The fi rst law of thermodynamics states that while energy can be changed from one form to another, the total quantity of energy, E, in the universe is constant.1 This statement can be quantitatively expressed as follows:
DEuniv50 (2.1)
However, it is very inconvenient to consider the entire universe every time we need to do a calculation. As we have seen, we can break down the universe into the region in which we are interested (the system) and the rest of the universe (the surroundings). The system is separated from the surroundings by its boundary. We can now restate the fi rst law by saying that the energy change of the system must be equal to the energy transferred across its boundaries from the surroundings. Energy can be transferred by heat, Q, by work, W, and, in the case of open systems, by the energy associated with the mass that fl ows into and out of the system. In essence, the fi rst law then lets us be accountants of the energy in the system, by tracking the “deposits” to and “withdrawals” from the surroundings in much the same way as you would account for the balance of money in your bank account.
We will consider explicit forms of the fi rst law for closed and open systems shortly.
► 2.1 THE FIRST LAW OF THERMODYNAMICS
Forms of Energy
The energy within a system can be transformed from one form to another.
►EXERCISE Name the three common forms that energy is divided into. See if you can defi ne each form:
Energy is classifi ed according to three specifi c forms: (1) The macroscopic kinetic energy, EK is the energy associated with the bulk (macroscopic) motion of the system as a whole. For example, an object of mass m moving at velocity VS has a kinetic energy given by:
EK51
2mVS2 (2.2)
(2) The macroscopic potential energy, EP, is the energy associated with the bulk (macro- scopic) position of the system in a potential fi eld. For example, an object in the Earth’s gravitational fi eld has a potential energy given by:
EP5mgz (2.3)
where z is the height above the surface of the Earth and g is the gravitational con- stant.2 (3) The internal energy, U, is the energy associated with the motion, position, and chemical-bonding confi guration of the individual molecules of the substances within the system.
Energy is not an absolute quantity but rather is only defi ned relative to a reference state, so we must be careful to identify the particular reference state that we are using.
As you read this text, what is your kinetic energy (assuming you are not riding the bus)?
1 Nuclear reaction presents an interesting case where energy and mass are coupled. However, we will not address this case in this text.
2 Potentials due to surface tension or electric or magnetic fi elds can also be included.
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38 ► Chapter 2. The First Law of Thermodynamics
If you answer zero, you are correct in the context of a well-defi ned reference state, the Earth. However, if instead we had considered the sun to be the reference state, the answer would be quite different. The Earth is in motion with a velocity of 30,000 m/s around the sun, and your kinetic energy is on the order of 106 J! In the case of kinetic and potential energy, we usually defi ne EK (i.e., VS ) as zero when there is no motion relative to the Earth and EP (i.e., z) as zero at the surface of the Earth. In fact, these reference states are so obvious they are sometimes implicitly assumed. In this text, we will be care- ful to identify references states explicitly. This effort will become useful in the case of U, where there is more than one convenient reference state. What is the reference state used for U in the steam tables?
In your introductory physics course, you focused primarily on changes associated with the fi rst two forms of energy. Since solving fi rst-law problems involves profi ciency in relating different forms of energy, it is instructive to review a typical example from mechanics that you may have seen in introductory physics. It is presented in the context in which we will approach problems in this text. You should realize that, as chemical engineers, the form of energy that we will primarily focus on is internal energy, which is not covered in the following example.
If a large stone is dropped from a cliff 10 m high, how fast will it be going when it hits the ground?
SOLUTION From Equation (2.1), we have:
DE5 DEK1 DEP50 (E2.1A)
We can defi ne the process as throwing the stone off the cliff. Typically, we set up our problem by labeling the thermodynamic states between which our process is going. We can defi ne state 1 as the initial state when the stone is at the top of the cliff and state 2 as when the stone hits the ground. Using Equations (2.2) and (2.3), Equation (E2.1A) becomes:
a1 2mVS22
21 2mVS12
b1 1mgz22mgz1250
By convention, we defi ne the change in a property, D, as “fi nal − initial.” Now, using the reference states described above, we get:
a1 2mVS22
2 1 2mVS12
b 1 1mgz22mgz12 50 or,
1
2mVS222mgz150 Finally, solving for the fi nal velocity yields:
VS25"2gz15"219.83m/s24 2 1103m4 25143m/s4
This value is equivalent to 31 miles/hr. Ouch! Note that our reference state of energy is arbitrary and if we had chosen different reference states, we would still get the same answer.
EXAMPLE 2.1 Typical Energy Problem in Mechanics
0 0
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2.1 The First Law of Thermodynamics ◄ 39
As we mentioned, internal energy, U, is an important form of energy for chemical engi- neering applications. Originally internal energy was viewed simply as all forms of energy that are not associated with bulk motion or bulk position. However, it is instructive to take advantage of our knowledge of chemistry and consider a molecular perspective on internal energy. Internal energy encompasses all forms of molecular energy, includ- ing the kinetic and potential energies of the molecules themselves. A change in inter- nal energy can present itself in several macroscopic manifestations; that is, molecular energy is noticed in the real world in different ways. Changes in internal energy can result in the following:
1. changes in temperature, for example, Tlow h Thigh
2. changes in phase, for example, solid h gas
3. changes in chemical structure, that is, chemical reaction 1N213H2 h 2NH32 A change in internal energy that leads to a change in temperature is often termed sen- sible heat. Likewise, we often refer to a change in internal energy that results in phase transformations as latent heat.
Let’s now examine how we can relate changes in molecular, chemical energy to the three macroscopic features described above. There are two general components of inter- nal energy—molecular potential energy and molecular kinetic energy. Molecular poten- tial energy can be either intermolecular (between different molecules) and intramolecular (within the same molecule) in character. Remember that we use the term molecular when we are describing what the species are doing on the atomic scale while we use the term macroscopic to describe behavior in the bulk (or molar) scale of the world in which we live.
Like macroscopic kinetic energy, by molecular kinetic energy we mean motion; in this case, the motion of the individual molecules in a system. The type of motion depends on the phase the species are in. In the gas phase, for example, the molecules are fl ying around at signifi cant speeds. This motion is referred to as translational motion because the individ- ual molecules are going somewhere—that is, translating—even though the bulk of the gas may not be. Not all molecules have the same speed, but, at equilibrium, their speeds vary according to a Maxwell–Boltzmann distribution. Do you know how fast the average oxy- gen molecule that you are now breathing is moving? [ANSWER: Oxygen molecules at room temperature are moving, on average, as fast as a jet plane, that is approximately 450 m/s.]
Additionally, diatomic and polyatomic molecules (as opposed to atoms) can vibrate and rotate, which provide additional modes of molecular kinetic energy—vibrational and rotational motion. As we saw in Chapter 1, the measured macroscopic property
One philosophical comment: Energy is inherently a very abstract quantity; it is very hard to say exactly what energy is. However, you (hopefully!) are comfortable using it in the context of the problem above. The ability to apply the abstract property energy to solve engineering problems lies in your experience with it. This experience typically translates into relative comfort in including internal energy in the energy balance to solve fi rst-law problems. Keep in mind that soon we will introduce thermodynamic properties with which you have less experience, such as entropy, S, and Gibbs energy, G. These properties are fundamentally no more challenging than energy to learn to work with; however, you may have an initial period of discomfort as you gain experience. The trick is that when you work with any thermodynamic property enough, you get used to it and become profi cient at solving the type of problems for which it is useful.
Ways We Observe Changes in U
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40 ► Chapter 2. The First Law of Thermodynamics
temperature is representative of how fast the gas molecules are moving in the system (formally temperature is proportional to the mean-square velocity). However, the mol- ecules’ speed is directly related to the molecular kinetic energy, which, in turn, is one part of the internal energy. Hence, as a gas increases in temperature, the average velocity of its molecules increases (the molecules move faster), so it has greater internal energy (see point 1 above). In contrast, solids do not have translational motion; their main mode of molecular kinetic energy is in the form of vibrations. The vibrations of the atoms in a solid are called phonons. Again, on the one hand, phonons represent part of the internal (or molecular) energy, and, on the other hand, they are directly related to the tempera- ture of the solid. So the faster the solid is vibrating, the greater the temperature and the greater the internal energy.
Consider next a phase change, such as the sublimation of a solid into a vapor. An example with which you may have experience is the case with CO2 (dry ice) at atmospheric temperature and pressure. The solid is held together by bonds between the molecules.
Often the bonding in a solid results from electrostatic attraction of the molecules, that is, molecular potential energy. The attraction between molecules adds stability and decreases the molecular energy of the system. On the other hand, the molecules in the vapor are much farther away from one another and have little or no attraction. Thus, the vapor phase is representative of higher internal energy relative to the solid at the same temperature.
To sublimate, the molecular energy of the attraction of the bonds must be overcome; that is, energy must be added to the system, resulting in higher internal energy. Similar argu- ments hold for the melting of solids and evaporation of liquids (see point 2 above).
Finally, consider chemical reaction. In this case, the chemical bonds between the atoms in the molecules of the reactants are broken and replaced by the bonds of the products. For example, ammonia is produced by the reaction:
N213H2 h 2NH3
A triple bond between N atoms and three single bonds between H atoms are replaced by six NiH bonds (three each for two molecules produced). The strength of a chemi- cal bond is determined by the overlap of the valence electrons of the constituent atoms.
Thus the energetics change as the atoms are rearranged, resulting in a change in the thermodynamic property, U. In this case, the product is lower in energy (a lot more sta- ble), so U is reduced (see point 3 above).3
Internal Energy of an Ideal Gas
We next explore the property dependence for the internal energy for an ideal gas. As we learned with the discussion of point 1 earlier, internal energy consists of two com- ponents, molecular kinetic energy and molecular potential energy, and temperature is directly related to one of them, the molecular kinetic energy. Because an ideal gas exhib- its no intermolecular forces (see Section 1.3), its molecular potential energy is constant (assuming no chemical reactions). Therefore, barring chemical reaction, the internal energy, u, depends only on motion of the molecules, or the temperature. Hence,
uideal gas5f1T only2 (2.4)
Said another way, the internal energy for an ideal gas is independent of the position of the molecules. In Chapter 4, we will consider the thermodynamic properties of real
3 In Chapter 3, we will learn that thermodynamic entropy also plays a role in determining how far a reaction will proceed.
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gases, where the molecules are close enough to experience the effect of intermolecular forces. In this case, two independent intensive properties are needed to specify the state of a system with constant composition, such as:
ureal gas5u1T, v2 or,
ureal gas5u1T, P2
EXAMPLE 2.2 Equivalent Energy Stored in u
In Example 2.1, we considered the potential energy of a stone at the top of a 10-m cliff. When it fell, it gained kinetic energy, resulting in a velocity around 31 miles/hr. Consider now an equivalent mass of water initially at 25ºC. How hot would the water end up if its internal energy increased by the same amount?
SOLUTION If we write Equation (2.2) on a per-mass basis, we have:
De^K51
2VS22598 3J/kg450.098 3kJ/kg4
We have used specifi c energy and converted the units to be consistent with the steam tables.
Again we will use state 1 to denote the initial state and state 2 to denote the fi nal state. Liquid water is subcooled at 25ºC and 1 atm; however, we do not expect the properties of a liquid to be signifi cantly affected by pressure. Therefore, we can use the temperature tables for saturated liquid water at 25ºC (which is technically at a pressure of 0.03 atm).4 From Appendix B.1:
u^l,15104.86 3kJ/kg4
The problem statement says the internal energy of the water increases by the same amount as the energy of the stone, that is,
Du^ 5u^l,22u^l,15 De^k50.098 3kJ/kg4 So for the fi nal state of water, we have:
u^l,25 Du^ 1u^l,15104.96 3J/kg4
We can now go to the steam tables and determine at which temperature saturated water has this energy. Again we neglect the effect of the pressure difference between the subcooled state and the saturated state. Interpolating, we get:
u^l,2 1at T22 2u^1 1at 253°C4 2
u^l 1at 303°C4 22u^l 1at 253°C4 2 5104.962104.86 125.772104.8650.005 Finally, solving for the fi nal temperature yields:
T2525110.00525525.02°C
The temperature of the water barely changes! Thus the energy stored in a stone 10 m up a cliff corresponds to a negligible amount of internal energy. This example illustrates that a large amount of energy is stored in u relative to the other forms of energy, and, consequently, why we are so interested in internal energy. As engineers, it provides us a large resource to be harvested.
2.1 The First Law of Thermodynamics ◄ 41
4 We often use this trick to fi nd the properties of subcooled water (or other substances) when the pressure is not appreciably different from the saturation pressure. It will serve you well to catalog your experiences of such tricks for reference for solving problems in the future.
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42 ► Chapter 2. The First Law of Thermodynamics
Work and Heat: Transfer of Energy Between the System and the Surroundings
In the sciences we need to be very careful about how we use language and defi ne terms such as work and heat. Both terms refer to the transfer of energy between the sur- roundings and system. In a closed system, the transfer of energy between the surround- ings and the system can only be accomplished by heat or by work. Heat is the transfer of energy by a temperature gradient, whereas all other forms of energy transfer in a closed system occur via work. We generally associate work with something useful being done by (or to) the system. We will examine these terms in more detail below.
Work
There are many forms of work, for example, mechanical (expansion/compression, rotat- ing shaft), electrical, and magnetic. The most common case of work in engineering ther- modynamics is when a force causes a displacement in the boundary of a system. In the case of expansion, for example, the system needs to push the surroundings out of the way to increase the boundary; in this process, the system expends energy. Thus, the sys- tem exchanges energy with the surroundings in the form of work. The work, W, can be described mathematically by the line integral of the external force, FE, with respect to the direction of displacement, dx:
W53FE#dx (2.5)
In contrast to thermodynamic properties, the work on a system depends not only on the initial state, 1, and the fi nal state, 2, of the system, but also on the specifi c path that it takes.
Whenever we calculate the work, we must account for the real path that the system takes.
Since work refers to the transfer of energy between the system and the surround- ings, it has the same units as energy, such as joules, ergs, BTU, and so on. To complete the defi nition, we need to choose a sign convention for work. In this text, we will say work is positive when energy is transferred from the surroundings to the system and work is negative when energy is transferred from the system to the surroundings. The defi nition given by Equation (2.5) is consistent with this sign convention. You should be aware that this sign convention is arbitrary. We choose this convention to be consistent with today’s convention. However, when the fi rst and second laws of thermodynamics were originally formulated, in the context of powering the steam engine, the opposite sign convention was used: Work from the system to the surroundings was defi ned as positive (since the engineering objective was to get work out of the system to power a train!). When you go to other sources, be careful to note which sign convention is chosen for work or you may get tripped up.
A plot of FE vs. x for a general process is shown in Figure 2.1a. The work associated with the process in Figure 2.1a can be obtained from the area under the curve [which is equivalent to graphically integrating the expression in Equation (2.5)]. If the boundary of the system does not move, no work has been done, no matter how large the force is.
If the external force is acting on a surface of cross-sectional area A, we can divide and multiply the terms on the right hand side of Equation (2.5) by A as follows:
W53FE
A #d1Ax2 53PE#dV53PEdV cos u 5 23PEdV (2.6) where PE is the external pressure to the surface. The negative sign in Equation (2.6) results, since the external force and displacement vectors are in opposite directions.
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