15.4

Thermal Processes

A system can interact with its surroundings in many ways, and the heat and work that come into play always obey the first law of thermodynamics. This section introduces four common thermal processes. In each case, the process is assumed to be quasi-static, which means that it occurs slowly enough that a uniform pressure and temperature exist throughout all regions of the system at all times.
An isobaric process is one that occurs at constant pressure. For instance, Figure 15-5 shows a substance (solid, liquid, or gas) contained in a chamber fitted with a frictionless piston. The pressure P experienced by the substance is always the same and is determined by the external atmosphere and the weight of the piston and the block resting on it. Heating the substance makes it expand and do work W in lifting the piston and block through the displacement . The work can be calculated from W = Fs (Equation 6.1), where F is the magnitude of the force and s is the magnitude of the displacement. The force is generated by the pressure P acting on the bottom surface of the piston (area = A), according to F = PA (Equation 10.19). With this substitution for F, the work becomes W = (PA)s. But the product A·s is the change in volume of the material, , where Vf and Vi are the final and initial volumes, respectively. Thus, the expression for the work is

| (15.2)

Consistent with our sign convention, this result predicts a positive value for the work done by a system when it expands isobarically (Vf exceeds Vi). Equation 15.2 also applies to an isobaric compression (Vf less than Vi). Then, the work is negative, since work must be done on the system to compress it. Example 3 emphasizes that applies to any system, solid, liquid, or gas, as long as the pressure remains constant while the volume changes.

Figure 15-5 The substance in the chamber is expanding isobarically because the pressure is held constant by the external atmosphere and the weight of the piston and the block.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 3 | | | Isobaric Expansion of Water

One gram of water is placed in the cylinder in Figure 15-5, and the pressure is maintained at . The temperature of the water is raised by 31 C°. In one case, the water is in the liquid phase, expands by the small amount of , and has a specific heat capacity of 4186 J/(kg · C°). In another case, the water is in the gas phase, expands by the much greater amount of , and has a specific heat capacity of 2020 J/(kg · C°). Determine the change in the internal energy of the water in each case.

Reasoning The change in the internal energy is given by the first law of thermodynamics as (Equation 15.1). The heat Q may be evaluated as (Equation 12.4). Finally, since the process occurs at a constant pressure (isobaric), the work W may be found using (Equation 15.2).
Knowns and Unknowns The following table summarizes the given data:

Description | Symbol | Value | Comment
Mass of water | m | 1.0 g | 0.0010 kg
Pressure on water | P |  | Pressure is constant.
Increase in temperature |  | 31 C° |
Increase in volume of liquid |  |  | Expansion occurs.
Specific heat capacity of liquid | cliquid | 4186J/(kg·C°) |
Increase in volume of gas |  |  | Expansion occurs.
Specific heat capacity of gas | cgas | 2020 J/(kg·C°) |
Unknown Variables
Change in internal energy of liquid |  | ? |
Change in internal energy of gas |  | ? |

Modeling the Problem

| Step 1 The First Law of Thermodynamics
| The change in the internal energy is given by the first law of thermodynamics, as shown at the right. In Equation 15.1, neither the heat Q nor the work W is known, so we turn to Steps 2 and 3 to evaluate them.

| (15.1)

| Step 2 Heat and Specific Heat Capacity
| According to Equation 12.4, the heat Q needed to raise the temperature of a mass m of material by an amount is

where c is the material’s specific heat capacity. Data are available for all of the terms on the right side of this expression, which can be substituted into Equation 15.1, as shown at the right. The remaining unknown variable in Equation 15.1 is the work W, and we evaluate it in Step 3.

| Step 3 Work Done at Constant Pressure
| Under constant-pressure, or isobaric, conditions, the work W done is given by Equation 15.2 as

where P is the pressure acting on the material and is the change in the volume of the material. Substitution of this expression into Equation 15.1 is shown at the right.

Solution Combining the results of each step algebraically, we find that

Applying this result to the liquid and to the gaseous water gives

For the liquid, virtually all the 130 J of heat serves to change the internal energy, since the volume change and the corresponding work of expansion are so small. In contrast, a significant fraction of the 63 J of heat added to the gas causes work of expansion to be done, so that only 49 J is left to change the internal energy.
Related Homework: Problem 12
It is often convenient to display thermal processes graphically. For instance, Figure 15-6 shows a plot of pressure versus volume for an isobaric expansion. Since the pressure is constant, the graph is a horizontal straight line, beginning at the initial volume Vi and ending at the final volume Vf. In terms of such a plot, the work is the area under the graph, which is the shaded rectangle of height P and width .

Figure 15-6 For an isobaric process, a pressure-versus-volume plot is a horizontal straight line, and the work done [] is the colored rectangular area under the graph.

Another common thermal process is an isochoric process, one that occurs at constant volume. Figure 15-7a illustrates an isochoric process in which a substance (solid, liquid, or gas) is heated. The substance would expand if it could, but the rigid container keeps the volume constant, so the pressure–volume plot shown in Figure 15-7b is a vertical straight line. Because the volume is constant, the pressure inside rises, and the substance exerts more and more force on the walls. Although enormous forces can be generated in the closed container, no work is done (), since the walls do not move. Consistent with zero work being done, the area under the vertical straight line in Figure 15-7b is zero. Since no work is done, the first law of thermodynamics indicates that the heat in an isochoric process serves only to change the internal energy: .

Figure 15-7 (a) The substance in the chamber is being heated isochorically because the rigid chamber keeps the volume constant. (b) The pressure-volume plot for an isochoric process is a vertical straight line. The area under the graph is zero, indicating that no work is done.

A third important thermal process is an isothermal process, one that takes place at constant temperature. The next section illustrates the important features of an isothermal process when the system is an ideal gas.
Last, there is the adiabatic process, one that occurs without the transfer of heat. Since there is no heat transfer, Q equals zero, and the first law indicates that . Thus, when work is done by a system adiabatically, W is positive and the internal energy decreases by exactly the amount of the work done. When work is done on a system adiabatically, W is negative and the internal energy increases correspondingly. The next section discusses an adiabatic process for an ideal gas.
A process may be complex enough that it is not recognizable as one of the four just discussed. For instance, Figure 15-8 shows a process for a gas in which the pressure, volume, and temperature are changed along the straight line from X to Y. With the aid of integral calculus, a useful fact can be proved, as stated in the following Problem solving insight: the area under a pressure-volume graph is the work for any kind of process. Thus, the area representing the work has been colored in Figure 15-8. The volume increases, so that work is done by the gas. This work is positive by convention, as is the area. In contrast, if a process reduces the volume, work is done on the gas, and this work is negative by convention. Correspondingly, the area under the pressure-volume graph would be assigned a negative value. In Example 4, we determine the work for the case shown in Figure 15-8.

Figure 15-8 The colored area gives the work done by the gas for the process from X to Y.

| Example 4 | | | Work and the Area Under a Pressure–Volume Graph

Determine the work for the process in which the pressure, volume, and temperature of a gas are changed along the straight line from X to Y in Figure 15-8. Reasoning The work is given by the area (in color) under the straight line between X and Y. Since the volume increases, work is done by the gas on the surroundings, so the work is positive. The area can be found by counting squares in Figure 15-8 and multiplying by the area per square.
Solution We estimate that there are 8.9 colored squares in the drawing. The area of one square is , so the work is

Check Your Understanding 2

The drawing shows a pressure-versus-volume plot for a three-step process: . For each step, the work can be positive, negative, or zero. Which answer below correctly describes the work for the three steps? (The answer is given at the end of the book.)

Work Done by the System
a. | Positive | Negative | Negative
b. | Positive | Positive | Negative
c. | Negative | Negative | Positive
d. | Positive | Negative | Zero
e. | Negative | Positive | Zero

Background: For any kind of process, the work done can be evaluated by using a pressure-versus-volume plot. Recall that work is positive, negative, or zero, depending on how the volume changes.
For similar questions (including calculational counterparts), consult Self-Assessment Test 15.1. This test is described at the end of Section 15.6.

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.