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Chapter 19: Problem 8

An ideal gas undergoes a process during which the pressure is kept directly proportional to the volume, so that \(p=\alpha V\) where \(\alpha\) is a positive constant. If the volume changes from \(V_{1}\) to \(V_{2}\), how much work is done by the gas? Express your answer in terms of \(V_{1}, V_{2},\) and \(\alpha\)

Short Answer

Expert verified
The work done by the gas is \(W = \frac{1}{2} \alpha (V_{2}^2 - V_{1}^2)\)

Step by step solution

01

Writing the Work Formula

The work done \(W\) on/by the gas can be given by the integral of pressure \(p\) with respect to volume \(V\), from the initial volume \(V_{1}\) to the final volume \(V_{2}\). The formula is \(W = \int_{V_{1}}^{V_{2}} pdV\).
02

Substituting the Pressure-Volume Relationship

Since the pressure is directly proportional to the volume, \(p = \alpha V\). By substituting this into the work formula, the equation becomes: \(W = \int_{V_{1}}^{V_{2}} \alpha V dV\)
03

Calculating the Integral

The integral can be calculated by the power rule: \(\int x^n dx = \frac{1}{n+1}x^{n+1}\). Therefore, \( \int_{V_{1}}^{V_{2}} \alpha V dV = \frac{1}{2} \alpha (V_{2}^2 - V_{1}^2)\)
04

Final Answer

The work done by the gas is given by: \(W = \frac{1}{2} \alpha (V_{2}^2 - V_{1}^2)\)

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Key Concepts

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

Work Done by Gas
When we talk about the work done by a gas, we're looking at how much energy the gas has expended while changing its volume. This is a fundamental concept in thermodynamics where gases do work on their surroundings by expanding. For an ideal gas undergoing a transformation, the work done can be calculated by the integral of pressure with respect to volume change. In simpler terms, if you imagine a piston filled with gas, as the gas expands and pushes the piston outward, it's doing work against the external pressure.

In our specific scenario, the gas鈥� pressure increases proportionally to its volume, which makes for an interesting case as work calculation involves integrating a function where pressure is a variable that depends on volume. Remember that the work done by a gas can be both positive and negative鈥攑ositive when the gas expands (doing work on the surroundings) and negative when it's compressed (work is done on the gas).
Pressure-Volume Relationship
The pressure-volume relationship, often depicted on a graph as a curve, is crucial in understanding how gases behave under different conditions. For an ideal gas, the product of pressure and volume is a constant when temperature remains unchanged (Boyle鈥檚 Law). However, in our exercise, we鈥檙e looking at a direct proportionality; as the volume, denoted by V, increases, so does the pressure, represented by p. This relationship can be mathematically expressed as p = \(\alpha V\), where \(\alpha\) is a constant.

This linear relationship between pressure and volume means that as the gas expands in a container, the pressure it exerts on the container walls increases at the same rate. This sets the stage to calculate work done by using integral calculus, as the work is essentially the area under the curve represented by this linear relationship in a pressure-volume graph.
Integral Calculus in Physics
The application of integral calculus in physics is a powerful tool for solving a variety of problems, particularly those involving changes over an interval. Integrals allow you to calculate the total effect of something that changes across a range, like how the area under a pressure-volume curve represents the work done by a gas during expansion or compression.

In this exercise, integral calculus is used to determine the work done by the gas when the pressure and volume are directly proportional. A definite integral from V1 to V2 is needed to find the exact amount of work done over the change in volume. The integral of the function p = \(\alpha V\) from the initial to the final volume is established and, with the application of the power rule, we can compute the answer. By integrating, we sum up infinitely small amounts of work done by the gas during each infinitely small increase in volume, giving us the total work done over the entire process.

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Most popular questions from this chapter

A pneumatic shock absorber consists of a cylinder with a radius of \(1.50 \mathrm{~cm}\) and a length that varies from a maximum of \(30.0 \mathrm{~cm}\) to a minimum of \(30.0 \mathrm{~cm} / v,\) where \(v\) is the compression ratio. For ease of manufacturing, \(v\) must be an integer. When the cylinder is fully extended and cooled, the air inside has an ambient pressure of \(101 \mathrm{kPa}\) and ambient temperature of \(30.0^{\circ} \mathrm{C}\). A quick adiabatic compression heats the air. The cylinder then expands isothermally until it reaches its maximum length. At that time the length remains fixed and the cylinder cools isochorically, returning to the ambient temperature. (a) How much work is done by the air during a full cycle, in terms of \(v ?\) (b) If the air temperature cannot exceed \(400^{\circ} \mathrm{C}\), what is the maximum integer value of \(v ?\) (c) If at least \(25.0 \mathrm{~J}\) of work must be done by the gas in a full-compression cycle, what is the minimum allowable integer value of \(v ?\) (d) What is the only permissible value of \(v ?\) (e) How much heat leaves the air in the cylinder during the isochoric process if \(v\) has the value calculated in (d)?

Comparing Thermodynamic Processes. In a cylinder, \(1.20 \mathrm{~mol}\) of an ideal monatomic gas, initially at \(3.60 \times 10^{5} \mathrm{~Pa}\) and \(300 \mathrm{~K},\) expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a \(p V\) -diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

Engine Turbochargers and Intercoolers. The power output of an automobile engine is directly proportional to the mass of air that can be forced into the volume of the engine's cylinders to react chemically with gasoline. Many cars have a turbocharger, which compresses the air before it enters the engine, giving a greater mass of air per volume. This rapid, essentially adiabatic compression also heats the air. To compress it further, the air then passes through an intercooler in which the air exchanges heat with its surroundings at essentially constant pressure. The air is then drawn into the cylinders. In a typical installation, air is taken into the turbocharger at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{~Pa}\right),\) density \(\rho=1.23 \mathrm{~kg} / \mathrm{m}^{3},\) and temperature \(15.0^{\circ} \mathrm{C}\). It is compressed adiabatically to \(1.45 \times 10^{5} \mathrm{~Pa}\). In the intercooler, the air is cooled to the original temperature of \(15.0^{\circ} \mathrm{C}\) at a constant pressure of \(1.45 \times 10^{5} \mathrm{~Pa}\). (a) Draw a \(p V\) -diagram for this sequence of processes. (b) If the volume of one of the engine's cylinders is \(575 \mathrm{~cm}^{3},\) what mass of air exiting from the intercooler will fill the cylinder at \(1.45 \times 10^{5} \mathrm{~Pa}\) ? Compared to the power output of an engine that takes in air at \(1.01 \times 10^{5} \mathrm{~Pa}\) at \(15.0^{\circ} \mathrm{C},\) what percentage increase in power is obtained by using the turbocharger and intercooler? (c) If the intercooler is not used, what mass of air exiting from the turbocharger will fill the cylinder at \(1.45 \times 10^{5} \mathrm{~Pa}\) ? Compared to the power output of an engine that takes in air at \(1.01 \times 10^{5} \mathrm{~Pa}\) at \(15.0^{\circ} \mathrm{C},\) what percentage increase in power is obtained by using the turbocharger alone?

During an adiabatic expansion the temperature of \(0.450 \mathrm{~mol}\) of argon (Ar) drops from \(66.0^{\circ} \mathrm{C}\) to \(10.0^{\circ} \mathrm{C}\). The argon may be treated as an ideal gas. (a) Draw a \(p V\) -diagram for this process. (b) How much work does the gas do? (c) What is the change in internal energy of the gas?

A monatomic ideal gas expands slowly to twice its original volume, doing \(450 \mathrm{~J}\) 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.
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