19 The First Law of Thermodynamics - PDF

**19 THE FIRST LAW DF THERMDDYNAMICS** **LEARNING GOALS** By studying this chapter, you will learn: - How to represent heat transfer and work done in a thermodynamic process. - How to calculate the work done by a thermodynamic system when its volume changes. - What is meant by a p...

**19 THE FIRST LAW DF THERMDDYNAMICS** **LEARNING GOALS** By studying this chapter, you will learn: - How to represent heat transfer and work done in a thermodynamic process. - How to calculate the work done by a thermodynamic system when its volume changes. - What is meant by a path between thermodynamic states. - How to use the first law of thermodynamics to relate heat transfer, work done, and internal energy change. - How to distinguish among adiabatic, isochoric, isobaric, and isothermal processes. - How we know that the internal energy of an ideal gas depends only on its temperature. - The difference between molar heat capacities at constant volume and at constant pressure, and how to use these quantities in calculations. - How to analyze adiabatic processes in an ideal gas.\ 19.1 The popcorn in the pot is a thermodynamic system. In the thermodynamic process shown here, heat is added to the system, and the system does work on its surroundings to lift the lid of the pot. A steam locomotive operates using the first law of thermodynamics: Water is heated and boils, and the expanding steam does work to propel the locomotive. Would it be possible for the steam to propel the locomotive by doing work as it condenses? Every time you drive a car, turn on an air conditioner, or cook a meal, you reap the practical benefits of thermodynamics, the study of relationships involving heat, mechanical work, and other aspects of energy and energy transfer. For example, in a car engine heat is generated by the chemical reaction of oxygen and vaporized gasoline in the engine\'s cylinders. The heated gas pushes on the pistons within the cylinders, doing mechanical work that is used to propel the car. This is an example of a thermodynamic process. The first law of thermodynamics, central to the understanding of such processes, is an extension of the principle of conservation of energy. It broadens this principle to include energy exchange by both heat transfer and mechanical work and introduces the concept of the internal energy of a system. Conservation of energy plays a vital role in every area of physical science, and the first law has extremely broad usefulness. To state energy relationships precisely, we need the concept of a thermodynamic system. We\'ll discuss heat and work as two means of transferring energy into or out of such a system. **19.1 Thermodynamic Systems** We have studied energy transfer through mechanical work (Chapter 6) and through heat transfer (Chapters 17 and 18). Now we are ready to combine and generalize these principles. We always talk about energy transfer to or from some specific system. The system might be a mechanical device, a biological organism, or a specified quantity of material, such as the refrigerant in an air conditioner or steam expanding in a turbine. In general, a thermodynamic system is any collection of objects that is convenient to regard as a unit, and that may have the potential to exchange energy with its surroundings. A familiar example is a quantity of popcorn kernels in a pot with a lid. When the pot is placed on a stove, energy is added to the popcorn\ by conduction of heat. As the popcorn pops and expands, it does work as it exerts an upward force on the lid and moves it through a displacement (Fig. 19.1). The state of the popcorn changes in this process, since the volume, temperature, and pressure of the popcorn all change as it pops. A process such as this one, in which there are changes in the state of a thermodynamic system, is called a thermodynamic process. In mechanics we used the concept of system with free-body diagrams and with conservation of energy and momentum. For thermodynamic systems, as for all others, it is essential to define clearly at the start exactly what is and is not included in the system. Only then can we describe unambiguously the energy transfers into and out of that system. For instance, in our popcorn example we defined the system to include the popcorn but not the pot, lid, or stove. Thermodynamics has its roots in many practical problems other than popping popcorn (Fig. 19.2). The gasoline engine in an automobile, the jet engines in an airplane, and the rocket engines in a launch vehicle use the heat of combustion of their fuel to perform mechanical work in propelling the vehicle. Muscle tissue in living organisms metabolizes chemical energy in food and performs mechanical work on the organism\'s surroundings. A steam engine or steam turbine uses the heat of combustion of coal or other fuel to perform mechanical work such as driving an electric generator or pulling a train. **Signs for Heat and Work in Thermodynamics** We describe the energy relationships in any thermodynamic process in terms of the quantity of heat [*Q*]{.math.inline} added to the system and the work [*W*]{.math.inline} done by the system. Both [*Q*]{.math.inline} and [*W*]{.math.inline} may be positive, negative, or zero (Fig. 19.3). A positive value of [*Q*]{.math.inline} represents heat flow into the system, with a corresponding input of energy to it; negative [*Q*]{.math.inline} represents heat flow out of the system. A positive value of [*W*]{.math.inline} represents work done by the system against its surroundings, such as work done by an expanding gas, and hence corresponds to energy leaving the system. Negative [*W*]{.math.inline}, such as work done during compression of a gas in which work is done on the gas by its surroundings, represents energy entering the system. We will use these conventions consistently in the examples in this chapter and the next. CAUTION Be careful with the sign of work [*W*]{.math.inline} Note that our sign rule for work is opposite to the one we used in mechanics, in which we always spoke of the work done by the forces acting on a body. In thermodynamics it is usually more convenient to call [*W*]{.math.inline} the work done by the system so that when a system expands, the pressure, volume change, and work are all positive. Take care to use the sign rules for work and heat consistently! Test Your Understanding of Section 19.1 In Example 17.8 (Section 17.6), what is the sign of [*Q*]{.math.inline} for the coffee? For the aluminum cup? If a block slides along a horizontal surface with friction, what is the sign of [*W*]{.math.inline} for the block? **19.2 Work Done During Volume Changes** A simple but common example of a thermodynamic system is a quantity of gas enclosed in a cylinder with a movable piston. Internal-combustion engines, steam engines, and compressors in refrigerators and air conditioners all use some version of such a system. In the next several sections we will use the gas-incylinder system to explore several kinds of processes involving energy transformations. We\'ll use a microscopic viewpoint, based on the kinetic and potential energies of individual molecules in a material, to develop intuition about thermodynamic quantities. But it is important to understand that the central principles of thermodynamics can be treated in a completely macroscopic way, without reference to microscopic models. Indeed, part of the great power and generality of thermodynamics is that it does not depend on details of the structure of matter.\ 19.2 (a) A rocket engine uses the heat of combustion of its fuel to do work propelling the launch vehicle. (b) Humans and other biological organisms are more complicated systems than we can analyze fully in this book, but the same basic principles of thermodynamics apply to them.\ (a)  (b) 19.3 A thermodynamic system may exchange energy with its surroundings (environment) by means of heat, work, or both. Note the sign conventions for [*Q*]{.math.inline} and [*W*]{.math.inline}.  19.4 A molecule striking a piston (a) does positive work if the piston is moving away from the molecule and (b) does negative work if the piston is moving toward the molecule. Hence a gas does positive work when it expands as in (a) but does negative work when it compresses as in (b). 19.5 The infinitesimal work done by the system during the small expansion [*dx*]{.math.inline} is [*dW* = *pAdx*]{.math.inline}.  First we consider the work done by the system during a volume change. When a gas expands, it pushes outward on its boundary surfaces as they move outward. Hence an expanding gas always does positive work. The same thing is true of any solid or fluid material that expands under pressure, such as the popcorn in Fig. 19.1. We can understand the work done by a gas in a volume change by considering the molecules that make up the gas. When one such molecule collides with a stationary surface, it exerts a momentary force on the wall but does no work because the wall does not move. But if the surface is moving, like a piston in a gasoline engine, the molecule does do work on the surface during the collision. If the piston in Fig. 19.4a moves to the right, so that the volume of the gas increases, the molecules that strike the piston exert a force through a distance and do positive work on the piston. If the piston moves toward the left as in Fig. 19.4b, so the volume of the gas decreases, then positive work is done on the molecule during the collision. Hence the gas molecules do negative work on the piston. Figure 19.5 shows a system whose volume can change (a gas, liquid, or solid) in a cylinder with a movable piston. Suppose that the cylinder has cross-sectional area [*A*]{.math.inline} and that the pressure exerted by the system at the piston face is [*p*]{.math.inline}. The total force [*F*]{.math.inline} exerted by the system on the piston is [*F* = *pA*]{.math.inline}. When the piston moves out an infinitesimal distance [*dx*]{.math.inline}, the work [*dW*]{.math.inline} done by this force is \ [*dW* = *Fdx* = *pAdx*]{.math.display}\ But \ [*Adx* = *dV*]{.math.display}\ where [*dV*]{.math.inline} is the infinitesimal change of volume of the system. Thus we can express the work done by the system in this infinitesimal volume change as \ [\$\$\\begin{matrix} \\mathit{d}\\mathit{W} = \\mathit{p}\\mathit{d}\\mathit{V}\\\#(19.1) \\\\ \\end{matrix}\$\$]{.math.display}\ In a finite change of volume from [*V*~1~]{.math.inline} to [*V*~2~]{.math.inline}, \ [\$\$\\begin{matrix} \\mathit{W} = \\int\_{\\mathit{V}\_{1}}\^{\\mathit{V}\_{2}}\\mspace{2mu}\\mspace{2mu}\\mathit{p}\\mathit{d}\\mathit{V}\\ \\mathrm{\\ (work\\ done\\ in\\ a\\ volume\\ change)\\ }\\\#(19.2) \\\\ \\end{matrix}\$\$]{.math.display}\ In general, the pressure of the system may vary during the volume change. For example, this is the case in the cylinders of an automobile engine as the pistons move back and forth. To evaluate the integral in Eq. (19.2), we have to know how the pressure varies as a function of volume. We can represent this relationship as a graph of [*p*]{.math.inline} as a function of [*V*]{.math.inline} (a [*pV*]{.math.inline}-diagram, described at the end of Section 18.1). Figure 19.6 a shows a simple example. In this figure, Eq. (19.2) is represented\ 19.6 The work done equals the area under the curve on a [*pV*]{.math.inline}-diagram. graphically as the area under the curve of [*p*]{.math.inline} versus [*V*]{.math.inline} between the limits [*V*~1~]{.math.inline} and [*V*~2~]{.math.inline}. (In Section 6.3 we used a similar interpretation of the work done by a force [*F*]{.math.inline} as the area under the curve of [*F*]{.math.inline} versus [*x*]{.math.inline} between the limits [*x*~1~]{.math.inline} and [*x*~2~]{.math.inline}.) According to the rule we stated in Section 19.1, work is positive when a systam expands. In an expansion from state 1 to state 2 in Fig. 19.6a, the area under the curve and the work are positive. A compression from 1 to 2 in Fig. 19.6b gives a negative area; when a system is compressed, its volume decreases and it does negative work on its surroundings (see also Fig. 19.4b). CAUTION Be careful with subscripts 1 and 2 When using Eq. (19.2), always remember that [*V*~1~]{.math.inline} is the initial volume and [*V*~2~]{.math.inline} is the final volume. That\'s why the labels 1 and 2 are reversed in Fig. 19.6b compared to Fig. 19.6a, even though both processes move between the same two thermodynamic states. If the pressure [*p*]{.math.inline} remains constant while the volume changes from [*V*~1~]{.math.inline} to [*V*~2~]{.math.inline} (Fig. 19.6c), the work done by the system is \ [\$\$\\begin{matrix} \\mathit{W} = \\mathit{p}\\left( \\mathit{V}\_{2} - \\mathit{V}\_{1} \\right)\\\#(19.3) \\\\ \\end{matrix}\$\$]{.math.display}\ (work done in a volume change at constant pressure) In any process in which the volume is constant, the system does no work because there is no displacement. **Example 19.1 Isothermal expansion of an ideal gas** As an ideal gas undergoes an isothermal (constant-temperature) expansion at temperature [*T*]{.math.inline}, its volume changes from [*V*~1~]{.math.inline} to [*V*~2~]{.math.inline}. How much work does the gas do? **solution** IDENTIFY and SET UP: The ideal-gas equation, Eq. (18.3), tells us that if the temperature [*T*]{.math.inline} of [*n*]{.math.inline} moles of an ideal gas is constant, the quantity [*pV* = *nRT*]{.math.inline} is also constant: [*p*]{.math.inline} and [*V*]{.math.inline} are inversely related. If [*V*]{.math.inline} changes, [*p*]{.math.inline} changes as well, so we cannot use Eq. (19.3) to calculate the work done. Instead we must use Eq. (19.2). To evaluate the integral in Eq. (19.2) we must know [*p*]{.math.inline} as a function of [*V*]{.math.inline}; for this we use Eq. (18.3).\ EXECUTE: From Eq. (18.3), \ [\$\$\\mathit{p} = \\frac{\\mathit{n}\\mathit{R}\\mathit{T}}{\\mathit{V}}\$\$]{.math.display}\ We substitute this into the integral of Eq. (19.2), take the constant factor [*nRT*]{.math.inline} outside, and evaluate the integral: \ [\$\$\\begin{matrix} \\mathit{W} & \\mathit{\\ } = \\int\_{\\mathit{V}\_{1}}\^{\\mathit{V}\_{2}}\\mspace{2mu}\\mspace{2mu}\\mathit{p}\\mathit{d}\\mathit{V} \\\\ & \\mathit{\\ } = \\mathit{n}\\mathit{R}\\mathit{T}\\int\_{\\mathit{V}\_{1}}\^{\\mathit{V}\_{2}}\\mspace{2mu}\\mspace{2mu}\\frac{\\mathit{d}\\mathit{V}}{\\mathit{V}} = \\mathit{n}\\mathit{R}\\mathit{T}\\ln\\frac{\\mathit{V}\_{2}}{\\mathit{V}\_{1}}\\mathrm{\\ (ideal\\ gas,\\ isothermal\\ process)\\ } \\\\ \\end{matrix}\$\$]{.math.display}\ We can rewrite this expression for [*W*]{.math.inline} in terms of [*p*~1~]{.math.inline} and [*p*~2~]{.math.inline}. Because [*pV* = *nRT*]{.math.inline} is constant, \ [\$\$\\mathit{p}\_{1}\\mathit{V}\_{1} = \\mathit{p}\_{2}\\mathit{V}\_{2}\\ \\mathrm{\\text{\~or\~}}\\ \\frac{\\mathit{V}\_{2}}{\\mathit{V}\_{1}} = \\frac{\\mathit{p}\_{1}}{\\mathit{p}\_{2}}\$\$]{.math.display}\ SO \ [\$\$\\mathit{W} = \\mathit{n}\\mathit{R}\\mathit{T}\\ln\\frac{\\mathit{p}\_{1}}{\\mathit{p}\_{2}}\\ \\mathrm{\\ (ideal\\ gas,\\ isothermal\\ process)\\ }\$\$]{.math.display}\ EVALUATE: We check our result by noting that in an expansion [*V*~2~ \> *V*~1~]{.math.inline} and the ratio [*V*~2~/*V*~1~]{.math.inline} is greater than 1. The logarithm of a number greater than 1 is positive, so [*W* \> 0]{.math.inline}, as it should be. As an additional check, look at our second expression for [*W*]{.math.inline} : In an isothermal expansion the volume increases and the pressure drops, so [*p*~2~ \ 1]{.math.inline}, and [*W* = *nRT*ln (*p*~1~/*p*~2~)]{.math.inline} is again positive. These results also apply to an isothermal compression of a gas, for which [*V*~2~ \ *p*~1~]{.math.inline}. Test Your Understanding of Section 19.2 A quantity of ideal gas undergoes an expansion that increases its volume from [*V*~1~]{.math.inline} to [*V*~2~ = 2*V*~1~]{.math.inline}. The final pressure of the gas is [*p*~2~]{.math.inline}. Does the gas do more work on its surroundings if the expansion is at constant pressure or at constant temperature? (i) constant pressure; (ii) constant temperature; (iii) the same amount of work is done in both cases; (iv) not enough information is given to decide.\ 19.7 The work done by a system during a transition between two states depends on the path chosen.\ (a)  (c) (d)  **19.3 Paths Between Thermodynamic States** We\'ve seen that if a thermodynamic process involves a change in volume, the system undergoing the process does work (either positive or negative) on its surroundings. Heat also flows into or out of the system during the process if there is a temperature difference between the system and its surroundings. Let\'s now examine how the work done by and the heat added to the system during a thermodynamic process depend on the details of how the process takes place. **Work Done in a Thermodynamic Process** When a thermodynamic system changes from an initial state to a final state, it passes through a series of intermediate states. We call this series of states a path. There are always infinitely many different possibilities for these intermediate states. When they are all equilibrium states, the path can be plotted on a [*pV*]{.math.inline}-diagram (Fig. 19.7a). Point 1 represents an initial state with pressure [*p*~1~]{.math.inline} and volume [*V*~1~]{.math.inline}, and point 2 represents a final state with pressure [*p*~2~]{.math.inline} and volume [*V*~2~]{.math.inline}. To pass from state 1 to state 2 , we could keep the pressure constant at [*p*~1~]{.math.inline} while the system expands to volume [*V*~2~]{.math.inline} (point 3 in Fig. 19.7b), then reduce the pressure to [*p*~2~]{.math.inline} (probably by decreasing the temperature) while keeping the volume constant at [*V*~2~]{.math.inline} (to point 2 on the diagram). The work done by the system during this process is the area under the line [1 → 3]{.math.inline}; no work is done during the constantvolume process [3 → 2]{.math.inline}. Or the system might traverse the path [1 → 4 → 2]{.math.inline} (Fig. 19.7 c ); in that case the work is the area under the line [4 → 2]{.math.inline}, since no work is done during the constant-volume process [1 → 4]{.math.inline}. The smooth curve from 1 to 2 is another possibility (Fig. 19.7d), and the work for this path is different from that for either of the other paths. We conclude that the work done by the system depends not only on the initial and final states, but also on the intermediate states-that is, on the path. Furthermore, we can take the system through a series of states forming a closed loop, such as [1 → 3 → 2 → 4 → 1]{.math.inline}. In this case the final state is the same as the initial state, but the total work done by the system is not zero. (In fact, it is represented on the graph by the area enclosed by the loop; can you prove that? See Exercise 19.7.) It follows that it doesn\'t make sense to talk about the amount of work contained in a system. In a particular state, a system may have definite values of the state coordinates [*p*, *V*]{.math.inline}, and [*T*]{.math.inline}, but it wouldn\'t make sense to say that it has a definite value of [*W*]{.math.inline}. **Heat Added in a Thermodynamic Process** Like work, the heat added to a thermodynamic system when it undergoes a change of state depends on the path from the initial state to the final state. Here\'s an example. Suppose we want to change the volume of a certain quantity of an ideal gas from 2.0 L to 5.0 L while keeping the temperature constant at [*T* = 300 *K*]{.math.inline}. Figure 19.8 shows two different ways in which we can do this. In Fig. 19.8a the gas is contained in a cylinder with a piston, with an initial volume of 2.0 L. We let the gas expand slowly, supplying heat from the electric heater to keep the temperature at 300 K. After expanding in this slow, controlled, isothermal manner, the gas reaches its final volume of 5.0 L ; it absorbs a definite amount of heat in the process. Figure 19.8 b shows a different process leading to the same final state. The container is surrounded by insulating walls and is divided by a thin, breakable partition into two compartments. The lower part has volume 2.0 L and the upper part has volume 3.0 L. In the lower compartment we place the same amount of the same gas as in Fig. 19.8a, again at [*T* = 300 *K*]{.math.inline}. The initial state is the same as before. Now we break the partition; the gas undergoes a rapid, uncontrolled expansion, with no heat passing through the insulating walls. The final volume is 5.0 L , the same as in Fig. 19.8a. The gas does no work during this expansion\ because it doesn\'t push against anything that moves. This uncontrolled expansion of a gas into vacuum is called a free expansion; we will discuss it further in Section 19.6. Experiments have shown that when an ideal gas undergoes a free expansion, there is no temperature change. Therefore the final state of the gas is the same as in Fig. 19.8a. The intermediate states (pressures and volumes) during the transition from state 1 to state 2 are entirely different in the two cases; Figs. 19.8a and 19.8b represent two different paths connecting the same states 1 and 2. For the path in Fig. 19.8b, no heat is transferred into the system, and the system does no work. Like work, heat depends not only on the initial and final states but also on the path. Because of this path dependence, it would not make sense to say that a system \"contains\" a certain quantity of heat. To see this, suppose we assign an arbitrary value to the \"heat in a body\" in some reference state. Then presumably the \"heat in the body\" in some other state would equal the heat in the reference state plus the heat added when the body goes to the second state. But that\'s ambiguous, as we have just seen; the heat added depends on the path we take from the reference state to the second state. We are forced to conclude that there is no consistent way to define \"heat in a body\"; it is not a useful concept. While it doesn\'t make sense to talk about \"work in a body\" or \"heat in a body,\" it does make sense to speak of the amount of internal energy in a body. This important concept is our next topic. Test Your Understanding of Section 19.3 The system described in Fig. 19.7a undergoes four different thermodynamic processes. Each process is represented in a [*pV*]{.math.inline}-diagram as a straight line from the initial state to the final state. (These processes are different from those shown in the [*pV*]{.math.inline}-diagrams of Fig. 19.7.) Rank the processes in order of the amount of work done by the system, from the most positive to the most negative. (i) [1 → 2]{.math.inline}; (ii) [2 → 1]{.math.inline}; (iii) [3 → 4]{.math.inline}; (iv) [4 → 3]{.math.inline}. **19.4 Internal Energy and the First Law of Thermodynamics** Internal energy is one of the most important concepts in thermodynamics. In Section 7.3, when we discussed energy changes for a body sliding with friction, we stated that warming a body increased its internal energy and that cooling the body decreased its internal energy. But what is internal energy? We can look at it in various ways; let\'s start with one based on the ideas of mechanics. Matter consists of atoms and molecules, and these are made up of particles having kinetic and potential energies. We tentatively define the internal energy of a system as the sum of the kinetic energies of all of its constituent particles, plus the sum of all the potential energies of interaction among these particles. CAUTION Is it internal? Note that internal energy does not include potential energy arising from the interaction between the system and its surroundings. If the system is a glass of water, placing it on a high shelf increases the gravitational potential energy arising from the interaction between the glass and the earth. But this has no effect on the interaction between the molecules of the water, and so the internal energy of the water does not change. We use the symbol [*U*]{.math.inline} for internal energy. (We used this same symbol in our study of mechanics to represent potential energy. You may have to remind yourself occasionally that [*U*]{.math.inline} has a different meaning in thermodynamics.) During a change of state of the system, the internal energy may change from an initial value [*U*~1~]{.math.inline} to a final value [*U*~2~]{.math.inline}. We denote the change in internal energy as [*ΔU* = *U*~2~ − *U*~1~]{.math.inline}.\ 19.8 (a) Slow, controlled isothermal expansion of a gas from an initial state 1 to a final state 2 with the same temperature but lower pressure. (b) Rapid, uncontrolled expansion of the same gas starting at the same state 1 and ending at the same state 2.\ (a) System does work on piston; hot plate adds heat to system [(*W* \> 0]{.math.inline} and [*Q* \> 0)]{.math.inline}. \(b) System does no work; no heat enters or leaves system ( [*W* = 0]{.math.inline} and [*Q* = 0]{.math.inline} ).  **Mastering PHYSI®s** ActivPhysics 8.6: Heat, Internal Energy, and First Law of Thermodynamics\ 19.9 In a thermodynamic process, the internal energy [*U*]{.math.inline} of a system may (a) increase [(*ΔU* \> 0)]{.math.inline}, (b) decrease ( [*ΔU* \ 0]{.math.inline}. Your body also warms up during exercise; by perspiration and other means the body rids itself of this heat, so [*Q* \