Chapter Review
|
Chapter 18: The Laws of Thermodynamics Chapter Review |
|
Chapter Review
In this final chapter on thermodynamics we bring all four laws together hoping to provide a basic understanding of the importance of thermodynamics and many of its applications.
18-1 The Zeroth Law of Thermodynamics
The zeroth law of thermodynamics was introduced in chapter 16; we review this law here for the sake of completeness. The zeroth law of thermodynamics lays important groundwork for all of thermodynamics, especially for providing a working definition of the concept of temperature. The zeroth law says the following:
If system A is in thermal equilibrium with system B, and system C is also in thermal equilibrium with system B, then systems A and C will be in thermal equilibrium if brought into thermal contact.
The property that determines whether or not two systems will be in thermal equilibrium is temperature. When two systems are at the same temperature they will be in thermal equilibrium if brought into thermal contact.
18-2 The First Law of Thermodynamics
The first law of thermodynamics is essentially just an application of the conservation of energy to situations involving heat flow. Any system has a certain amount of internal energy U; this energy consists of all the potential and kinetic energy contained within the system. Changes in the internal energy result from either heat flow into (positive Q) or out of (negative Q) the system, and/or work done by (positive W) or on (negative W) the system. The mathematical expression of this law is
DU = Q - W.
Notice that this expression is consistent with the intuitive notion that the internal energy will increase when either heat is added to the system and/or work is done on the system.
An important distinction between internal energy, heat, and work is that changes in internal energy only depend on the initial and final states of the system (which are determined by pressure, temperature, and volume) and U is therefore called a state function. Both heat and work depend not only on the states involved, but also on the process by which a system is changed from one state to another.
Example 18.1 The First Law: If 4530 J of work is done on a 0.750 kg piece of copper while its temperature rises from 18.2 oC to 31.2 oC, what is the change in internal energy of the piece of copper? (Assume a constant pressure of 1 atm.)
Picture the Problem The picture shows the piece of copper in question.
Strategy Since we are given the amount of work W, if we can determine the heat flow Q, we can use the first law to determine DU.
Solution
| 1. We can use the specific heat, Table 16-2 in the text, to determine Q: |  |
| 2. The work is done on the copper so: | W = -4530 J |
| 3. According to the first law of thermodynamics: | DU = Q - W = 3802.5 J - (-4530 J) = 8330 J |
Insight Remember when using the first law of thermodynamics that it is important to keep track of whether the heat flows is into or out of the system and if the work is done on or by the system.
Practice Quiz
|
|
|
|
 |
 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
18-3 Thermal Processes
As mentioned at the end of the previous section, some quantities, such as Q and W, depend on the process by which a system is altered. Because of this fact, we need to understand the basic types of processes used in the study and application of thermodynamics.
In this chapter we only consider processes that are quasi-static and free from disspative forces. Quasi-static means that a process takes place so slowly that we can consider the system to be in thermal equilibrium with its surroundings throughout the process. These conditions allow processes to be reversible, which means that both the system and its environment can be returned to their precise states at the beginning of the process. Processes for which these conditions do not apply are called irreversible.
(A) Constant Pressure and Constant Volume
For a gas that expands or contracts during a process while held at constant pressure (sometimes called an isobaric process), the work done by the gas during the process is found to be
W = P(DV) (for constant pressure),
where DV = Vf - Vi is the change in the volume of the gas. The above result can be interpreted graphically as the area under the curve of a pressure versus volume plot (see below). In fact, the area under the curve of a pressure versus volume plot for an expanding (or contracting) gas equals the work done by (or on) the gas for any process, not just processes at constant pressure.
Physlet Illustration: Work Done at Constant Pressure | |||
|---|---|---|---|
|
|
|||
| A piston floats atop a cylinder filled with 0.02 moles of an ideal gas. The cylinder is surrounded by a constant temperature bath. A digital temperature probe, which reads the gas temperature in °C, is attached to the cylinder. A digital pressure sensor, which reads the pressure in kPa, is also attached. Play the animation to heat the temperature bath, and watch the volume and pressure on the accompanying PV plot and in the data table below it. The table also displays the total work done by the gas since the process began. What can you conclude? Start | |||
Hints:
|
At constant volume (an isochoric process), no work is done by the gas during a reversible process. This fact is consistent with the above expression because DV = 0 when the volume is held constant. Also, intuition suggests that there should be no work done because work results from force acting through distance and if the gas does not expand or contract through any distance you would expect the net work done by the gas to be zero. Thus,
W = 0. (for constant volume)
Isothermal Processes
An isothermal process is one that takes place at constant temperature. For an ideal gas, the relationship between the pressure and the volume during an isothermal process is
,
where the constant is NkT. The work done by the gas during an isothermal process can be derived from the ideal-gas equation of state to be
Physlet Illustration: Work Done in an Isothermal Process | |||
|---|---|---|---|
|
|
|||
| A piston sits atop a cylinder filled with 0.04 moles of an ideal gas. The cylinder is surrounded by a constant temperature bath. A digital temperature probe, which reads the gas temperature in °C, is attached to the cylinder. A digital pressure sensor, which reads the pressure in kPa, is also attached. Play the animation to push the piston down into the cylinder, and see how the volume and pressure change on the accompanying PV plot and in the data table below. The table also displays the total work done by the gas since the process began. What can you conclude? Start | |||
Hints:
|
Adiabatic Processes
If no heat flows into or out of a system during a process the process is called adiabatic. This type of process occurs when a system is well insulated or when the process takes place so rapidly that heat doesn't have time to flow. During adiabatic processes, the pressure, volume, and temperature may all change. The pressure versus volume curve for a system undergoing an adiabatic process is called an adiabat.
Example 18.2 Expanding Gas: A monatomic ideal gas consisting of 6.32 moles of atoms expands from a volume of 14.1 m3 to 27.6 m3. How much work is done by the gas if the expansion is (a) at a constant pressure of 133 kPa, and (b) isothermal at T = 303 K?
Picture the Problem The picture shows the gas at its initial volume on the left and its final expanded volume on the right.
Strategy For both part (a) and (b) we can use the above expressions to calculate the work done.
Solution
Part (a)
| 1. Use the expression for the work done at constant pressure: | W = P(DV) = (133 x 103 Pa)(13.5 m3) = 1.80 x 106 J |
Part (b)
| 1. Use the expression for the work done during an isothermal processes: | |
 |
Insight Don't forget that the natural logarithm, ln, is the same as the ordinary logarithm, log, base e.
Practice Quiz
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
A Pearson Company Distance Learning at Prentice Hall Legal Notice |
