16-1. Sections Covered in the Text: Chapter 17. Example Problem 16-1 Estimating the Thermal Energy of a gas. Energy Revisited

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1 Heat and Work Sections Covered in the Text: Chapter 17 In this note we continue our study of matter in bulk. Here we investigate the connection between work and heat in bulk matter. Work and heat are both examples of the transfer of energy between a system and the environment. We shall see how the state of a system changes in response to work and heat. We shall see that the transfer of energy and the change in the state of a system are related by a powerful statement of energy conservation called the first law of thermodynamics. For the most part we shall concentrate on fluid systems, and in particular, on gases. Energy Revisited We saw in Notes 08 and 09 that the most general statement of the work-kinetic energy theorem is ΔK = W C + W diss + W ext, [16-1] where 1 K is the change in kinetic energy of the system 2 W C is the work done by conservative forces on the system. Since eq[16-1] already includes the change in kinetic energy, we shall write W C as the negative of the change in potential energy, that is, as W C = U. 3 W diss is the work done by friction-like dissipative forces within the system. This work has the effect of increasing the system s thermal energy: W diss = E th. 4 W ext is the work done on the system by external forces that originate in the environment. An example of an external force is the tension in a rope. With these definitions we can recast eq[16-1] as ΔK + ΔU + ΔE th = W ext. [16-2] Now if we put the sum of the kinetic and potential energies equal to the total mechanical energy E mech = K + U and the sum of the mechanical and thermal energies equal to the total system energy E sys = E mech + E th we can recast eq[16-2] as ΔE sys = ΔE mech + ΔE th = W ext. [16-3] If the system is an isolated one then W ext = 0 (and of course E sys = 0). In this note we investigate nonisolated systems, or the consequences when W ext is not zero. Before we get underway it is useful to actually estimate the thermal energy of a system. This is not, in general, easy. It can be done if the system is a gas, which is the simplest macroscopic system. Example Problem 16-1 Estimating the Thermal Energy of a gas Estimate the thermal energy of 1 mol of nitrogen gas at 20 C. Solution: The thermal energy of a gas is the total kinetic energy of its moving molecules. That is, E th = K micro. We have seen in a previous note that the average speed of an N 2 molecule at 20 C is v avg = 550 m.s 1. Consequently, an average molecule in the gas has kinetic energy K avg = (1/2)mv 2 avg. Nitrogen atoms have atomic mass number A = 14, so the mass of each molecule is Thus K avg = 1 2 mv 2 avg m(n 2 ) = 28u = 4.65 x kg. 1 2 (4.65x10 26 kg)(550m.s 1 ) 2 = 7.0 x J. This is the kinetic energy of a single molecule. Since 1 mol of gas contains N A molecules, the thermal energy of 1 mol is E th = K micro = N A K avg = 4200 J. This is a lot of energy. It is the energy required to raise a 1 kg mass 420 meters above the ground! 1 Work Revisited We saw in Note 08 that work done on a system can be positive or negative. If the work done is positive (Figure 16-1a) then energy is transferred from the environment to the system, and the system s energy increases. If the work done is negative (Figure 16-1b) then energy is transferred from the system to the environment, and the system s energy decreases. 1 We shall see in Notes 17 and 18 that it is impossible to convert this much thermal energy entirely into work. 16-1

2 Figure 16-2a. An example of work done going into increasing a system s mechanical (potential) energy. Figure Work can be positive or negative. The source of the force is part of the environment. Energy Transfer Revisited Two ways in which an external force may do work on a system are illustrated in Figures Figure 16-2a shows the work done going into increasing the system s mechanical energy, potential energy to be precise. Figure 16-2b shows the work done going into increasing the system s thermal energy (when a force of friction is present). The point is that when an external force does work on a system, where the energy ends up going when transferred depends on the manner in which the work is done. To reinforce these ideas we look again at a problem involving friction we have seen before. Example Problem 16-2 Work and Thermal Energy Revisited A child moves a 1.0 kg block a distance 1.0 m across the floor by pulling on a rope (Figure 16-3). The coefficient of kinetic friction between the block and the floor is 0.1. Calculate (a) the work W ext done by the child, (b) the work W C done by conservative forces, (c) the work W diss done by friction, (d) the change in thermal energy E th of the system and (e) the change in kinetic energy K of the system Figure 16-2b. An example of work done going into increasing a system s thermal energy when a force of friction is present. fk block n mg 1.0 m 2.0 N Figure A block on a surface. Solution: The system is just the block. The surface and the child pulling on the rope are part of the environment. The

3 system is therefore not an isolated one. The force of friction is f k = µn = (kg) 9.8(m.s 2 ) = 0.98 N. (a) The child applies a constant force of 2.0 N through a distance of 1.0 m and so does an amount of work W child = W ext = FΔx = 2.0(N) 1.0(m) = 2.0 J. We can say in convenient language that the work is done by the tension in the rope. More realistically, the work is done by the child who exerts the force that gives rise to the tension. (b) In this example, conservative forces (such as the force of gravity) do no work on the system so W C = 0. (c) In convenient language the work done by dissipative forces (friction) is W diss = f k Δx = µnδx = (kg) 9.8 (m.s 2 ) 1.0(m) = 0.98 J. The work W diss is negative, meaning that this amount of positive mechanical energy is transferred out of the system. That the work done by friction is negative means that the real source of the work lies elsewhere in this case with the child. (d) W diss is the negative of the change in thermal energy of the system: ΔE th = W diss = 0.98 J. Thus the thermal energy of the system (block) increases. Where does this energy come from? It comes from the child. (d) Now that we know W C, W diss and W ext we can calculate the change in kinetic energy of the system from eq[16-1] ΔK = W C + W diss + W ext = (J) (J) = 1.02 J. You can see that the change in kinetic energy of the system plus the change in thermal energy of the system equals the total work done by the child, namely 2.0 J. Thus as the child pulls on the rope a part of the energy the child transfers goes into increasing the block s kinetic energy, another part goes into increasing the block s thermal energy. What About Heat? Eq[16-3] is, in fact, incomplete. We shall see in this note that energy can also be transferred between a system and the environment via a thermal interaction. (Friction is not a thermal interaction as we shall see. The heating of a quantity of water on a stove is an example of a thermal interaction.) The energy transferred in a thermal interaction is called heat. To include heat, usually denoted Q, we modify eq[16-3] to read ΔE sys = ΔE mech + ΔE th = W + Q, [16-4] where we now write W = W ext, dropping the subscript. Both work and heat are forms of energy transferred between the system and the environment. We shall investigate heat in more detail later in this note. For now we shall focus on work on a macroscopic system. The simplest macroscopic system is an ideal gas. Work Done in Ideal Gas Processes We examined ideal-gas processes in Note 15. The emphasis there was on calculating an unknown volume, pressure or temperature. In this section we investigate the same ideal-gas processes from the point of view of work. We have seen that the work done by a force F s on a system as it moves from position s i to position s f is defined as the integral W = s f F s ds. s i Let us calculate the work done on a gas as it expands or compresses (Figure 16-4). Suppose the gas is in a container closed at one end. At the other end a tightlyfitting moveable piston enables us to apply an external force F ext to the piston. Clearly, F ext = F gas = pa. Suppose the piston moves a small distance dx to the right shown in Figure 16-4b. As it does so, the external force (the environment) does an amount of work on the piston and therefore on the gas of dw = F ext dx = padx. [16-5] If dx is positive (the gas expands) then dw is negative; if dx is negative (the gas is compressed) then dw is 16-3

4 positive. As the piston moves the distance dx, the volume of the gas changes by an amount dv = Adx. In the meantime let us examine what the work done by the external force (Figures 16-5) means geometrically. Eq[16-7] states that the work done on a gas is the negative of the area under the gas s pv curve as the volume changes from to. Figure 16-5a shows a gas expansion. The area under the pv curve is positive. This means that the environment does a negative amount of work. Figure 16-5b shows a contraction. The area is negative (the integration is performed from a larger to a smaller volume) so the work done by the environment on the gas is positive. Let us consider a numerical example of these geometrical interpretations. Figure Illustration of the work done by an external force on a piston, and therefore on a gas. Consequently, we can write eq[16-5] as dw = pdv. [16-6] If in the process the volume changes from to the total work done by F ext is the integral W = pdv. [16-7] To evaluate this integral we need to know the dependence of p on V (the function p(v)). We shall return to this issue presently. Figure The work done on a gas by an external force is the negative of the area under the gas s pv curve. The direction of the process is shown by the arrow on the trajectory. 16-4

5 Example Problem 16-3 Calculating the Work Done by an External Force on an Expanding Gas The pv diagram for an ideal gas process is drawn in Figure How much work is done on the gas by the environment as the gas s state is changed from i to f? expanding gas is, indeed, negative. pv diagrams for isochoric, isobaric and isothermal processes are redrawn in Figures We can calculate the work done by the environment in each case. Figure The pv diagram for an ideal gas process. Solution: The direction of the process is towards increasing volume, and therefore the area under the curve is positive. The work done by the environment is therefore negative. The process consists of two segments. In the first segment the area consists of two parts, a rectangular part and a triangular part. The second segment consists of just a rectangular part. First Process: The area of the small rectangular part is ( ) 10 6 (m 3 ) ( )(Pa) = 50 J. The area of the triangular part is (1/2)( ) 10 6 (m 3 ) ( )(Pa) = 50 J, making for a total area of A 1 = 100 J. Second Process: The second process consists of a single rectangular area of A 2 = ( ) 10 6 (m 3 ) ( )(Pa) = 150 J. The total work done by the environment is the negative of the sum of these areas: W = (A 1 + A 2 ) = 250 J. Thus the work done by the environment on the Figure Calculating the work done during ideal gas processes: (a) isochoric, (b) isobaric and (c) isothermal. Isochoric Process In the isochoric process (Figure 16-7a) the volume does not change. The area under the curve is zero and consequently, the work done is zero: W = 0 (isochoric process) [16-8] Isobaric Process In the isobaric process (Figure 16-7b) the volume changes from to. The area under the curve and the work done is W = pδv. [16-9] If s positive (the gas expands) then W is negative. If s negative (the gas is compressed) then W is 16-5

6 positive. Isothermal Process An isothermal process is shown in Figure 16-7c. Substituting p = nrt/n eq[16-7] we have for the work done as the volume changes from to : W = pdv = nrt V dv = nrt dv, V [16-10] where we have taken T outside the integral because it is constant. Thus eq[16-10] becomes W = nrt lnv = nrt ln V f. = nrt( ln ln ) [16-11] Because nrt = p i = p f during an isothermal process we can write eq[16-11] three ways: W = nrt ln V f = p i ln V f = p f ln V f [16-12] Having three expressions for the same process means that which expression you use to solve a problem will depend on what information you are given. greater than the area under the curve for process B. Both areas are positive. Thus the work done in process A is more negative than the work done in process B. Thus the work done depends on the path. The Nature of Heat We now come to the nature of heat. As late as the 18 th century, heat was regarded by natural scientists as a substance, called caloric, that was thought to be stored in a body and capable of flowing from a hotter to a cooler body on contact. Count Rumford (1798) showed that when heat is produced by friction, there is no limit to the amount of heat that can be generated. Rumford concluded that the caloric stored in a body cannot be exhausted and therefore heat is not a fluid but rather a manifestation of microscopic motion. This idea has since been confirmed experimentally and theoretically. 2 Joule's Experiments The British physicist James Joule ( ) studied the relationship between heat and work in a famous series of experiments. He did a certain amount of work on a system contained in an adiabatic (thermally isolated) calorimeter and measured the resulting temperature rise (Figure 16-9). He found that the temperature rise is the same whether the work is done by churning water mechanically, by heating electrically, by using a friction belt or compressing a gas. His results constitute the first experimental proof that heat is a form of energy. Work Depends on the Path In any process taking a gas from state i to state f the work done depends on the path taken between the two states. This is easily seen with reference to Figure Figure An illustration of what is called the mechanical equivalent of heat. The temperature of a quantity of water is raised by the same amount by heating or doing an equivalent amount of mechanical work on it Figure The work done depends on the path. In the figure, the area under the curve for process A is 2 Rumford used the term heat in the way in which it was understood in the 18 th century, not in the way it is understood today. He was mistaken in thinking heat is produced by friction. Friction is work that results in an increase in thermal energy. However, he was correct in his conclusion that heat is a manifestation of microscopic motion.

7 It is important to be clear on what heat is not, as much as on what heat is. Heat is the energy transferred between a system and the environment as a consequence of a temperature difference between the system and the environment. When you rub your hands together to warm them heat is not involved at all only work. There is no temperature difference between your hands. But heat is involved if you warm your hands over the element of a stove or an open flame. Only in the latter is there a temperature difference and a thermal interaction. Let us now examine thermal interactions in terms of heat exchange. Thermal Interactions The idea of heat flow in thermal interactions is illustrated in Figures When energy is transferred from the environment into the system (Figure 16-10a) Q is positive. This implies that the temperature of the environment is greater than the temperature of the system. When energy is transferred from the system to the environment (Figure 16-10b) Q is negative. This requires that the temperature of the system is greater than the temperature of the environment. The system is in thermal equilibrium with the environment (Figure 16-10c) when no energy is transferred either way. In this case Q = 0 and the system and the environment are at the same temperature. Units of Heat Joule s experiments demonstrated that heat and work are equivalent, being aspects of the same thing, namely energy. It follows that the most logical unit for heat is the unit of energy, namely the joule. However, historically, at a time when the nature of heat was still unknown, heat was expressed in the unit calorie (small c ). The calorie is still used today (though its usage is declining). It is defined as the quantity of heat required to change the temperature of 1 g of water by 1 C at 15 C. 3 The conversion between the calorie and the joule is 1 cal = J. A calorie unit is also used in the food industry. There it is called the food calorie and is written Cal with a capital C (to avoid confusion with the calorie with a small c). One Calorie is equivalent to 1000 calories, or 1 kcal. To summarize, 1 food calorie = 1 Cal = 1000 cal = 1 kcal = 4186 J. From this point onwards in our study of heat we shall assume that the change in mechanical energy of the system is zero ( E mech = 0). That is to say, we shall assume that the object as a whole is at rest. With this assumption the law of conservation of energy becomes particularly simple, as we shall see next. First Law of Thermodynamics Joule was the first to realize that if we can assume E mech = 0, we can write eq[16-4] as ΔE th = W + Q. [16-13] This equation is called the first law of thermodynamics. It is another way of expressing the law of conservation of energy when heat Q is part of the equation and when E mech = 0. Thus the first law of thermodynamics joins the zeroth law of thermodynamics as useful tools in our study of heat. Figure Illustrating the sign of heat. 3 In fact, the quantity of heat required to raise the temperature of 1 g of water by 1 C is temperature-dependent. There is therefore the need to define the calorie at an agreed-upon specific temperature. That temperature is 15 C. 16-7

8 Thermal Properties of Matter In this section we examine the consequences when the thermal energy of a system is changed ( E th 0). When the thermal energy of a system changes two things can happen: 1 the temperature of the system changes 2 the system undergoes a phase change, such as melting or freezing To quantify change in thermal energy we define another attribute of matter called specific heat. The specific heat determines the value E th for a temperature change T. Specific Heat Specific heat may be defined in these words: The specific heat of a substance is that amount of energy required to raise the temperature of 1 kg of the substance by 1 K. The symbol for specific heat is c. Specific heat depends only on the material from which an object is made. Specific heats and molar specific heats for a selection of solids and liquids are listed in Table Molar specific heat is denoted by the capital C. Q = ΔE th = McΔT. [16-15] Let us now consider an example involving water. Example Problem 16-4 A Problem in Heating Water How much heat is required to raise the temperature of 5.0 kg of water from 25 C to 100 C? Solution: The change in temperature is the same in C as in K, namely (100 25) = 75 K. From eq[16-15], and using data from Table 16-1, the heat required is Q = 5.0 (kg) 4190(J.kg 1.K 1 ) 75(K) = 1.57 x 10 6 J. The amount of heat required is 1.57 megajoules (MJ)! When heat energy is transferred to a substance the substance s temperature may rise or the substance may undergo a change in phase. This latter possibility we examine next. Table Specific heats and molar specific heats of a selection of solids and liquids Substance c (J.kg 1 K 1 ) C (J.mol 1 K 1 ) Solids Copper Iron Gold Ice Liquids Water Phase Change and Heat of Transformation When a substance is heated its temperature may rise and/or it may undergo a change in phase. Let us look again at the process of a quantity of ice being heated in an insulated container (Figures 15-2 and 16-11). The specific heat means that if the temperature of an object of mass M and specific heat c is changed by an amount T then the thermal energy of the object is changed by ΔE th = McΔT. [16-14] If T is positive then E th is positive; if it is negative then E th is negative. NOTE: T can be expressed in either C or K because both temperature scales have the same step size. If W = 0 then we can write from eqs[16-13] and [16-14]: Figure The temperature of a system that is heated at a steady rate. 16-8

9 At first the temperature of the solid increases linearly. This is consistent with eq[16-15] rearranged as ΔT Q = 1 Mc = slope of T vs Q graph [16-16] The slope is constant because M and c are constant. This also holds when the ice has been entirely converted to water. Indeed, the specific heat of water can be determined experimentally from such a graph. 4 But there are times during the melting and boiling segments of the graph, for example when heat is being transferred to the system, but the temperature doesn t change. These segments occur when the substance is undergoing changes of phase. During a phase change the thermal energy continues to increase but the energy added goes into effecting the change, breaking molecular bonds rather than speeding up the molecules. Thus a phase change is characterized by a change in thermal energy without a change in temperature. Heat of Transformation The heat of transformation L of a substance is defined as the amount of heat that causes 1 kg of a substance to undergo a change in phase. Therefore, the heat required for a mass M of a substance to undergo a change in phase is Q = ML (phase change). [16-17] Heat of transformation is a generic term that includes the heat of fusion L f, which is the heat of transformation between a solid and a liquid, and the heat of vaporization L v, which is the heat of transformation between a liquid and a gas. To summarize, the heat needed for these phase changes is undergo phase changes. Let us consider the problem of this being done for a quantity of ice. Table Melting/boiling temperatures and heats of transformation Substance T m ( C) L f (J.kg 1 ) T h ( C) L v (J.kg 1 ) Nitrogen (N 2 ) x x 10 5 Ethyl alcohol x x 10 5 Mercury x x 10 5 Water x x 10 5 Lead x x 10 5 Example Problem 16-5 A Problem Involving Phase Changes and Heats of Transformation How much heat is required to change 200 ml of ice at 20 C into steam? Solution: There are four steps involved in this procedure as you can see in Figure 16-11: (1) the temperature of ice is raised from 20 C to 0 C, (2) the ice is melted to liquid water at 0 C, (3) the temperature of the water is raised from 0 C to 100 C, (4) the water is boiled to steam at 100 C. We are given the volume of the ice. We must find its mass. We assume that the mass of the system, whether ice, water or steam, does not change over the four steps. The density of ice is 920 kg.m 3 and its volume is V = 200 ml = 2.00 x 10 4 m 3. Thus the mass is Q = ±ML f melt / freeze ±ML v boil /condense [16-18] M = ρv = 920(kg.m 3 ) (m 3 ) = kg. The heat needed for the four steps is where the ± indicates that heat must be added to the system during melting or boiling but removed from the system during freezing or condensing. Heats of transformation for a selection of substances are listed in Table If we know the heats of transformation we can calculate the total heat required to effect a temperature change in a system when the system may also 4 This is precisely what you will be doing in the experiment Temperature and Heat in the physics lab at about this time in the course. (1) Heating Ice Q 1 = Mc ice T ice = 0.184(kg) 2090(J.kg 1 K 1 ) 20 (K) = 7700 J. (2) Melting Ice Q 2 = ML f = 0.184(kg) (J.kg 1 ) = J. 16-9

10 (3) Heating Water Q 3 = Mc water T water = (kg) 4190(J.kg 1 K 1 ) 100 (K) (4) Boiling Water = J. Q 4 = ML v = (kg) (J.kg 1 ) = J. Thus the total heat required is Q = Q 1 + Q 2 + Q 3 + Q 4 = J. Notice that of the two phase changes more heat is required to boil water than to melt ice. The Specific Heats of Gases Notice that Table 16-1 lists specific heats for solids and liquids, but not gases. Gases are harder to characterize by a specific heat because the heat required to produce a specified temperature change in a gas depends on the process by which the gas changes state. For example, Figure shows the pv diagram for a gas system. Two isotherms are shown. Processes A and B, which start on the T i isotherm and end on the T f isotherm, have the same temperature change T = T f T i. But process A, which takes place at constant volume, requires a different amount of heat than does process B, which takes place at constant pressure. The reason is that work is done in process B but not in process A. Figure Processes A and B have the same T and the same E th, but they require different amounts of heat. It is conventional therefore to define a specific heat for a constant volume (isochoric) process and a specific heat for a constant pressure (isobaric) process. We shall define them as molar specific heats. Thus the quantity of heat required to change the temperature of n moles of gas by T in these two processes is Q = nc V ΔT (constant volume) [16-19] Q = nc P ΔT (constant pressure) where C s the molar specific heat at constant volume and C P is the molar specific heat at constant pressure. Experimental values of C V and C P for a selection of gases are listed in Table Table Experimental measurements of molar specific heats (J.mol 1 K 1 ) of a selection of gases. Gas Cp Cv Cp Cv Monatomic He Ne Ar Diatomic H N O You can see that for all the gases listed C P > C V, reflecting the fact that more heat is required for a constant pressure process than for a constant volume process. This difference in values of specific heats can only be explained by the kinetic theory of gases. We therefore postpone a specific calculation of them until Note 17. Cp Cv But look more closely at Table Notice that the molar specific heats of monatomic gases have the same value! Also, the molar specific heats of diatomic gases, while different from monatomic gases, have also very nearly the same value. Finally, and most astonishingly, notice that for both monatomic and diatomic gases the difference C P C s the same, namely 8.3 J.mol 1 K 1. This value appears to be equal to R, the universal gas constant. Can we explain these facts with the physics of heat and work that we now know? To attempt this let us focus again on processes A and B in Figure Both processes have the same T and therefore the same E th. Process A is an isochoric process in which W = 0. Thus the first law applied to 16-10

11 this process is ( ΔE th ) A = W + Q = 0 + Q ConstVol = nc V ΔT. [16-20] Process B is an isobaric process in which W = p V. Thus ( ΔE th ) B = W + Q = pδv + Q Const Pr ess = pδv + nc P ΔT. [16-21] Since ( E th ) A = ( E th ) B we have, from eqs[16-20] and [16-21] pδv + nc P ΔT = nc V ΔT. [16-22] From the gas law pv = nrt we can write Δ( pv ) = Δ(nRT). [16-23] In an isobaric process p is constant so pδv = nrδt. [16-24] Substituting this expression for p nto eq[16-22] gives nrδt + nc P ΔT = nc V ΔT. Cancelling n T we are left with thermodynamics reduces to E th = W. This means that compressing a gas adiabatically (W > 0) increases E th. Thus an adiabatic compression raises the temperature of a gas. On the other hand, a gas that expands adiabatically (W < 0) cools as its thermal energy decreases. Thus an adiabatic expansion lowers E th and the temperature of a gas. The work done on a gas in an adiabatic process goes entirely into changing the thermal energy of the gas. But we have just seen that E th = nc V T for any process. Thus W = nc V ΔT (adiabatic process) [16-27] We have already seen how gas processes (isochoric, isobaric, isothermal) can be represented as trajectories on a pv diagram. How can an adiabatic process be represented on a pv diagram? We shall look quickly at the answer. First we define the specific heat ratio γ: γ = C 1.67 monatomic gas P = [16-28] C V 1.40 diatomic gas All monatomic gases have the same γ as do all diatomic gases. The answer is that an adiabatic process is one in which 5 pv γ = const C P C V = R. [16-25] or p f γ = p i γ. [16-29] Thus we see that with the physics of heat and work of this note we can explain the data in column four of Table Finally, since E th is the same for all gas processes that have the same T we can write from eq[16-20]: ΔE th = nc V ΔT (any process). [16-26] The curves found by graphing p = const/v γ are called adiabats (Figure 16-13). Adiabats resemble isotherms but are in fact steeper than isotherms. It is clear from the figure that the temperature falls during an adiabatic expansion and rises during an adiabatic compression. We shall find this result useful in the next section. Adiabatic Processes Thus far we have encountered the isochoric process, the isobaric process and the isothermal process. We now add to that list the adiabatic process. An adiabatic process is an ideal-gas process in which no heat energy is transferred between the gas and the environment (Q = 0). An adiabatic process is one that takes place faster than heat can be transferred. For an adiabatic process, with Q = 0, the first law of 5 The textbook goes through the process of proving eq[16-29]. Due to a lack of time we shall omit this

12 Using the ideal gas law we can solve for the final temperature: p T f = T f i = (303K)(25.1) 1 10 p i = 761 K = 488 C. ( ) (b) The pv diagram is shown in Figure (c) The work done is W = nc V T with T = 458 K. We can find the number of moles from the ideal gas law: Figure Adiabats have a steeper curve than do isotherms. Let us consider a numerical example. Example Problem 16-6 An Adiabatic Compression n = p 1 V 1 RT 1 Thus the work done is = mol. W = nc V ΔT = (0.0201mol)(20.8J.mol 1 K 1 )(458K) = 192 J. Air containing gasoline vapor is admitted into the cylinder of an internal combustion engine at 1.0 atm pressure and 30 C. The piston rapidly compresses the gas from 500 cm 3 to 50 cm 3, a compression ratio of 10. (a) What are the final temperature and pressure of the gas? (b) Show the compression process on a pv diagram. (c) How much work is done to compress the gas? Solution: The compression is described as being rapid so we shall assume it to be adiabatic. We also assume the gas is 100% air. Now air is a mixture of the diatomic gases O 2 and N 2 (mostly) so we shall take γ = 1.40 from eq[16-28]. (a) We solve for the final pressure using eq[16-29]: p f = p i γ = (1.0atm)(10) 1.40 = 25.1 atm. Figure The adiabatic compression of the gas in an internal combustion engine. The work done has gone entirely into thermal energy of the gas

13 To Be Mastered Definitions: thermal energy, heat, isochoric process, isobaric process, isothermal process, thermal interaction Definitions: calorie, first law of thermodynamics, specific heat Definitions: C P, C V Calculation of: molar specific heats of a monatomic gas Definitions: heat of transformation, heat of fusion L f, heat of vaporization L v Definition: adiabatic process Typical Quiz/Test/Exam Questions 1. 1 mole of an ideal gas is expanded isothermally at temperature 300 K to twice its initial volume. By how much does its thermal energy change? Justify your answer