Thermodynamics Study Guide (page 3)
The purpose of this lesson is to take this knowledge one step forward and put into law some of the "natural" behaviors of fluids when they exchange heat with their surroundings–the law of thermodynamics. Thermodynamics is a compound word meaning change of heat, in Greek. We will relate heat, work, and internal energy, and then study the laws of thermodynamics. We will exemplify these laws through simple processes that are the foundation of applications such as engines and refrigerators.
From a microscopic point of view, we have learned that particle—atoms, molecules, and ions—are in a continuous motion. They also interact with their neighbors. These facts are measured by two types of energies that can characterize the atomic components of each object, be it in a solid, a liquid, or a gas phase. The two types of energies are called kinetic and potential energies. Kinetic energy measures the translational, rotational, and vibrational motion, whereas the potential energy deals with the interaction between particles. The two energies together constitute the internal energy of the system regardless of its phase (solid, liquid, or gas).
The question we will be asking now is how to change this state of motion and interaction? The answer is through energy exchange. Then, the next step is to realize what forms of energy are appropriate for this goal. One evident answer, from the past lesson, is heat. Heat transferred to or from an object will change the state of the object, its temperature, and its internal energy. Particles will move faster or slower with the input or release of heat (respectively).
But is heat the only way to accomplish the change in internal energy? The answer: With objects, you can always use mechanical interaction, or work. The mechanical parameters characterizing the object are the pressure, P, and the volume, V. Both of them will affect the internal state of the system.
The mechanical parameters depend on each other through an equation called the equation of state. In the special case of the ideal gas, where particles are considered identical and independent of each other, these two parameters are connected through the expression called the ideal gas law:
P · V = n · R· T
Where T is the temperature in Kelvin, n is the number of moles of substance studied, and R is the universal gas constant and is equal to:
R = 8.314472 J. K1 · mol–1
Consider now a thermodynamic system on which we apply mechanical work (for instance, the piston in a car), and assume that there is a gas in the piston. Next, assume that we compress the volume by applying an external force. What we are doing is performing work on the gas; so the gas receives energy, this time in the form of mechanical work. We can use the definition of work we studied a few lessons back and calculate this work by replacing the force with the corresponding constant pressure (force exerted on an area A, as shown in Figure 12.1).
W = F · d = (P · A) · d
But A · d = ΔV is the change of volume for the gas. You can see that because the piston is pushed in, the volume decreases, and then the change in volume is negative. In order to maintain a positive work performed on the system, the previous work equation has to be modified by inlcluding a minus sign.
W = –P · ΔV
The work done on a gas at constant pressure is equal to the product of pressure and the change in volume.
As we can see from the above definition, work is dependent on a mechanical change. Therefore, a certain state has no characteristic work; rather the change in state can be triggered by work being exerted on the system or by the system exerting work on its surroundings.
Positive or Negative Work
The work performed by a thermodynamic system on its surroundings is considered negative (as in the case of an exploding pop bottle). The work performed by the surroundings on the system is considered positive (as in the previous example where the gas is compressed from outside).
Consider a piston like the ones discussed previously with a radius of 8 cm. The pressure inside the pistons is about equal to the atmospheric pressure, and, due to some internal reactions, the system expands by about 5 cm. Find the volume change and the work done by the system on the surrounding.
Start by converting to SI. Next, set up your equation to solve the problem.
r = 8 cm = 0.08 m
ΔL = 5 cm = 0.05 m
P = 1.013 · 105
ΔV = ?
The change in volume, for a cylinder, is the area of the section, a circle, multiplied by the height.
Δ V = A · ΔL
A = π · r2
Δ V = π · r2 · ΔL = π · (0.08 m)2 · 0.05 m
ΔV = 1 · 10–3 m3
W = –P · ΔV = –1.013 · 105 · 1 · 10 –3 m3
W = –1.013 · 102N · m
or with one number of significant figures, as the problem gives us: W = – 1.102 J.
The work is negative because the system makes work on its surrounding, or, in other words, is losing some of its energy.
A geometric approach to interpreting the formula for work offers a second method to measure work performed in a thermodynamic process. Because work is proportional to pressure and volume, we can represent the two parameters and graph their dependence. In the previous case where volume is increasing, the graph will look like the one in Figure 12.2.
The arrow shows us the direction of the process (the volume is increasing; there is work done by the system on its surroundings to push the air out and make room for itself). The definition of work also considers the product of the pressure and the change in volume. According to the geometry of the graph, this is exactly the area darkened between the initial and final volumes and the constant pressure.
The graph is called a PV-diagram, and the work can be calculated by finding the area under the graph in the PV-diagram.
Consider the graph in Figure 12.2. Reading the measurements from the graph, calculate the amount of work performed by the system on the surroundings.
To our advantage, all data is already in SI. The area is a simple geometrical shape: a rectangle. The area under the graph will be:
width · height = (201 · 10–3m3) · 1· 105 N/m2 = 1 · 102 · J
The decision regarding the sign will have to come from interpreting the graph according to the convention: Expanding objects yield a negative value of work. So, our final result is:
W = –1 · 102J
This is exactly the same as the previous answer, with one error. The error stems from the fact that the graph does not give enough detail on the pressure measurements.
Zeroth Law of Thermodynamics
We have seen that pressure and volume are mechanical parameters. There are also other properties of a system that are dependent on the motion of the constituent atoms and molecules. Characterizing this motion is not an easy job though, because the number of particles in anyone material is astronomic. In one
mole of every substance, you can count a certain number of particles; that number is called Avogadro's number, or NA = 6.022 . 1023 particles/mol. Given the number of particles in a substance, N, one can find the number of moles from:
n = N/NA
Also, the mass of the substance can be determined based on the number of particles and the mass per particle:
m = mparticle · N
With this, we can define the parameters necessary to completely characterize a thermodynamic state: pressure, P, volume, V, number of moles, n, and mass, m.
A first law that summarizes one of these parameters, temperature, is the zeroth law of thermodynamics. In all natural phenomena, two or more systems are in thermal contact. They can exchange energy through heat and, after a period of time, will have the same temperature: The colder object warms up and the warmer object cools down. In other words, they will ultimately be in thermal equilibrium. To generalize this behavior, the zeroth law states: Two systems, each in thermal equilibrium with a third system (1 with 2, and 2 with 3), are in thermal equilibrium with each other (l with 3).
Hence, each part of this system has the same temperature. The idea of equilibrium is so important that it was determined that it was necessary for this statement to preceede the first and second laws of thermodynamics, thus the name zeroth. The contact between two or more objects when they exchange heat but no mass is called thermal contact.
Zeroth Law of Thermodynamics
Two systems, each in thermal equilibrium with a third system (1 with 2, and 2 with 3), are in thermal equilibrium with each other (1 with 3).
First Law of Thermodynamics
As one defines the parameters in mechanics that characterize the mechanical state, position, and speed, then the interactions (force and torque) and the energies, we will similarly define parameters for the next concept. We have introduced the thermodynamic parameters previously (pressure, volume, temperature, and moles). We have talked about heat and internal energy exchange between systems. Now it is time to apply the principle of conservation of energy to this subject. The statement of the first law of thermodynamics does exactly this.
Any change in the system is due to exchange of energy with its surroundings. Energy is not a vector quantity, and therefore, we need a convention to establish the sign of these energies because of the different consequences of exchanging energy. For confirmation of this need, read ahead to the next example. At the beginning of this lesson, we talked about the convention for work; now it is time to define a convention for heat.
First Law of Thermodynamics
The first law of thermodynamics, also called conservation of energy, says that the internal energy of a system changes from an initial value Ui to a final value Uf, due to heat exchange Q and work performed on or by the system, W.
ΔU = Uf – Ui = Q + WConvention for Work
- Work is a positive quantity when work is performed on the system.
- Work is a negative quantity when work is performed by the system.
You have two completely identical containers, and the same quantity of liquid is in both of them. You put one of them in contact with a colder object and it loses 100 J of heat. The other one you put in contact with a warm source and it gains 100 J. The energy is the same. Is the final state of the two containers the same?
The state is not similar because the system absorbing heat will have atoms and molecules moving faster: It has larger temperature, while the system that cools down will have slower particles and smaller temperature. Because the thermodynamic state is characterized by P, V, T, and m, even if the rest of the parameters are the same, the temperature is different, so the states are different.
Second Law of Thermodynamics
The second and final law of thermodynamics refers to a "natural" process—the flow of heat. Numerous circumstances in your life have allowed you to experience the essence of this law.
There are other ways to express the meaning of this law. These other expressions require the introduction of machines, such as engines and refrigerators, or concepts, such as entropy. The formulation of these expressions varies with the field of study. So don't be surprised if you open a book geared toward engineering professionals and find the law expressed with engines or if you open a more advanced textbook and find a discussion about entropy. Each and every field of study introduces the law based on previous knowledge of that field of study.
Second Law of Thermodynamics
Heat flows from a substance at a higher temperature to a substance at a lower temperature and does not flow spontaneously in the reverse direction.
Simple Thermodynamic Processes
In nature, rarely is there complete and self-sustained thermal isolation of a system. That means that in most cases, the states of a system have at least one variable parameter, if not more. We will consider simple thermodynamic processes where all but one parameter vary.
Isobar process: This is a process in which pressure is constant (P = constant). In the section on thermodynamic work, we studied an isobar process and its PV-diagram, and we were able to find a simple way to calculate work from a geometrical interpretation. The work in such a process is the area under the PV-diagram.
Isochoric process or isoyolumic process: This is a process in which the volume is constant while the pressure varies, as in Figure 12.5.
At a constant volume:
ΔV = 0
W = – P · ΔV = 0 J
ΔU = Q + W = Q
The internal energy increases (ΔU > 0) if the system absorbs heat, and it decreases (ΔU < 0) if the system loses heat. As you can see in the figure, there is no area under the PV-curve, so there is no work.
Isothermal process: This is the process in which the temperature is constant. For an ideal gas, internal energy can be shown to be proportional with the temperature, hence in this process:
T = constant
ΔU = 0 J
ΔU = Q + W = 0
Q = – W
We interpret this expression as follows: If a system evolves without changing its internal energy, work will be done by the system only if it receives heat, and heat will be released by the system only if work will be done on it.
Adiabatic process: This is a process where no exchange of heat happens. When there is no heat exchange with the surroundings (thermal isolated system), the work can be done only at the expense of the changing internal energy.
If you compare the PV-diagram of the adiabatic and isothermal processes, you might mistakenly consider them identical. Although both of them are curves, in a PV-diagram, the slope of the adiabatic transformation is steeper than the slope of the isotherm at the same volume, as shown in Figure 12.6.
The two isothermals, shown by the symbol in the graph, intersect the adiabatic curve (Δ) evolving between T1 and T2 temperatures and the same volumes V1 and V2.
Let's study the first law:
Q = O J
ΔU= W = –P · ΔV
If the system expands, then the work is negative and the change in internal energy is negative also. This means the energy of the constituent particles and the temperature of the system decreases.
Based on these processes, machines such as heat engines and refrigerators have been constructed. A heat engine is a device that converts internal energy into work by heat flow from a high temperature source to a low temperature source. Because these machines have to go through multiple repetitions of the same process, we call the sequence of process a thermodynamic cycle. A refrigerator uses mechanical work to take heat from a cold source and release it to a hot source, effectively lowering the temperature of a system.
Practice problems of this concept can be found at: Thermodynamics Practice Questions
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