0.210 \ \mathrm{m}^{3} \]Plug the values into the formula:\[ W = 1.80 \times 10^{5} \times 0.210 = 3.78 \times 10^{4} \ \mathrm{J} \]
03

Use the First Law of Thermodynamics

The first law of thermodynamics states \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added, and \( W \) is the work done by the system. We calculated \( W \) in the previous step, and \( Q \) is given.\[ \Delta U = 1.15 \times 10^{5} - 3.78 \times 10^{4} \]\[ \Delta U = 7.72 \times 10^{4} \ \mathrm{J} \]
04

Consider if Gas Type Matters

For this problem, whether the gas is ideal does not matter. The calculations are purely thermodynamic and do not depend on the specific properties of the gas beyond its behavior of expanding at constant pressure.

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.

First Law of Thermodynamics
The First Law of Thermodynamics is an essential principle in physics that provides a clear understanding of how energy interactions occur within a thermodynamic system. This law is essentially about energy conservation.
It states that the change in internal energy of a system, denoted as \( \Delta U \), is equal to the heat added to the system, \( Q \), minus the work done by the system, \( W \). This relationship is mathematically expressed as:
\[ \Delta U = Q - W \]
  • \( \Delta U \): Change in internal energy
  • \( Q \): Heat added
  • \( W \): Work done by the system
This equation helps us analyze energy changes during processes like heating, cooling, and work done by gases. If energy enters a system as heat, it can either be stored within the system as internal energy or leave as work done by the system. Understanding this basic idea helps us solve problems in thermodynamics with greater ease.
Constant Pressure Process
In a constant pressure process, the pressure of the system remains unchanged throughout the entire process. This is also known as an isobaric process. Such processes are common in real-world scenarios and are essential for studying the behavior of gases in dynamic environments.
During this type of process, as seen in the exercise, the gas expands while maintaining a fixed pressure. We use the formula for work done at constant pressure:\[ W = P \Delta V \]
where:
  • \( P \): Constant pressure
  • \( \Delta V \): Change in volume, \( V_{final} - V_{initial} \)
This formula allows us to calculate the work done by the gas as it expands against the constant pressure in the cylinder. The work required for expansion depends not only on the pressure but also significantly on the change in volume.
Internal Energy Change
Internal energy change, \( \Delta U \), is a core concept in thermodynamics and relates to the energy stored within a system. It signifies how much energy is held inside the system due to factors like temperature and molecular interaction.
When analyzing any thermodynamic process, it's crucial to determine how much the internal energy changes. Using the first law of thermodynamics, \( \Delta U = Q - W \), enables us to compute it.
In this exercise, after the system undergoes an expansion at constant pressure, the calculation of \( \Delta U \) involves:
  • Identifying \( Q \) (heat added, \( 1.15 \times 10^{5} \) J)
  • Determining \( W \) (work done by the gas, \( 3.78 \times 10^{4} \) J)
  • Applying them in the equation to find \( \Delta U \)
These calculations show how much energy has shifted within the gas system as part of the internal energy change, helping us to better understand energy flow and transformation in thermodynamic 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

A Thermodymamic Process in a Liquid. A chemical engineer is studying the properties of liquid methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) She uses a stecl cylinder with a cross-sectional. The cylindor is and containing \(1.20 \times 10^{-2} \mathrm{m}^{3}\) of methanol. The cylinder is equipped with a tightly fitting piston that supports a load of \(3.00 \times 10^{4} \mathrm{N}\) . The temperature of the system is increased from \(20.0^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C}\) . For methanol, the coefficient of volume expansion is \(1.20 \times 10^{-3} \mathrm{K}^{-1}\) , the density is 791 \(\mathrm{kg} / \mathrm{m}^{3}\) , and the specific heat capacity at constant pressure is \(c_{p}=2.51 \times 10^{3} \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . You can ignore the expansion of the stecl cylinder. Find (a) the increase in volume of the methanol; (b) the mechanical work done by the methanol against the \(3.00 \times 10^{6} \mathrm{N}\) force; (c) the amount of heat added to the methanol; (d) the change in internal cnergy of the methanol. (e) Based on your results, explain whether there is any substantial difference between the specific heat capacities \(c_{p}\) (at constant pressure) and \(c_{V}\) (at constant volume) for methanol under these conditions.

Two moles of carbon monoxide (CO) start at a pressure of 1.2 atm and a volume of 30 liters. The gas is then compressed adiabatically to \(\frac{1}{3}\) this volume. Assume that the gas may be treated as ideal. What is the change in the internal energy of the gas? Does the internal energy increase or decrease? Does the temperature of the gas increase or decrease during this process? Explain.

An ideal gas expands while the pressure is kept constant. During this process, does heat flow into the gas or out of the gas? Justify your answer.

Two moles of an ideal gas are compressed in a cylinder at a constant temperature of \(85.0^{\circ} \mathrm{C}\) until the original pressure has tripled. (a) Sketch a \(p V\) -diagram for this process. (b) Calculate the amount of work done.

A cylinder with a movable piston contains 3.00 \(\mathrm{mol}\) of \(\mathrm{N}_{2}\) gas (assumed to behave like an ideal gas). (a) The \(\mathrm{N}_{2}\) is heated at constant volume until 1557 \(\mathrm{J}\) of heat have been added. Calculate the change in temperature. (b) Suppose the same amount of heat is added to the \(\mathrm{N}_{2}\) , but this time the gas is allowed to expand while remaining at constant pressure. Calculate the temperature change. (c) In which case, (a) or \((b),\) is the final internal energy of the \(\mathbf{N}_{2}\) higher? How do you know? What accounts for the difference between the two cases?
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Mechanics and Materials

Read Explanation

Physics of Motion

Read Explanation

Electricity and Magnetism

Read Explanation

Modern 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.