16 Chapter 12 - Doane Universityphysics.doane.edu/hpp/Resources/Fuller3/pdf/F3Chapter_12.pdf ·...

Physics Including Human Applications

253

Chapter 12 Thermodynamics GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition:

PV diagram efficiency of a heat engine isochoric process Carnot cycle isobaric process refrigerator isothermal process coefficient of performance of a refrigerator adiabatic process heat engine

Laws of Thermodynamics State three laws of thermodynamics, and explain the operation of a physical system in terms of these laws. Thermodynamics Problems Solve problems consistent with the laws of thermodynamics. PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 5, Energy, and Chapter 10, Temperature and Heat.


254

Chapter 12 Thermodynamics

12.1 Introduction Much of your travel is made possible by a device that we call a heat engine. We can think of a heat engine as being any device which operates by heat energy input, does work, and has a heat energy output. The human body is a heat engine. The refrigerator in your home is a type of heat engine. In this chapter you will be introduced to the fundamental principles of these and other forms of heat engines. In Chapter 10, it was indicated that when the temperature of an object is changed many different effects may take place - such as change in length or volume, change in resistance, change in pressure, change in electromotive force, change in color, and so on. In such processes there is usually a transfer of heat energy and also a performance of work, a force acting through a displacement. The study of the phenomena that result from energy changes produced by a transfer of heat and performance of work is called thermodynamics. These phenomena involve heat energy, temperature, and work. Thermodynamics is the study of the laws that govern thermal processes. When you have completed this study of thermodynamics, you will be able to explain the significance of these laws as they apply to both living and nonliving systems. Thermodynamics concerns itself with a well-defined system, which interacts directly with its surroundings, and by this interaction performs some useful function. Thermodynamics is not concerned with internal effects by themselves. 12.2 The Zeroth Law of Thermodynamics You know that when two equal amounts of ice are added to two identical containers of hot tea to make iced tea, after the ice has melted both containers of hot tea will be at the same final temperature. This is an example of the zeroth law of thermodynamics. The zeroth law of thermodynamics can be stated as follows: If system A is in thermal equilibrium with system B, and system B is in thermal equilibrium with system C, then A, B, and C are in equilibrium with each other, and they are at the same temperature. Thermal equilibrium is defined as the condition in which there is no net energy exchange between the systems in equilibrium. The zeroth law is seemingly trivial, but it is the basis of thermometers since it defines operationally temperature as the property of a system that determines its thermal equilibrium with another system. The zeroth law of thermodynamics clearly points out that temperature, as measured by thermometers, involves systems in equilibrium; that is, the temperature being measured by the thermometer in the system must be constant in time, indicating equilibrium, before a measurement is made. In this connection it is important to point out that the thermal inertia of the thermometer determines the time required for the thermometer to reach equilibrium with the system, as well as the effect of the thermometer on the system being monitored. The larger the thermal inertia (mass x specific heat of the thermometer), the longer the response time, and the more energy transferred to or from


255

the thermometer in attaining equilibrium. In this chapter we will use the symbol T to refer to absolute temperature in degrees Kelvin (0øC = 273øK). 12.3 External Work Consider as a system a cylinder containing a working substance (a gas) and closed by a movable piston (Figure 12.1). If the gas pushes the piston out, work is done by the system. If an outside agent pushes in the piston, work is done on the system. In either case the work is called external work. Work that is done by one part of the system on another part of the same system is called internal work. The study of internal work as such is not part of thermodynamics.

Returning to our example system, the cylinder containing gas enclosed by a movable piston, assume the piston has an area A, and a pressure P is acting on the piston. The total force acting on the piston is given by the product of the pressure times area, PA. If the piston is displaced a small distance Δx, then work is done, and the amount of work ΔW is given by the product of the force times the displacement, ΔW = (force)(displacement) = (PA) (Δx) (12.1) We notice that the product of area times displacement, A Δx, is equal to the change in the volume ΔV of the system as the piston moves. Then we can express the amount of work done by the system when it pushes the piston out a small distance Δx by the following equation ΔW = P ΔV (12.2) If the pressure remains constant over a whole series of small displacements, then the total amount of work done is given by the pressure times the change in volume. W = P (V2 -V1) (12.3) where the difference between the final and initial volumes (V2 -V1) is the change in volume. If the pressure is not constant, we have to use calculus methods, as we discuss in the Enrichment section of this chapter, Section 12.15. EXAMPLE During normal breathing the volume of the lungs increases by about 500 ml in each inspiration. What is the amount of work done on the lungs during inspiration? The standard pressure of the atmosphere is about 1.00 x 10 5 N/m2. W = (1.00 x 105 N/m2) (500 ml x 10-6 m3/ 1 ml) W = 50 J We can interpret the meaning of Equation 12.3 by studying a PV diagram, the graph of the pressure versus the volume of a confined gas (Figure 12.2). If we allow the system to expand, the pressure of the gas will decrease as the volume increased from Vi to Vf.


256

The amount of work done by the gas for a small displacement Δx is given by Equation 12.2 and is shown as the shaded area in the figure since PΔV is equal to the area of the shaded portion of the graph. This area on the graph has the units of joules.

If we add the work done by many of these small displacements as the system expands from V1 to V2, we find that the work done by the gas is represented by the area under the curve from C to D between the vertical lines at V1 and V2 (shown as the large shaded portion of the graph). If we allow the gas to expand, the system goes from C to D, the gas pushes the piston out, and we say the system does work. By convention we call this positive work. If we compress the gas so that the volume of the system goes from V2 to V1 (the pressure changes from D to C), we say work is done on the system. We designate this as negative work. Let us consider a cyclic system as shown in Figure 12.3. First, the working substance expands from V1 to V2 in accordance with line CD. The work done is positive and represented by the area between the curve CD and the volume axis. Then the working substance is compressed from V2 to V1 in a process represented by the curve DEC. The work of compression is represented by the area under DEC, DV2V1CE, and is negative. For this cycle, the area under CD is greater than the area under DEC. The total amount of work done in one cycle is equal to the enclosed area between CD and DEC. This area represents the net work that the system does in one cycle. It is evident that the amount of net work done per cycle depends upon the path that is followed from C to D and back again. That is, it depends upon the processes used during the expansion and compression portions of the cycle. 12.4 The First Law of Thermodynamics You have probably had the experience of doing mechanical work on a system and thereby changing the temperature of the system. When rapidly pumping up the tires of a bicycle with a hand pump, you may have noticed that both the piston of the pump and the tire become hot. The quantitative relationship between the work done of a system and its thermodynamic properties is expressed by the first law of thermodynamics.


257

The first law of thermodynamics which is a formulation of the conservation of energy states that the change in the internal energy of a system is equal to the heat added to the system minus the work done by the system. In equation form this becomes ΔU = ΔQ - ΔW (12.4) or ΔQ = ΔU + ΔW or ΔW = ΔQ - ΔU where ΔU is the change in internal energy, ΔQ is the change in heat energy, and ΔW is the positive work done by the system. Let us look at this equation carefully. The internal energy of a system depends only on the state of the system. For this reason it is called a state function. A state function is dependent only on the variables defining the state of the system such as the pressure, temperature, and volume for an ideal gas. A state function is independent of the process involved in getting the system to a given state. When a system undergoes a change and moves from one thermodynamic state to another, the change in the internal energy of the system will depend only on the final and initial states. The change in internal energy will be the same regardless of the processes involved in changing states. In Chapter 5 we had an analogous case in which the work, or energy change, was independent of path and depended only upon the beginning and ending points. In that situation we defined the system as conservative, and we said we were dealing only with conservative forces. In the first law of thermodynamics the change in internal energy depends only upon the change in absolute temperature. The heat and work terms in Equation 12.4 are not state functions; they are dependent on the process involved in the change of state. For example, heat added at constant pressure produces a different change than the same amount of heat added at a constant volume as we shall see later in this chapter. Likewise, the work done on the system depends on the process involved. For example, work at constant temperature produces a different effect than the same amount of work applied at constant pressure. These terms are parallel to work against nonconservative force as discussed in Chapter 5. The importance of the first law of thermodynamics is that the sum of the work done and the heat added in moving between two different thermodynamic states always produces the same change in the internal energy of the system. The work done on the system may be mechanical, electrical, or chemical. All three of these forms of work occur in living systems. EXAMPLE Consider a diet problem. A person undertakes a diet program that provides 2000 kcal per day, and the person expends energy in all forms to a total of 3000 kcal per day. In order to do this much work on this diet, the person must obtain energy stored as internal energy in the body. Loss of body fat will occur as this energy is used. Using the first law of thermodynamics, we see that the heat energy input is represented by the


258

2000 kcal (ΔQ = 2000 kcal) from the food when burned in the body. The body does work amounting to 3000 kcal (ΔW = + 3000 kcal). This work takes many forms, mechanical work in terms of body motion, chemical changes in the muscles, and electrical energy in nerve activity. From the first law it follows that the internal energy must decrease by 1000 kcal per day, ΔU = ΔQ -ΔW = 2000 kcal - 3000 kcal = - 1000 kcal. If this energy were stored as adipose (fat) tissue (7500 kcal/kg), then the person would lose about 1 kg in a week. 12.5 Isochoric Processes An isochoric process is one that takes place at a constant volume. An isochoric process is represented on a PV diagram by a vertical line (Figure 12.4). You see that the area under the curve is 0. We know that the pressure times the change in volume is the work done. The volume does not change, so no work is done: ΔW = PΔV

where ΔV = 0. Therefore ΔW = 0. For an isochoric process the first law of thermodynamics reduces to an equality between the change in internal energy and the change in heat energy. ΔU = ΔQ (12.5) If the heat is added to the system, the internal energy is increased, and the temperature rises. If heat is given up by the system, there is a decrease in internal energy and a decrease in the temperature. For a confined ideal gas the heat added to the gas during an isochoric process is given by the product of the number of moles of gas, the specific heat of the gas at constant volume, and the change in temperature, ΔU = ΔQ = ncv ΔT (12.6) where n is the number of moles, (See footnote 1), cv, the molar specific heat at constant volume, is the amount of heat required to raise the temperature of one mole of gas one degree during an isochoric process see Table 12.1 , and ΔT is the change in temperature.


259

EXAMPLE Find the amount of heat to be added to raise the pressure of 1.50 moles of helium from 75.0 cm Hg to 100 cm Hg at constant volume. How much is the temperature raised? The original temperature is 27.0øC. For a gas P1V1/T1 = P2V2/T2 For this case, V1 = V2, so P1/T1 =P2/T2 75.0 cm hg/300øK = 100 cm Hg/T2 T2 = 400øK temperature increase = 100øK ΔQ = ncv ΔT cv = 3.00 cal /mole-øK ΔQ = (1.50 moles)(3.00 cal/mole-øK)(100øK) = 450 cal 12.6 Isobaric Processes An isobaric process is one that takes place at a constant pressure and can be represented by a horizontal line in a PV diagram (Figure 12.5). The area under the curve CD is equal to the product of the pressure times the change in volume, PΔV, and represents the work that is done during the isobaric process.

Let us consider a system that is a confined ideal gas. In Section 10.9 on the gas laws, we learned that if the pressure is constant and the volume changes, there is also a change in temperature. A change in temperature means that there has been a change in internal energy. Hence if heat which is added to the system causes work to be done and an increase in internal energy, the first law of thermodynamics for an isobaric process is given by, ΔQ = ΔU + ΔW = ΔU + PΔV (12.7) As an example, consider the case of changing one gram of water at 100øC into one gram of steam at 100øC. The heat added goes into external work (change in volume) and internal energy. The heat which is added during an isobaric process is given by


260

ΔQ = ncpΔT (12.8) where n is the number of moles, cp, the molar specific heat at constant pressure, is the amount of heat required to raise the temperature of one mole one degree during an isobaric process, and ΔT is the change in temperature. The change in internal energy, which is independent of the process, is the same as would have occurred by an isochoric process, ΔU = ncv ΔT (12.6) Now we can rewrite the first law of thermodynamics for an isobaric process, ncpΔT = ncv ΔT +PΔV (12.9) This equation shows that the molar specific heat of a gas at constant pressure is always greater than the molar specific heat of a gas at constant volume. By rearranging the above equation, and using the relationship that PΔV = (constant) ΔT for an ideal gas, we can show that the difference between the two different molar specific heats is a constant number, independent of the kind of gas that is used in the system (see Table 12.1), cp - cv = constant = R (12.10) where R is the universal gas constant and has a value of 8.31 J/mole-øK. EXAMPLE Helium (3.00 moles) originally at 273øK and a volume of 0.067 m 3 expands at a constant pressure of 1.01 x 105 N/m2 to a volume of 0.134 m3. Find the change in temperature, the change in internal energy, the work done, and the heat added to the system. From the general gas law P1V1/T1 = P2V2/T2. Because P1 =P2 V1/T1 =V2/T2 0.0067 m3/273øK = 0.134 m3/ T2 T2 = 546øK T2 - T1 = 273øK The change in internal energy ΔU, ΔU = ncvΔT = (3.00 moles) (3.00 cal/mole-øK) (273øK) = 2460 cal The work done ΔW, ΔW = PΔV = (1.01 x 105 N/m2) (0.134 -0.067)m3 = 6.77 x 103 J. The heat added ΔQ, ΔQ = ncpΔT = (3.00 moles) (4.98 cal/mole-øK) (273øK) = 4080 calories In order to verify the first law of thermodynamics, you must express all energy quantities in the same units. The work done in calories is work done in joules divided by 4.19 since the mechanical energy equivalent of 1 calorie of heat energy is 4.19 joules (see Chapter 10).


261

From the first law of thermodynamics, ΔW =PΔV = ΔQ - ΔU. Now expressing each term in joules, we have ΔQ - ΔU = (4080 - 2460) cal = 6.79 x 103 joules ΔW = PΔV = 6.77 x 103 J We find that we have a discrepancy of less than one percent, which is not surprising. We may introduce a discrepancy of that size by using only three significant figures. 12.7 Isothermal Processes An isothermal process is one that takes place at a constant temperature. For a system of a confined ideal gas, we find that at constant temperature the pressure and volume of the system are given by Boyle's law (Section 10.9), PV = constant. On the PV diagram an isothermal process for an ideal gas is represented by a hyperbolic curve whose equation is PV = constant See Figure 12.6.

Since the internal energy of a system is a state variable dependent only upon the temperature of the system, for an isothermal process the internal energy of the system remains constant, ΔU = 0. The first law of thermodynamics for an isothermal process states the equality between the work done and the change in heat energy: ΔQ = ΔW =PΔV (12.11) For an isothermal process the heat added equals the work done by the system and is represented by the shaded area under the curve in Figure 12.6. If the system consists of an ideal gas that expands from a volume of V1 at a pressure P1 to a volume of V2 at a constant temperature, the work done by the gas is given by W = P1V1 lnV2/V1 (12.12) (A derivation of this result is given in Section 12.16)


262

EXAMPLE If 50 cal of heat are added to an ideal gas in a cylinder under the condition of a constant temperature, what is the work done by the expanding gas? The internal energy of an ideal gas depends only on its temperature. In this problem the temperature is constant (i.e., ΔU = 0). Therefore, ΔQ = +ΔW, the heat input is equal to the work done by the gas. Note that ΔW is positive for work done by the gas. Under the constant temperature conditions the work done by the gas equal to 50 cal, the same as the heat input energy. 12.8 Adiabatic Processes Any process that takes place very rapidly is essentially adiabatic. An adiabatic process occurs if no heat is added to or taken from the system – that is, ΔQ = 0. The curve GH in Figure 12.7 represents an adiabatic process for an ideal gas. The equation for the curve GH is PVγ = constant, where γ is the ratio of the molar specific heats cp/cv (Table 12.1).

The constant γ has a maximum value of 5/3 for a monoatomic ideal gas, and is always greater than 1. For the adiabatic process in Figure 12.7 , the first law of thermodynamics can be expressed by the following equation, ΔU = -ΔW = -PΔV = ncvΔT = (P1V1 - P2V2)/(1 - γ) (12.13) A derivation of this result is given in Section 12.16. For an adiabatic process if the system does work, there is a decrease in internal energy; if work is done on the system, there is an increase in internal energy - the heating that occurs when you rapidly compress the air in a tire pump, for example.


263

EXAMPLE The air in an automobile tire is released. What happens to the temperature of the air as it expands into the atmosphere? Calculate the values of the changes assuming that the original air pressure in the tire was 3.00 atmospheres (atm) and the temperature was 20.0øC. This expansion is a rapid process and can be approximated by an adiabatic process. The expanding gas does work on the atmosphere so ΔW is positive, and thus ΔU is negative. The internal energy of the gas is reduced; that is, its temperature is lowered. For a unit volume of air (1.00 m3) in the tire, we can use the expression for an adiabatic process 3(1) γ = 1(Vf)

γ where the γ for air is 1.40. Vf = 3(1/1.4) = 2.19 m3 Now we can use the ideal gas law to find the change in temperature, PoVo/To =PfVf/Tf (3.00)(1.00)/293 = (1.00)(2.19)/Tf Tf = 214øK, or - 59øC 12.9 The Second Law of Thermodynamics The first law of thermodynamics is necessary, but not sufficient, to explain the thermal phenomena that you observe every day. When you mix two glasses of water of different temperatures, what is the final state of the water? Is this final state the only possible state consistent with the first law of thermodynamics? There are an infinite number of ways in which energy could be conserved for this system, but the situation actually occurring is the one in which the final temperature of the water is a definite temperature falling somewhere between the initial temperatures of the two glasses of water. This final state is result of the second law of thermodynamics. There are several equivalent ways of stating the second law of thermodynamics. The Clausius statement of the second law is: Heat energy cannot pass spontaneously from a system at a lower temperature to one at a higher temperature. Your experiences with hot and cold objects may make this statement of the second law seem trivial. You have never seen an isolated object at room temperature become hot. Why not? We know the total heat energy available in all the objects in the room is enough warm up one object. But nature does not behave in that fashion. To transfer heat energy from a lower temperature object to a higher temperature we must do work on the system. If the Clausius statement were not true, we would be able to transfer heat energy from inside our homes to the hotter outside air in the summer to air-condition our homes without using any other energy. The Kelvin-Planck statement of the second law of thermodynamics is that it is not possible to perform a process which only extracts heat energy from a constant temperature object and performs an equivalent amount of work. You may think that an isothermal process would violate this statement of the second law because the change in heat energy is equal to the work done by the system,


264

Equation 12.11. However, an isothermal expansion of the system leaves the final system at a different volume than at the beginning of the process. To test this statement of the second law we must return the system to its original volume since the only process that can occur is the change of heat energy into work. But it is just this return process that is impossible without changing the way nature performs. The operational equivalence of the Kelvin-Planck and Clausius statements of the second law is shown by the following example. Suppose the Kelvin-Planck statement can be violated. Then we could construct a heat engine that only takes heat energy from a reservoir of heat energy at constant temperature and converts the heat energy into work. We can then use this engine to operate a refrigerator that extracts heat from the inside of the refrigerator and delivers it to a hotter object. The net result of the heat engine-refrigerator combination is the transfer of heat energy from a cold object to a hot object without doing any net work on the total system, a violation of the Clausius statement. 12.10 Heat Engines The internal combustion gasoline engine and the steam engine are examples of engines that makes use of a combustion process to produce motion and to do work. These are both examples of heat engines. A heat engine absorbs a quantity of heat energy Qh from a hot reservoir of heat at temperature Th, does work W, rejects a quantity of heat energy Qc to a cold heat reservoir at a temperature Tc, and returns to its original state. For example, your automobile engine burns gasoline at a high temperature Th from which the engine extracts heat energy Qh to turn the crank shaft and wheels of the automobile to do work W. The rejected heat energy Qc is returned, via the exhaust gases and other thermal transport processes, to the atmosphere at a temperature Tc. Part of the work is used to return the engine to its original state, ready to begin the cycle again. Heat engines operate in cycles, using the heat absorbed during each cycle to do work. A schematic diagram of a heat engine is shown in Figure 12.8 .

We can combine this cyclic nature of a heat engine with the first law of thermodynamics to develop an expression for the efficiency of a heat engine. Since the initial and final states of a cycle of the heat engine are the same, the internal energy of the engine must be the same at the beginning and end of each cycle, ΔU = 0; then ΔW = ΔQ, or for the whole cycle W = Qh - Qc (12.14)


265

The efficiency of a heat engine is defined as the useful work output divided by the energy input. For the heat engine the energy input is given by the heat energy absorbed from the high temperature reservoir, so efficiency = work output/energy input = W/Qh = (Qh - Qc)/Qh = 1 - Qc/Qh (12.15) EXAMPLE What is the efficiency of a person whose daily diet is equivalent to 4.0 kg of milk and who does useful work at the rate of 50 watts for an eight hour day? Heat of combustion of milk = 650 cal/g. Qh = 650 x 4.0 x 103 x 4.19 = 11 x 106 J W = (50 J/sec)(8 hr x 60 min/1 hr x 60 sec/1 min) = 1.4 x 106 J efficiency = 1.4 x 106 J/11 x 106 = .13 = 13 percent. Where does all the energy go? 12.11 Carnot's Engine Both the first and second law of thermodynamics are basic to the study and understanding of heat engines. How do you make the heat engine more efficient? Is there a limit to the efficiency of a heat engine? In 1824 Sadi Carnot, a French engineer, was the first to approach the problem of heat engines from fundamental considerations. Carnot's approach was an entirely theoretical one. He disregarded the mechanical operation and details and considered the fundamentals. In fact, our discussion in the previous section followed Carnot's approach. Carnot also introduced the concept of a reversible cycle which we now know as Carnot's cycle. The theoretical Carnot cycle is for an ideal engine that has a cylinder and piston made of perfect insulating materials and has three separate parts - one part is a perfect heat conductor and a high temperature heat reservoir at temperature Th, one part is a perfect insulator, and the third part is a perfect conductor and cold heat reservoir at temperature Tc (Figure 12.9). The cylinder was filled with an ideal working substance, such as an ideal gas.


266

The steps in the Carnot cycle are as follows: 1. The high temperature part Th is connected to the cylinder and the piston moves, giving the isothermal expansion ΔQ = ΔW along curve AB 2. The insulating part replaces the heat reservoir Th, and the gas is further expanded as an adiabatic process, ΔQ = 0. As work is done, the temperature of the gas is lowered to temperature Tc along curve BC. 3. The lower temperature Tc reservoir replaces the insulating part. The gas is then compressed at a constant temperature Tc. As the gas is compressed, the part at Tc must absorb some heat, and the amount absorbed is equal to the work of compression (curve CD). 4. Part Tc is replaced by the insulating head, and the gas is adiabatically compressed to the starting point. The work done in compressing the gas increases the internal energy of the gas until it reaches temperature Th (curve DA). A summary of the cycle shows that heat Qh is put into the system in step 1 and that heat Qc is taken from the system in step 3. The difference between Qh and Qc represents the amount of work done during the cycle, W = Qh - Qc. Follow the Carnot cycle around the curves on the PV diagram of Figure 12.9. Starting at point A (P1V1Th), step 1 is represented by the curve AB, an isothermal expansion at temperature Th to P2V2. The heat input is Qh, work is equal to area under the curve AB and bounded by the corners ABV2V1. Step 2 is represented by BC an adiabatic expansion from P2V2 Th to P3V3Tc. The work done is equal to the area under the curve BC and is bounded by the corners BCU3V2. There is a decrease in the internal energy so the temperature drops from Th to Tc. Step 3, represented by curve CD, is an isothermal compression at temperature Tc, heat Qc is exhausted from the system. Work is done on the system in compressing the ideal gas. It is negative work equal in magnitude to the area under the curve CD and bounded by the corners CV3V4D. Step 4, represented by curve DA, is an adiabatic compression from P4V4Tc to P1V1Th. Work is again done on the system. This is negative work equal in magnitude to the area under the curve DA, which is bounded by DV4V1A. This work increases the internal energy. Notice that if all the work done on and by the system is added for one cycle the net work is represented by the area inside of the curves that join the points ABCD. As a result of his theoretical work, Carnot postulated a theorem which we can state in the following way: No engine working between two heat reservoirs can be more efficient than a Carnot engine operating between those two reservoirs. In other words the Carnot cycle defines the best possible heat engine permitted by the laws of nature. What is the efficiency of a Carnot engine. We can use Equation 12.15 and our knowledge of isothermal processes to calculate this efficiency in terms of the temperatures of the reservoirs Th and Tc, efficiency = 1 - Qc/Qh (12.15) For isothermal processes we know that the change in heat energy must be equal to the work done (Equation 12.11), and we have stated that the work done by the system during an isothermal expansion is given by Equation 12.12, Qh = W =P1V1 ln (V2/V1) (12.12)


267

for the expansion AB. For the compression CD, Qc must be the negative of the work done by the system since the system does negative work, Qc = -WCD = -P3V3 ln (V4/V3) =P3/V3 ln (V3/V4). efficiency = 1 - P3V3 ln (V3/V4) /P1V1 ln (V2/V1) (12.16) Next we use the ideal gas laws and the relationship for adiabatic processes for a confined gas. P1V1/Th =P3V3/Tc so P3V3/P1V1 =Tc/Th For the adiabatic processes

P2V2γ =P3V3

γ and P1V1γ = P4V4

γ Dividing the first equation by the second, we obtain

P2V2γ /P1V1

γ =P3Vγ /P4V

γ For the isothermal processes P2/P1 =V1/V2 and P3/P4 =V4/V3 Then

(V2/V1)γ-1 = (V3/V4)

γ-1 or V2/V1 =V3/V4 Then we can reduce Equation 12.16 to efficiency = 1 - Tc/Th (12.17) where Tc and Th are the absolute temperatures in Kelvin degrees. The efficiency of the Carnot engine depends only upon the temperatures of the hot and cold heat reservoirs, and no engine can be more efficient than its equivalent Carnot engine. EXAMPLES

1. A person burns food at a temperature of 37øC and exhausts body heat to an environment of 20øC. What is the maximum efficiency possible for that person? efficiency = 1 - Tc/Th = 1 -293/310 = 5.48 x 10-2 = 5.48 percent

2. An automobile engine during combustion produces a source of heat energy with a temperature of 400øC. What is the maximum efficiency of this engine when the outside temperature is 20øC? efficiency = 1 - 293/673 = 565 x 10-1 = 56.5 percent The usual automobile engine has an efficiency of about 20 percent.


268

12.12 Refrigerators A refrigerator is, in thermodynamic terms, a heat engine that works in reverse. A refrigerator absorbs heat from a low-temperature heat reservoir and exhausts heat to a high-temperature reservoir. To accomplish this task, work must be done on the refrigerator system. Of course, we know from the Clausius statement of the second law that it is not possible to build a refrigerator that requires no work input. From the first law we can write an equation for the heat exhausted to the hot reservoir Qh as the sum of the work done on the refrigerator W and the heat absorbed from the cold reservoir Qc, Qh = W + Qc (12.18) Refrigerators are characterized by their coefficient of performance. The coefficient of performance h of a refrigerator is defined as the ratio of the amount of heat energy absorbed from the cold reservoir to the amount of work done on the refrigerator, η = Qc/Qh - Qc (12.19) Typical refrigerators have performance coefficients of about 5. The larger the performance coefficient, the better is the refrigerator. Is there a maximum possible coefficient of performance for a refrigerator? For our results for the efficiency of a Carnot engine we can transform Equation 12.19 into a relationship between the temperatures of the hot and cold reservoirs. We can combine the information from the equations for the efficiencies of heat engines, efficiency = 1 - Qc/Qh = 1 -Tc/Th to deduce that Qc/Qh =Tc/Th (12.20) We combine this equation with Equation 12.19 to express the coefficient of performance as the ratio of the cold temperature to the temperature difference between the reservoir temperatures, η = Tc/(Th -Tc ) (12.21) EXAMPLE A room air conditioner removes heat from a room at 20øC and exhausts it to the outside at 37øC. What is the maximum possible coefficient of performances for this air conditioner? η = 293ø/(310ø - 293ø) = 17.2 12.13 Entropy One way to characterize the properties of a thermodynamic system is in terms of its orderliness. A system where everything is lined up in rows is highly ordered. A system where everything is happening at random is highly disordered. One form of the second law of thermodynamics is based on the concept of the order of a system. In this form the second law states that real processes always involve an increase in disorder. To facilitate the formulation of this law, we define the entropy of a system. Entropy is a measure of the disorder of a system, and it is defined in the following equation,


269

ΔS = ΔQ/T (12.22) in which ΔS = change in entropy, ΔQ = change in heat energy, and T is the absolute temperature at which process takes place. Therefore, entropy, like internal energy, is a state function. Entropy depends only on the thermodynamic state of the system, and changes in entropy are independent of the changes involved in going from one state to another. Reversible changes are idealized processes. For reversible processes there is no increase in the disorder of the system. Irreversible processes are ones in which there is an increase of entropy and an increase in disorder. Every state of a system has a definite value of entropy. In terms of entropy, the second law of thermodynamics can be written for any isolated system as: ΔS ≥ 0 (12.23) that is, a process that starts in one equilibrium state and ends in another will go in a direction so the entropy of the system plus its environment increases or remains the same. For ΔS > 0, the process is irreversible; for ΔS = 0, the process is reversible (idealized); and for ΔS < 0, the process is impossible. The entropy of an isolated system never decreases in going from one state to another. It should be noted that the directedness of time (arrow of time) is associated with our experience with phenomena that obey the second law of thermodynamics. For example, if you noted a footprint appearing before your eyes on the beach, you would surely conclude that something strange was happening. In fact, you might have the feeling that time was running backward! Our experience has been conditioned by the increase in disorder (the vanishing of a footprint in the sand, for example) associated with the passage of time. Another example in this connection is the phenomena associated with a drop of ink added to a glass of water. The ink drop breaks up and diffuses throughout the water (going from a spatially ordered drop to a randomized distribution of ink particles of considerable disorder). And again, if an ink drop appeared in a glass of water before your eyes, you would certainly have reason to wonder about your sense of time. 12.14 Entropy and Living System Do living systems obey the second law of thermodynamics? Living systems can be characterized as systems that take in energy and create order. These statements certainly are counter to the second law of thermodynamics except that the second law applies only to isolated or closed systems. Upon reflection it becomes apparent that when a living system is isolated, it ceases to live, and then it certainly obeys the second law of thermodynamics. Could you accept the following statement as a definition of life? Any system that takes in energy and creates order is a living system. If this is an unacceptable definition, what would you add? 12.15 The Third Law of Thermodynamics The third law of thermodynamics is a statement about the changes in physical systems that occur as the temperature approaches absolute zero. In one form, the third law may be written, as T → 0, S → 0, or as the temperature approaches absolute zero, the entropy approaches zero. This gives us some insight into low-temperature phenomena. In


270

particular, the phenomena of superfluidity of liquid helium and superconductivity of aluminum are examples of systems that show increased ordering as absolute zero is approached. In terms of atoms and electrons this means that the systems are in their lowest energy states. This allows exact knowledge of the state of the particles of the system and thus perfect order. This form of the third law of thermodynamics allows us to consider the realm of low temperature research in terms of increased order rather than in terms of energy. Indeed, an important question to ask about a system being considered for low temperature study might well be, "Will increased order make a difference in the behavior of the system?" ENRICHMENT 12.16 Calculus Derivations of Thermodynamics Equations Let us derive the equation for an adiabatic expansion of an ideal gas. To obtain this equation we need to obtain a general relationship between the two molar specific heats, cp and cv. We can do this using the calculus form of Equation 12.9 cpdT = n cvdT + PdV (12.24) for an isobaric process, but for an ideal gas PV = BT where B is a constant. For a constant pressure process P dV = B dT (12.25) Now we combine these two equations to obtain the following general result, n cp dT - n cv dT = B dT cp - cv =B/n (12.26) This result was introduced in Section 12.6, Equation 12.10. For an adiabatic process we can write the calculus form of Equation 12.13 n cv dT + P dV = 0 (12.27) For an ideal gas, P dV + V dp = B dT (12.28) Now we solve both equations for dT and set them equal to each other dT = P dV/n cv = P dV + V dP/B = dT (12.29) 0 = (n cv + B) P dV +ncv dP but B = n(cp - cv); so 0 = (n cv + n cp -n cv) P dV +n cv V dP 0 = cp P dV + cv V dP (12.30)

Let γ = cp / cv;

0 = γdV/V +dP/P Now we integrate this equation

constant = γ In V + ln P = lnVγ + ln P = lnPVγ


271

PVγ = constant (12.31) This equation was introduced in Section 12.8. In Section 12.3 you learned that external work done by a thermodynamic system is ΔW = PΔV. In calculus notation we can express this relationship as dW = P dV (12.32) If we have a curve for a given process plotted on a PV diagram and we wish to determine the work done in going along the curve from A to B, then we integrate this expression,

W = ∫VAVB P dV (12.33)

If the pressure is a function of volume, P = f(V), then the total work done by the change in the volume of the system from VA to VB is,

W = ∫VAVB f(V) dV (12.34)

EXAMPLES ISOTHERMAL PROCESS

1. PV = constant = C P = f (V) =C/V C =PAVA =PBVB The work done in an isothermal expansion for an ideal gas is given by

W = ∫VAVB C/V dV = C ln (V)|VA

VB =C ln (VB/VA) (12.35) This equation was introduced in Section 12.7, Equation 12.12.

ADIABATIC PROCESS

2. PVγ = constant where γ = cp/cv.

PVγ = C1

P =C1/( Vγ)

C1 = PAVAγ =PBVB

γ The work done by an adiabatic expansion for an ideal gas is given by

W = ∫VAVBC1dV/ Vγ = {(C1)/(1 - γ)}(V1-γ)|VA

VB = PB VBγ VB

1-γ /(1 - γ) - PAVAγ VA

1-γ/(1 - γ)

W = (PBVB - PAVA)/ (1 - γ) (12.36) This equation was introduced in Section 12.8, Equation 12.13.


272

ENTROPY 3. ΔS = ΔQ/T, or in calculus notation dS = dQ/T.

∫S1S2 dS = ∫dQ/ T

In many processes the change in the quantity of heat dQ is a function of the temperature, dQ =f(T)dT. Thus,

S2 - S1 = ∫S1S2 f(T)dT/T

For example for an isochoric process Q = ncvT, dQ = ncv dT S2 - S1 =ncv ln (T2/T1) (12.37)

SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of this summary with the number of the section where you can find related content material. Definitions Write the correct word or phrase in the blank.

1. A ______ is a system that ______ heat from a ______ temperature reservoir, does work upon the universe, and ______ heat to a temperature reservoir, and whose performance is characterized by its ______.

2. A ______ is a system that _______ heat from a temperature reservoir, has work done upon it by an external agent, and _______ heat to a ______ temperature reservoir, and whose performance is characterized by its ______.

3. The _______ is an idealized process that can be used to calculate the greatest possible ______ of a heat engine.

4. A PV diagram is a ______ on which the horizontal axis represents the ______ of the system, the vertical axis represents the _______ of the system and the area between a curve and the ______ axis represents the _______.

5. Define each of the following processes, describe the representation of the process on a PV diagram and write the first law of thermodynamics equation for the process: a. adiabatic b. isothermal c. isobaric d. isochoric


273

Laws of Thermodynamics 6. State the laws of thermodynamics, and describe the operation of a confined

volume of helium in a Carnot engine in terms of these laws. Thermodynamics Problems

7. A cylinder contains oxygen at a pressure of 2 atm. The volume is 3 liters and temperature is 300øK. The oxygen is carried through the following processes: a. Heating at constant pressure 500øK b. Cooling at constant volume to 250øK c. Cooling at constant pressure to 150øK d. Heating at constant volume to 300øK. Show the processes above in a PV diagram giving P and V at the end of each process. Calculate the net work done by the oxygen.

Answers

1. heat engine, absorbs, high, rejects, low, efficiency, the absolute temperatures of the reservoirs, efficiency (Section 12.10)

2. refrigerator, absorbs, low, rejects, high coefficient of performance (Section 12.12) 3. Carnot cycle, efficiency (Section 12.11) 4. graph, volume, pressure, volume axis, work done (Section 12.3)

5. a. constant heat energy, PVγ = constant curve, ΔQ = 0; ΔU = -PΔV; b. constant temperature, PV = constant, hyperbolic curve, ΔU = 0; ΔQ = P ΔV c. constant pressure, a horizontal line, ΔQ = ΔU +P ΔV or ncpΔT = ncvΔT + PΔV d. constant volume, a vertical line, ΔW = 0; ΔU = ΔQ = ncvΔT (Sections 12.5, 12.6,

12.7, 12.8) 6. zeroth law: helium reaches same temperature as high temperature reservoir; first

law: the helium expands and ΔU = ΔQ - ΔW; second law: the heat absorbed by the helium gas at high temperature is partially converted to work and the rest is rejected at a lower temperature (Section 12.9)

7. See Figure 12.10; net work = area within rectangle = 2 liter-atm (Section 12.3, 12.5, 12.6)


274

ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the equations will help you learn to translate words into equations and to solve single-concept problems. Equations ΔW = P ΔV (12.2) ΔU = ΔQ - ΔW (12.4) ΔU = ncv ΔT (12.6) ΔQ = ncpΔT (12.8) cp - cv = R (12.10) W =P1V1 ln (V2/V1) (12.12) efficiency = W/Qh = 1 - Qc/Qh = 1 - Tc/Th (12.15)(12.17) η = Qc/W = Qc/Qh - Qc = Tc/(Th -Tc ) (12.19, 12.21) ΔS = ΔQ/T (12.22) ΔS ≥ 0 (12.23)

PVγ = constant (12.31)

cp/cv = γ (definition) Problems

1. A cylinder is filled with a gas and 2-kg piston of 18 cm diameter closes the cylinder and falls 10 cm in compressing the gas. How much work is done?

2. If 200 cal of heat are added to a system which does 500J of work, how much is the internal energy of the system changed in this process?

3. It is known that the heat input into an engine to produce 45,000 J of work is 15 x 104 J. How much heat is lost through the exhaust?

4. What is efficiency of the engine of problem 3? 5. The intake temperature of a Carnot engine is 500øK, and the exhaust temperature

is 360øK. What is the efficiency of the engine? Answers

1.96 J 2. 338 J 3. 105,000 J 4. 30 percent 5. 28 percent


275

EXERCISES These exercises are designed to help you apply the ideas of a section to physical situations. When appropriate the numerical answer is given in brackets at the end of each exercise.

Section 12.3 1. A compressed gas is allowed to expand from a volume of 1.0 m3 to 2.5 m3 against

the pressure of the atmosphere (P = 1.02 x 105 N/m2). What work does the gas do? From where does the energy come to do this work? [1.50 x 105 J, from internal energy or applied heat energy]

Section 12.4

2. One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a pressure of 1.013 x 105 N/m2. The heat of vaporization at this pressure is 539 cal/g. Compute the external work, the increase in internal energy, and the amount of heat energy added to the system. [1.69 x 102 J, 2.09 x 103 J, 2.26 x 103 J]

3. When a enclosed system of an ideal gas is taken from A to C via the path ABC, 2000 cal of heat input into the system, and 750 cal of work are done (Figure 12.11).

a. How much heat is put into the system along the path ADC if the work is 250 cal? b. When the system returns from C to A along AC the work is 500 cal. Does the

system absorb or liberate heat, and how much? c. Assume that the internal energy at A is 250 cal and at C is 1500 cal. How much

heat is absorbed in process AB and DC? TB = 3 TA, TC = 3TB d. What is the change in internal energy for path AC. [a. 1500 cal; b. 1750 cal

liberated; c. 500 cal, 1000 cal; d. 1250 cal] Section 12.5

4. One mole of water vapor in a sealed pressure cooker is heated from 100øC so that the absolute pressure doubles. What is the final temperature? How much heat energy is added to the system? What is the change in the internal energy of the water vapor? [473øC, 2514 cal, 2514 cal]


276

Section 12.6 5. A confined quantity (9.6 moles) of superheated steam at 427øC drive a piston 30

cm in diameter a distance of 0.80 m at a pressure 5.00 x 105 N/m2. The final temperature of the steam is 77øC. What is the work done by the steam on the piston? How much heat energy is used during the process? What is the change in the internal energy of the steam? [2.83 x 104J, 1.23 x 105 J, -9.49 x 104 J]

6. A given quantity of gas at a constant pressure of 10 N/cm2 expands from a volume of 10 liters to a volume of 20 liters. How much work is done by the gas? If the original temperature was 27øC, what was the final temperature? What else would you need to know to calculate the heat added to the system? [1000 J, 600øK]

7. A steel cylinder of cross-sectional area of 20 cm2 contains 200 cm3 of mercury. The cylinder is equipped with a tightly fitting piston which supports a load of 30,000 N. The temperature is increased from 15øC to 65øC. Neglecting expansion of the steel cylinder, find a. the increase in volume of the mercury b. the mechanical work done against the force c. the amount of heat added d. the change of internal energy [a. 1.8 cm3; b. 27 J; c. 19,000 J; d. 19,000 J]

Section 12.7

8. Assume that you push down on the plunger of a hand tire pump slowly so that the temperature of the pump remains constant. At the end of your push, the volume of confined air has decreased from 1.05 liters to 0.150 liters. Assume the starting conditions were a pressure of 1.00 atm and a temperature of -13.0øC. What is the final pressure? How much work was done on the system? What is the change in heat energy of the system? What is the change in internal energy of the system? Where did the heat energy go? [7.0 atm, 2.04 atms-liters, 2.04 atms-liters, 0, the surroundings of the pump]

Section 12.8

9. Assume that you push down the plunger of a hand tire pump so rapidly that no heat escapes from the system during this action. At the end of your push the volume of the confined air has decreased from 1.05 liters to 0.150 liters. Assume the starting conditions were a pressure of 1 atm at a temperature of - 13.0øC. What is the final pressure? How much work was done on the system? What is the change in heat energy of the system? What is the change in the internal energy of the system? What is the final temperature of the air? [15.2 atm, 3.09 atm - liters, 0, +3.3 atm - liter, 293øC ]

10. A gas at 27øC at 1 atm is compressed until its volume is one-tenth of its original volume. This compression is done so fast as to be adiabatic (γ = 1.5). a. find the final pressure b. find the final temperature assuming the gas is ideal. [a. 32 atm; b. 960øK]


277

Section 12.11 11. A Carnot engine whose high temperature reservoir is 127øC takes in 200 cal of

heat at this temperature, and it gives up 160 cal to a low-temperature reservoir. What is the efficiency of this engine and the temperature of the exhaust reservoir? [20 percent, 320øK]

12. A Carnot engine is operating between the two temperatures of 450øK and 300øK. The heat furnished at a high temperature reservoir is 1350 cal. How much work is done by the engine, and how much heat is given out in exhaust? [450 cal, 900 cal]

Section 12.12

13. A refrigerator requires 450 J of work to exhaust 1350 J of heat to the outside air. How much heat energy is absorbed from the low-temperature reservoir? What is the coefficient of performance for this refrigerator? [900 J, 2]

14. A room air conditioner can remove 160 J of heat from the room for every 200 J of heat it exhausts to the outside air. What is the work that must be done on the air conditioner to accomplish this? What is the coefficient of performance of the air conditioner? [40 J, 4]

Section 12.13

15. Given the graph in Figure 12.12 for the molar specific heat of a system at constant pressure near T = 0, find the entropy per mole of this material at 4øK.[1.5 x 10-4 J/mol øK]

16. Compare the entropy change for 1 mole (16 g) of water going from ice to water at

0øC and from water to steam at 100øC. Does this answer support the idea that greater entropy changes accompany changes of greater disorder? [ΔSws/ΔSiw = 4.93]


278

PROBLEMS The following problems may involve more than one physical concept. The numerical answer is given in brackets at the end of the problem.

17. The cycle of a heat engine is described as follows: a. start with n moles of gas at P0V0T0 b. change to 2P0V0 at constant volume c. change to 2P02V0 at constant volume d. change to P02V0 at constant volume e. change to P0V0 at constant pressure Show this cycle on the PV diagram, and find the temperature for the end of each

process. What is the maximum possible efficiency for this engine? [2T0, 4T0, 2T0, T0, 75 percent]

18. Assume the gas in problem 17 is an ideal monatomic gas, γ = 5/3. The specific heat of the gas at constant volume is 3.00 cal/mole - K at T0 = 300øK. Assume you have 0.250 moles of the gas. Find a. the heat input b. the heat exhaust c. the efficiency of the engine [a. 975 cal; b. 825 cal; c. 15.4 percent]

19. A cylinder contains air at a pressure of 30 N/cm2. The original volume at 27øC is 4.0 liters. The air is carried through the following processes: a. heating at constant pressure to 227øC b. cooling at constant volume to -23øC c. cooling at constant pressure to -123øC d. heating at constant volume to 27øC Show these on a PV diagram, and give the coordinates at the end of each process.

Calculate the net work done by the gas. [a. 30 N/cm2, 4 liters; b. 30 N/cm2, 6 2/3 liters; c. 15 N/cm2, 6 2/3 liters; d. 15 N/cm2, 4 liters; W = 400 J]

20. Use the data given in problem 19. a. How many moles of air were in the cylinder? b. Find the heat input during heating at constant pressure to 227øC and heating at constant volume to 27øC given that cp = 7.00 cal/mole- øC and cv = 5.00 cal/mole-øC. c. Find the heat liberated during cooling at constant volume to -23øC and cooling at constant pressure to -123øC. d. What is the efficiency of this device as a heat engine? e. What would be the efficiency of a Carnot engine operating in the same temperature range? [a. 0.48 mole; b. Qp = 672 cal, Qv = 360 cal; c. Qv = 600 cal, Qp = -336 cal; d. 9 percent; e. 70 percent]

21. Assume that you had a Carnot engine working between the same temperature as an automobile motor. Estimate the temperatures. What efficiency did you get?


279

22. Given the following data from five actual air conditioners, rank them in the order of efficiency (1 British Thermal Unit (BTU) = 1054.8 joules). a. cooling capacity = 24,000 BTU, power = 3600 watts b. cooling capacity = 21,000 BTU, power = 2800 watts c. cooling capacity = 13,000 BTU, power = 1380 watts d. cooling capacity = 10,000 BTU, power = 1375 watts e. cooling capacity = 7,800 BTU, power = 850 watts [c, e, b, d, a]

23. A racing cyclist is capable of sustaining a power output of 300 watts for extended periods. The associated rate of change of internal energy is measured to be 1400 watts. Find the rate of heat production and the mechanical efficiency (defined as efficiency =Pw/Pu where Pw = power of work or output power and Pu = power of internal energy or input power). [1100 watts, 21.4 percent]

24. Given the graph in Figure 12.13 for the blood pressure during heart contraction (systole), find the rate of work done by the heart beating 72 beats per minute. P2 = 140 mm (systolic) Hg; P1 = 90 mm (diastolic) Hg; V2 - V1 = 80 cm3. What percentage of a metabolic rate of 85 kcal/hr is used for this heart action? [14.7 watts, 15 percent]

25. The enthalpy of a system is defined as H = U + PV. Thus, ΔH = ΔU +PΔV + VΔP.

For constant pressure processes most common for chemical reactions show that ΔH is the heat energy evolved or absorbed.

26. Given the three double containers in Figure 12.14 filled with water at the temperatures shown, which system has the greatest entropy? What conclusion can you make about the entropy and the enthalpy? (See the definition of enthalpy in problem 25.)

27. The Gibb's free energy is defined as G = H - TS. Show that the change in the

Gibb's free energy is a measure of the energy available for work in an isothermal and isobaric process. (Note that TΔS is the energy that goes into increasing disorder and measures the unavailable energy in the process.)

Footnote

1) A mole of any substance is the amount of that substance that contains 6.02 x 1023 molecules and has a mass in grams equal to the molecular of the substance in atomic mass units.

16 Chapter 12 - Doane Universityphysics.doane.edu/hpp/Resources/Fuller3/pdf/F3Chapter_12.pdf ·...

Documents

Transcript of 16 Chapter 12 - Doane Universityphysics.doane.edu/hpp/Resources/Fuller3/pdf/F3Chapter_12.pdf ·...