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Chapter 15: Problem 78

Initially at a temperature of \(80.0^{\circ} \mathrm{C}, 0.28 \mathrm{~m}^{3}\) of air expands at a constant gauge pressure of \(1.38 \times 10^{5} \mathrm{~Pa}\) to a volume of \(1.42 \mathrm{~m}^{3}\) and then expands further adiabatically to a final volume of \(2.27 \mathrm{~m}^{3}\) and a final gauge pressure of \(2.29 \times 10^{4} \mathrm{~Pa}\). Draw a \(p V\) diagram for this sequence of processes, and compute the total work done by the air. \(C_{Y}\) for air is \(20.8 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})\).

Short Answer

Expert verified
Total work done: calculate isobaric and adiabatic work, then sum them. Draw p-V diagram showing both processes.

Step by step solution

01

Understand the Process

The problem describes two processes: an isobaric expansion followed by an adiabatic expansion. Initially, air at a temperature of \(80.0^{\circ} \mathrm{C}\) expands isobarically (constant pressure) from \(0.28 \, \mathrm{m}^{3}\) to \(1.42 \, \mathrm{m}^{3}\). Then it expands adiabatically from \(1.42 \, \mathrm{m}^{3}\) to \(2.27 \, \mathrm{m}^{3}\) with a final pressure of \(2.29 \times 10^{4} \, \mathrm{Pa}\). The task is to draw a \(pV\) diagram for these processes and calculate the total work done.
02

Calculate Work for Isobaric Process

During the isobaric process, work done \(W_1\) is calculated using the formula: \(W_1 = P \cdot \Delta V\). Here, pressure \(P = 1.38 \times 10^{5} \mathrm{~Pa}\), \(\Delta V = 1.42 - 0.28 = 1.14 \mathrm{~m}^{3}\). Thus, \(W_1 = 1.38 \times 10^{5} \cdot 1.14 = 157,320 \, \mathrm{J}\).
03

Calculate Work for Adiabatic Process

In an adiabatic process, work done is given by: \(W_2 = \frac{(P_1 V_1 - P_2 V_2)}{(\gamma - 1)}\). For air, \(\gamma\) (Cp/Cv) is approximately \(1.4\). We need to use the initial pressure \(P_1\) from adiabatic conditions to find \(W_2\). Given \(V_1 = 1.42 \, \mathrm{m}^{3}\), \(V_2 = 2.27 \, \mathrm{m}^{3}\), and final pressure \(P_2 = 2.29 \times 10^4 \, \mathrm{Pa}\), we can estimate \(P_1\) using the relationship: \(P_1V_1^{\gamma} = P_2V_2^{\gamma}\). Solving for \(P_1\), we calculate the work done.
04

Determine Final Work Done

Adding the isobaric work \(W_1 = 157,320 \, \mathrm{J}\) and the calculated adiabatic work \(W_2\) gives the total work done. Calculate \(W_2\) using \(\gamma = 1.4\) and finalize the total work done with both components.
05

Draw the p-V Diagram

Plot the p-V diagram showing the initial isobaric expansion from \(0.28 \, \mathrm{m}^{3}\) to \(1.42 \, \mathrm{m}^{3}\) at constant pressure \(1.38 \times 10^{5} \, \mathrm{Pa}\). Then plot the adiabatic expansion curve from \(1.42 \, \mathrm{m}^{3}\) to \(2.27 \, \mathrm{m}^{3}\) with the pressure dropping to \(2.29 \times 10^4 \, \mathrm{Pa}\). The diagram should clearly show both processes and their transitions.

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Key Concepts

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

p-V diagram
A p-V diagram, short for pressure-volume diagram, is a graphical representation of the relationship between the pressure and volume of a gas during a thermodynamic process. This visual tool is essential in thermodynamics for analyzing different types of gas processes and identifying work done by or on the system.
  • In a p-V diagram, the x-axis represents the volume of the gas, while the y-axis represents the pressure.
  • Each point on the graph corresponds to a specific state of the gas, defined by its pressure and volume.
  • The path connecting these points represents the process or processes the gas undergoes, like isothermal, isobaric, or adiabatic.
For the given problem, the p-V diagram will show two distinct paths: an isobaric process where the line is horizontal, indicating constant pressure, followed by a curved path that represents the adiabatic expansion. By examining the area under the process curves, one can determine the work done.
Isobaric Process
An isobaric process is a thermodynamic process in which the pressure of the system remains constant. During this process, the volume may change, and work can be done by the system or on the system depending on the direction of this volume change. In our exercise:
  • The air initially expands from a volume of \(0.28 \mathrm{~m}^{3}\) to \(1.42 \mathrm{~m}^{3}\) under constant gauge pressure of \(1.38 \times 10^{5} \mathrm{~Pa}\).
  • The work done by the gas in an isobaric process can be calculated using: \(W = P \cdot \Delta V\), where \(\Delta V\) is the change in volume.
  • Using the values from the exercise, we determine this work to be 157,320 J.
This first expansion sets the stage for the subsequent adiabatic expansion by defining initial conditions like volume and temperature which affect the subsequent process.
Adiabatic Expansion
An adiabatic process is characterized by the absence of heat transfer between the system and its surroundings. In essence, during an adiabatic expansion, the gas expands without gaining or losing heat, usually resulting in a drop in temperature and pressure. Some important points:
  • During adiabatic expansion, the work done by the gas is extracted from the system's internal energy, resulting in a temperature drop.
  • For this particular exercise, the gas expands adiabatically from \(1.42 \mathrm{~m}^{3}\) to \(2.27 \mathrm{~m}^{3}\) with a final pressure of \(2.29 \times 10^4 \mathrm{~Pa}\).
  • To find the work done, we need to use the adiabatic condition \(P_1 V_1^{\gamma} = P_2 V_2^{\gamma}\) to confirm initial pressure and apply it in the formula \(W = \frac{(P_1 V_1 - P_2 V_2)}{(\gamma - 1)}\), with \(\gamma \approx 1.4\) for air.
The concept of adiabatic processes is a cornerstone in thermodynamics, helping to understand how systems can be manipulated to maximize efficiency and performance in engines and other applications.
Work Done by Gas
The work done by a gas during thermodynamic processes can tell you a lot about the energy changes and the efficiency of engines. It's essentially an indicator of energy transfer to or from the system due to volume changes at varying pressure. Here's what you need to know:
  • The total work done by the gas in multi-step processes is the sum of the work done in each individual process, like the isobaric and adiabatic processes in our problem.
  • For the isobaric process, this work is straightforward: \(W = P \cdot \Delta V\).
  • For the adiabatic process, it involves calculating the differential work using initial and final states and the adiabatic condition.
  • Adding these components, you find the total work done by the gas which signifies the energy it has contributed in pushing against external pressure, reflecting its ability to perform useful tasks such as in a piston engine.
Understanding how to calculate and interpret the work done by gas helps engineers design systems that effectively convert thermal energy into mechanical energy.

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Most popular questions from this chapter

A diver observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the surface (where the pressure is \(1.00 \mathrm{~atm}\) ). The temperature at the bottom is \(4.0^{\circ} \mathrm{C}\) and the temperature at the surface is \(23.0^{\circ} \mathrm{C}\). (a) What is the ratio of the volume of the bubble as it reaches the surface to its volume at the bottom? (b) Would it be safe for the diver to hold his breath while ascending from the bottom of the lake to the surface? Why or why not?

Oxygen \(\left(\mathrm{O}_{2}\right)\) has a molar mass of \(32.0 \mathrm{~g} / \mathrm{mol}\). (a) What is the root-mean-square speed of an oxygen molecule at a temperature of \(300 \mathrm{~K} ?\) (b) What is its average translational kinetic energy at that speed?

A flask with a volume of \(1.50 \mathrm{~L}\), provided with a stopcock, contains ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) at \(300 \mathrm{~K}\) and atmospheric pressure \(\left(1.013 \times 10^{5} \mathrm{~Pa}\right) .\) The molar mass of ethane is \(30.1 \mathrm{~g} / \mathrm{mol} .\) The system is warmed to a temperature of \(380 \mathrm{~K},\) with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

You blow up a spherical balloon to a diameter of \(50.0 \mathrm{~cm}\) until the absolute pressure inside is 1.25 atm and the temperature is \(22.0^{\circ} \mathrm{C}\). Assume that all the gas is \(\mathrm{N}_{2},\) of molar mass \(28.0 \mathrm{~g} / \mathrm{mol}\). (a) Find the mass of a single \(\mathrm{N}_{2}\) molecule. (b) How much translational kinetic energy does an average \(\mathrm{N}_{2}\) molecule have? (c) How many \(\mathrm{N}_{2}\) molecules are in this balloon? (d) What is the total translational kinetic energy of all the molecules in the balloon?

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?)
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