/*! 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 36 A gas under a constant pressure ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 36

A gas under a constant pressure of \(1.50 \times 10^{5} \mathrm{~Pa}\) and with an initial volume of \(0.0900 \mathrm{~m}^{3}\) is cooled until its volume becomes \(0.0600 \mathrm{~m}^{3}\). (a) Draw a \(p V\) diagram of this process. (b) Calculate the work done by the gas.

Short Answer

Expert verified
(a) Draw a horizontal line on the pV diagram at \(1.50 \times 10^{5} \, \text{Pa}\) from \(0.0900 \, \text{m}^3\) to \(0.0600 \, \text{m}^3\). (b) Work done by the gas is \(-4500 \, \text{J}\).

Step by step solution

01

Understand the Problem

We are dealing with a gas under constant pressure that undergoes a change in volume. We need to visualize this using a pressure-volume (\(pV\)) diagram and calculate the work done by the gas.
02

Draw the pV Diagram

To sketch the \(pV\) diagram, we plot volume on the x-axis and pressure on the y-axis. The pressure remains constant at \(1.50 \times 10^{5} \, \text{Pa}\), and the volume changes from \(0.0900 \, \text{m}^3\) to \(0.0600 \, \text{m}^3\). This gives us a horizontal line on the graph from \(V = 0.0900 \, \text{m}^3\) to \(V = 0.0600 \, \text{m}^3\), indicating a constant pressure process.
03

Calculate Work Done by the Gas

The work done by the gas during a change in volume at constant pressure is calculated using the formula \(W = p \, \Delta V\). Here, \(\Delta V = V_f - V_i\), where \(V_f = 0.0600 \, \text{m}^3\) and \(V_i = 0.0900 \, \text{m}^3\). Thus, \(\Delta V = 0.0600 \, \text{m}^3 - 0.0900 \, \text{m}^3 = -0.0300 \, \text{m}^3\). The pressure \(p\) is \(1.50 \times 10^{5} \, \text{Pa}\). Substituting these values into the formula gives: \(W = 1.50 \times 10^{5} \, \text{Pa} \times (-0.0300 \, \text{m}^3) = -4500 \, \text{J}\).
04

Interpret the Result

The negative sign in the work done indicates that the work is done on the gas by the surroundings as it is compressed. Hence, the gas loses energy to its surroundings.

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.

Pressure-Volume Diagram
A Pressure-Volume diagram, often known as a \(pV\) diagram, is a graph that represents the relationship between pressure and volume for a system such as a gas. It is an essential tool in thermodynamics to visualize processes and transitions experienced by a gas. In these diagrams:
  • The volume is plotted on the x-axis.
  • The pressure is plotted on the y-axis.
These diagrams can show different thermodynamic processes, including those under constant pressure, constant volume, and constant temperature.
The exercise specifically describes a situation where pressure remains constant while the volume decreases from \(0.0900 \, \text{m}^3\) to \(0.0600 \, \text{m}^3\). In such a scenario, the \(pV\) diagram will feature a horizontal line since the pressure does not change while volume decreases. This behavior evidences a constant pressure or isobaric process.
Isobaric Process
An isobaric process is a thermodynamic process in which the pressure remains constant. This type of process is quite common in engineering and natural science, and it's significant due to the simplicity it offers when studying the effects of heating and cooling gases.
In the situation given in the problem, the gas is cooled while maintaining a pressure of \(1.50 \times 10^{5} \, \text{Pa}\). This cooling leads to a reduction of volume but not a change in pressure. This is characteristic of isobaric processes, where some of the heat energy causes the gas to do work or being done on the gas, rather than changing the pressure.
One can easily understand an isobaric process by visualizing the horizontal line on a \(pV\) diagram, indicating that the pressure is stable while the volume can change. Unlike isochoric (constant volume) or isothermal (constant temperature) processes, the isobaric process allows both temperature and volume to vary, ensuring the pressure remains steady.
Work Done by Gas
The work done by a gas is a fundamental concept when dealing with thermodynamic processes. During isobaric processes, work can be calculated by examining the change in volume, since pressure remains constant. The formula to determine this is:\[ W = p \, \Delta V \]where \(W\) is the work done, \(p\) is the constant pressure, and \(\Delta V\) is the change in volume.
In the given exercise, the volume decreases from \(0.0900 \, \text{m}^3\) to \(0.0600 \, \text{m}^3\), which results in \(\Delta V = -0.0300 \, \text{m}^3\). With a constant pressure of \(1.50 \times 10^{5} \, \text{Pa}\), this yields:\[ W = 1.50 \times 10^{5} \, \text{Pa} \times (-0.0300 \, \text{m}^3) = -4500 \, \text{J} \]The negative sign indicates that work is being done on the gas rather than by it. This means energy is transferred from the surroundings into the gas, causing the gas to compress and lose energy.

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 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 \(1.80 \times 10^{5} \mathrm{~Pa}\) 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.

At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at \(20.0^{\circ} \mathrm{C}\) ? (Hint: The periodic table in Appendix C shows the molar mass (in \(\mathrm{g} / \mathrm{mol}\) ) of each element under the chemical symbol for that element. The molar mass of \(\mathrm{H}_{2}\) is twice the molar mass of hydrogen atoms, and similarly for \(\mathrm{N}_{2}\).)

The average temperature of the atmosphere near the surface of the earth is about \(20^{\circ} \mathrm{C}\). (a) What is the root-mean-square speed of hydrogen molecules, \(\mathrm{H}_{2}\), at this temperature? (b) The escape speed from the earth is about \(11 \mathrm{~km} / \mathrm{s}\). Is the average \(\mathrm{H}_{2}\) molecule moving fast enough to escape? (c) Compare the rms speeds of oxygen \(\left(\mathrm{O}_{2}\right)\) and nitrogen \(\left(\mathrm{N}_{2}\right)\) with that of \(\mathrm{H}_{2}\). (d) So why has the hydrogen been able to escape the earth's gravity, but the heavier gases (such as \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) ) have not, even though none of these gases has an rms speed equal to the escape speed of the earth? (Hint: Do all the molecules have the same rms speed, or are some moving faster? since the rms speed for \(\mathrm{H}_{2}\) is greater than that of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\), which gas would have a higher percentage of its molecules moving fast enough to escape?)

We have two equal-size boxes, \(A\) and \(B\). Each box contains gas that behaves as an ideal gas. We insert a thermometer into each box and find that the gas in box \(A\) is at a temperature of \(50^{\circ} \mathrm{C}\) while the gas in box \(B\) is at \(10^{\circ} \mathrm{C}\). This is all we know about the gas in the boxes. Which of the following statements must be true? Which could be true? (a) The pressure in \(A\) is higher than in \(B\). (b) There are more molecules in \(A\) than in \(B\). (c) \(A\) and \(B\) cannot contain the same type of gas. (d) The molecules in \(A\) have more average kinetic energy per molecule than those in \(B\). (e) The molecules in \(A\) are moving faster than those in \(B\). Explain the reasoning behind your answers.

A bicyclist uses a tire pump whose cylinder is initially full of air at an absolute pressure of \(1.01 \times 10^{5} \mathrm{~Pa}\). The length of stroke of the pump (the length of the cylinder) is \(36.0 \mathrm{~cm}\). At what part of the stroke (i.e., what length of the air column) does air begin to enter a tire in which the gauge pressure is \(2.76 \times 10^{5} \mathrm{~Pa}\) ? Assume that the temperature remains constant during the compression.
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Engineering Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Oscillations

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.