Statistical Physics PDF

STATISTICAL PHYSICS Mr. P.R.S.Tissera Department of Physical Sciences Faculty of Applied Sciences Sabaragamuwa University of Sri Lanka REVERSIBLE AND IRREVERSIBLE PROCESSES  Consider a typical system in thermodynamic equilibrium, say ‘n’ moles of a real gas confined in a cyli...

STATISTICAL PHYSICS Mr. P.R.S.Tissera Department of Physical Sciences Faculty of Applied Sciences Sabaragamuwa University of Sri Lanka REVERSIBLE AND IRREVERSIBLE PROCESSES  Consider a typical system in thermodynamic equilibrium, say ‘n’ moles of a real gas confined in a cylinder piston arrangement of volume ‘V’, the gas having a pressure ‘P’, and a temperature ‘T’  In an equilibrium state these thermodynamic variables remain constant with time  Suppose that the cylinder, whose walls are insulating but whose base conducts heats, is placed on a large reservoir maintained at this same temperature T, as in Figure 1 2 Figure 1 : A real gases is made go from initial state ‘i’ to final state ‘f’. The process can be carried out (a) irreversibly, by suddenly dropping a weight on the piston, or (b) reversibly, by adding sand to the piston a few grains at a time 3  Irreversible Processes  Depress the piston very rapidly  Then wait for equilibrium with the reservoir to be reestablished  During this process the gas is turbulent, and its pressure and temperature are not well defined  Can not plot the process as a continuous line on a PV diagram because we would not know what value for pressure (or temperature) to associate with a given volume  The system passes from one equilibrium state ‘i’ to another ‘f’ through a series of nonequilibrium states  This type of process is called Irreversible 4  Reversible Processes  Depress the piston exceedingly slowly  The pressure, volume, and temperature of the gases are, at all time, well defined quantities  During this entire process the system is always in a state differing only slightly from an equilibrium state  Can plot as a continuous line on a PV diagram  During this process a certain amount of heat Q is transferred from the system to the reservoir  This type of process is called Reversible 5  A reversible process is one that, by a differential change in the environment, can be made to trace its path  That is, if we add a few grains of sand to the piston when the piston is in particular state A, the volume decreases by dV and an equal amount of heat is transferred to the reservoir  If we next remove those few grains of sand an equal amount of heat is transferred from the reservoir returning the system to the original state A 6  Not all processes carried out very slowly are reversible  For example, if the piston in the example exerted a frictional force on the cylinder walls, it would not return to its previous state when we removed a few grains of sand  If we added sand slowly to the piston, the system would still evolve through a series of equilibrium states, but it would not do so reversibly  The word quisistate is used to describe processes that are carried out slowly enough so that the system passes through a continuous sequence of equilibrium states  A quisistate process may or may not be reversible 7  In an adiabatic process no heat is allowed to enter or to leave the system  An adiabatic process can be either reversible or irreversible  In a reversible adiabatic process we move the piston exceedingly slowly, perhaps using the sandloading technique  In an irreversible adiabatic process we shove the piston down quickly  The temperature of the gas will rise during an adiabatic compression  The Eint and the corresponding temperature change T are not the same for reversible and irreversible adiabatic processes 8 HEAT ENGINE AND THE SECOND LAW  A heat engine is a device for converting heat into useful work  That is, energy flows into a system in the form of heat, and some of this energy leaves the system in the form of work done on the environment  The ideal gas cylinder standing on the thermal reservoir at temperature T can serve as a representative example of a heat engine  If we remove a small amount of weight from the piston, the gas expands (isothermally), then heat enters the gas from the reservoir, and work is done by the gas 9  This heat engine would not be very useful over the long term, because it could not operate indefinitely  A more useful engine would be one that operated in a cycle, returning to its starting point after each unit work W was done, and retracing its steps continuously Figure 2 : A cyclical process, consisting of four steps, two (1 and 3) at constantvolume and two (2 and 4) at constantpressure 10  Sign Conventions for Heat and Work  Heat entering the system is considered to be positive; heat leaving a system is considered to be negative  Work done on a system, corresponding to a decrease in volume, is considered to be positive; work done by a system, corresponding to an increase in volume, is considered to be negative  Work done in a cyclical process is negative if the cycle is done in a clockwise sense on a PV diagram and positive if the cycle is done in counterclockwise 11  The four steps in our cycle are the following: Step 1 (ab)  Increase the temperature of the reservoir  And simultaneously add some additional weight to the piston  So that the pressure increases but the volume remains constant Step 2 (bc)  Increase the temperature of the reservoir  Allow the gas to expand at constantpressure  The piston does negative work on the gas 12 Step 3 (cd)  Decrease the temperature of the reservoir  And simultaneously remove some weight from the piston  So that the pressure decreases but the volume remains constant Step 4 (da)  Decrease the temperature of the reservoir  But keep the load on the piston constant  So the pressure remains constant while the volume decreases to its original value 13  The temperature increases for steps 1 and 2 and decreases for steps 3 and 4  Thus heat enters the system during steps 1 and 2 (Q10 and Q20), and heat leaves the system during steps 3 and 4 (Q30 and Q40)  Furthermore, note that for the entire cycle W0 because the cycle is done in the clockwise direction  The total heat Qn that enter the system Qin  Q1  Q 2  The total heat Qn that leave the system Q out  Q 3  Q 4  The net heat Q transferred Q  Q in  Q out 14  The change in internal energy for the cycle must be zero  Therefore, Q  W W  Q  Q in  Q out  Define the efficiency ‘e’ of any cycle to be the net amount of work done on the environment during the cycle, divided by the heat input Qin W Q in  Q out Q out e   1 Q in Q in Q in  We could make a perfectly efficient heat engine ( e = 1 ) if we could design a cycle to reduce Q out , the heat discharged, to zero; otherwise, the efficiency is always less than 1 15  One form of the second law of thermodynamics asserts that this goal of making a perfectly efficient heat engine is impossible ‘It is not possible in a cyclical process to convert heat entirely into work, with no other change taking place’  In our example, the “other change taking place” is the exhaust heat Q out , and thus the second law says that it is impossible to reduce Q out to zero  The above form of the second law, which is sometimes called the Kelvin-Planck form, states that ‘there are no perfect heat engine’ 16 Figure 3 : An engine is represented by the clockwise arrows around the central block. (a) In a perfect engine (b) In a real engine 17  EXAMPLE An automobile engine, whose thermal efficiency ‘e’ is 22%, operates at 95 cycles per second and does work at the rate of 120 hp ( 120 hp = 746 Js1 ). i. How much work per cycle is done on the system by the environment? ii. How much heat enters and leaves the engine in each cycle? 18 ANSWER 19 REFRIGERATORS AND THE SECOND LAW 20 Figure 4 : A refrigerator is represented by arrows pointing counterclockwise around the central block. (a) In a perfect refrigerator, (b) In a real refrigerator 21 REFRIGERATORS AND THE SECOND LAW  A refrigerator is basically a heat engine run in reverse  Like a heat engine, a refrigerator is considered to operate in a cyclical process  A more general refrigerator can be represented by the engine of Figure 3 run in reverse  Heat QL is extracted from the lowtemperature reservoir at temperature TL, and heat QH is discharged to the hightemperature reservoir at temperature TH 22  As in the case of the heat engine, there is no change of internal energy in a complete cycle, and therefore W  Q  QH  QL QL  W  QH  In the refrigerator, heat is input from the low-temperature reservoir, so Q L > 0, and is output to the high-temperature reservoir, so Q H < 0  Also, W > 0 since the environment does work on the working substance 23  In analogy with the efficiency of a heat engine, we evaluate a refrigerator in terms of the coefficient of performance K, defined by QL QL K   W QH  QL  In a perfect refrigerator, W = 0 (thus Q L = Q H ), and the coefficient of performance is infinite 24  An alternative statement of the second law of thermodynamics deals with the performance of a refrigerator: It is not possible in a cyclical process for heat to flow from one body to another body at higher temperature, with no other change taking place  In this statement, the “other change taking place” means that during the cycle external work must be done to cause the heat to move in this way, since it would prefer on its own to flow the other way  This statement of the second law is often called the Clausius form, and in effect it says that ‘there are no perfect refrigerators’ 25  Equivalence of the Clausius and KelvinPlanck Statements  The two statements we have presented for the second law are not independent and are, in fact, entirely equivalent  To show this, let us consider what would happen if the KelvinPlanck form were incorrect, and that we could build a perfect engine, converting heat QH entirely into work W  Let us use this work W to drive a real refrigerator, as shown in Figure 5 26 Figure 5 : (a) A real refrigerator, driven by a perfect engine, is equivalent to (b) a perfect refrigerator 27  Let us regard the combination of the perfect engine and the real refrigerator as a single device The heat taken by this device from the low-  Q1L temperature reservoir The heat transferred by this device to the  Q1H  Q H high-temperature reservoir  Q1H  W  Q1L  This example shows that, a violation of the KelvinPlanck statement of the second law implies a violation of the Clausius statement  Because a violation of either statement implies a violation of the other, the two statements are logically equivalent 28  EXAMPLE A household refrigerator, whose coefficient of performance K is 4.7, extracts heat from the cooling chamber at the rate of 250 J per cycle. a) How much work per cycle is required to operate the refrigerator? b) How much heat per cycle is discharged to the room, which forms the hightemperature reservoir of the refrigerator? 29 Answer 30  THE CARNOT CYCLE  We have seen that the second law of thermodynamic prevents us from building perfect heat engines and refrigerators  It can then logical to ask whether we can come as close to perfection as we like, or whether there is some other fundamental limitation on the performance of heat engines and refrigerators  It turns out that there is a fundamental limit, and to consider it we discuss an engine that operates on a particular cycle, called the Carnot Cycle 31  In the Carnot cycle, an ideal gas in our usual cylinder is the working substance  We use two thermal reservoirs, one at the high temperature TH and another at the low temperature TL  The cycle consists of four reversible processes, two isothermal and two adiabatic  The sequence, indicated schematically in Figure 6 and plotted on a PV diagram in Figure 7 32 Figure 6 : A Carnot cycle 33 Figure 7 : A PV diagram for the Carnot cycle illustrated in Figure 6. The working substance is assumed to be an ideal gas 34  The four steps in our cycle are the following: Step 1 (ab)  Put the cylinder on the high-temperature reservoir, with the gas in a state represented by point ‘a’ in Figure 7  Gradually, remove some weight from the piston, allowing the gas to expand slowly to point ‘b’  During this process, heat Q1 = Q H is absorbed by the gas from the high-temperature reservoir  Because this process is isothermal, the internal energy of the gas does not change  All added heat appears as the work done on the gas as the weight on the piston rises 35 Step 2 (bc)  Insulate the cylinder from the reservoir  By incrementally removing more weight from the piston, allow the gas slowly to expand further to point ‘c’ in Figure 7  This expansion is adiabatic because no heat enters or leaves the system (Q 2 = 0)  The piston does work W2 on the gas  The temperature of the gas drops to TL , because the energy to do work must come from the internal energy of the gas 36 Step 3 (cd)  Put the cylinder on the low-temperature reservoir  By gradually adding weight to the piston, compress the gas slowly to point‘d’ in Figure 7  During this process, heat Q 3 = − Q L is transferred from the gas to the reservoir  The compression is isothermal at temperature TL  Work is done on the gas by the descending piston and its load 37 Step 4 (da)  Insulate the cylinder from the reservoir  By adding still more weight, compress the gas slowly back to its initial point ‘a’ of Figure 7, thus complete the cycle  This compression is adiabatic because no heat enters or leaves the system  The piston does work W4 on the gas and its temperature rises to TH 38 Table 1 : Sign conventions Steps Q W Eint Steps 1 0 0 0 Steps 2 0 0 0 Steps 3 0 0 0 Steps 4 0 0 0 Cycle 0 0 0 39 Step 1 (ab, isothermal) W0 and E int  0 isothermal  Therefore, heat QH is absorbed from the hightemperature reservoir (TH) V Q H  W1  n R TH ln b Va Step 3 (cd, isothermal) W0 and E int  0 isothermal  Therefore, heat QL is transferred to the lowtemperature reservoir (TL) Vc Q L  W3  n R TL ln Vd 40   Vb  ln   QH  TH  Va  QL TL  Vc  ln    Vd  Step 2 (bc, adiabatic) and Step 4 (da, adiabatic) TH Vb  1  TL Vc  1 and TH Va  1  TL Vd  1 Vb  1 Vc  1  1  Va Vd  1 Vb Vc  Va Vd QH TH And therefore,  QL TL 41  The efficiency of a heat engine operating on a Carnot cycle: Q out QL e  1 1 Q in QH TL T  TL e  1  H TH TH  The efficiency of a Carnot engine depends only on the temperature of the two reservoirs between which it operates  The efficiency increases as TL decreases, approaching 1 as TL approaches 0 42  A Carnot cycle, because it is reversible, can be run backward to make a refrigerator  As done in the heat engine, it is possible to show that the coefficient of performance K for the refrigerator is given by QL K  QH  QL QL TL K   QH  QL TH  TL 43  Carnot Theorem and the Second Law The efficiency of any heat engine operating between two specified temperatures can never exceed the efficiency of a Carnot engine operating between the same two temperatures  That is, the Carnot efficiency is the upper limit for the performance of a heat engine  To show that violating Carnot’s theorem is also a violation of the second law 44 Figure 8 : (a) Engine X drives a Carnot refrigerator. If engine X were more efficient than a Carnot engine, then the combination would be equivalent to the perfect refrigerator shown in (b). 45  The net heat that flows from the lowtemperature  Q1L  Q L reservoir due to the combination of the two devices is  And the net heat delivered to the high-temperature  Q1H  Q H reservoir is  Applying the first law to each device separately, W  Q H  Q L  Q1H  Q1L  Defining Q to be the difference between QlH and Q H , we can have Q  Q1H  Q H  Q1L  Q L 46  Our hypothesis is that the efficiency of engine X can exceed the Carnot efficiency; that is, ex  e hypothesis W W  Q1H  Q H QH Q1H  And hence, Q  0  And also, Q  Q1H  Q H  Q1L  Q L  Thus the combination of engine X and the Carnot refrigerator is equivalent to the perfect refrigerator  This clearly violates the Clausius form of the second law, and so our original hypothesis must be false 47  If engine X operates on a cycle composed entirely of reversible processes, then its efficiency is equal to the Carnot efficiency e  e Carnot reversible  If the cycle is in part irreversible, then in effect a portion of the energy transferred in each cycle is lost, perhaps to friction, and can not be regained as useful work. And hence, e  e Carnot irreversible 48  Example The turbine in a steam power plant takes steam from a boiler at 520 C and exhausts it into a condenser at 100 C. What is its maximum possible efficiency? ANSWER  The maximum efficiency is the efficiency of a Carnot engine operating between the same two temperatures TH  TL 793 K  373 K e Max    53 % TH 793 K 49  The Thermodynamic Temperature Scale  The efficiency of a reversible engine is independent of the working substance and depends only on the two temperatures between which the engine works QL e 1 QH QL  Then, can depend only on the temperatures QH 50  This led Kelvin to suggest a new scale of temperature. If we let L and H represent these two temperatures, his defining equation is L QL  H QH  That is, two temperatures on this scale have the same ratio as the heats absorbed and rejected, respectively, by a Carnot engine operating between these temperatures  Such a temperature scale is called the Thermodynamic (or Kelvin) temperature scale 51  To complete the definition of the thermodynamic scale, we assign the standard value of 273.16 to the temperature of the triple point of water  Hence,  T rip  273.16 K  Then for a Carnot engine operating between reservoirs at the temperatures θ and θTrip , we have   Q    Trip 273.15 K Q Trip Q   273.16 K Q Trip 52  But according to the ideal gas temperature scale, X T  273.16 K X Trip  We see that on the thermometric scale Q plays the role of a thermometric property  However, does not depend on the characteristics of any substance because the efficiency of a Carnot engine is independent of the nature of the working substance  Therefore, we can obtain a scale of temperature that is free of the objection we can raise to the ideal gas temperature scale, and in fact we arrive at a fundamental definition of temperature 53  The efficiency of a Carnot engine using the thermometric temperature scale QL  e 1 1 L QH H  The efficiency of a Carnot engine using the ideal gas temperature scale  Therefore, L TL  T   H TH  Trip TTrip  And therefore,  T  Hence if an ideal gas were available for use in a constant-volume thermometer, the thermometer would yield the thermometric (or Kelvin) temperature 54  Absolute Zero Temperature  Suppose we have a system at a temperature T2 we wish to measure  We can take the system around a Carnot cycle, first doing adiabatic work on it to raise its temperature to T1, which is presumably known on the ideal gas scale, then transferring known heat isothermally, then doing adiabatic work to reduce its temperature back to T2, and finally transferring heat necessary to return the system to its original condition 55  b a c d Figure 9 : A series of Carnot cycles tending toward the absolute zero of temperature. The difference in slope between isothermal and adiabatic processes has been exaggerated for clarity 56  From the above arguments, we conclude Q2 T2  T1 Q1  Thus knowing T1 and measuring Q1 and Q 2 enables us to determine the thermodynamic temperature T2 directly  Regarding T2 as a known temperature, we can take the system around another Carnot cycle to determine a still lower temperature T3  In principle, we could continue this process to the absolute zero of temperature, however, the smaller the temperature, the smaller the heat Q transferred in an isothermal process between two given adiabatic processes 57  The fundamental feature of all cooling processes is that the lower the temperature, the more difficult it is to go still lower  This experience has led to the formation of the third law of thermodynamics ‘It is impossible by any procedure, no matter how idealized, to reduce any system to the absolute zero of temperature in a finite number of operations’  Hence, because we can not obtain a reservoir at absolute zero, a heat engine with 100% efficiency is a practical impossibility 58 ENTROPY: REVERSIBLE PROCESSES  Already you have studied that for a Carnot cycle QH QL  TH TL  Q L and Q H always have opposite signs. We can therefore write QH QL   0 TH TL  As a next step, we want to generalize the above equation to any reversible cycle, not just a Carnot cycle  To do this, we approximate any reversible cycle as an assembly of Carnot cycles 59 Figure 10 : A reversible cycle superimposed on a family of isotherms 60  1 4 2 3 Figure 11 : The isotherms are connected by adiabatic lines, forming an assembly of Carnot cycles that approximate the given cycle 61  By making the temperature interval between the isotherms in Figure 11 small enough, we can approximate the actual cycle as closely as we wish by an alternating sequence of isotherms and adiabatic lines  We can then write for the isothermal-adiabatic sequence of lines in Figure 11, Q  T  0 or, in the limit of infinitesimal temperature differences between the isotherms of Figure 11, dQ  T  0 62  ‫ׯ‬ indicates that the integral is evaluated for a complete traversal of the cycle, starting and ending at the same arbitrary point of the cycle  We call this new variable the entropy S, such that dQ dS  , T  And thus,  dS  0 63  If the integral of a variable around any close path in a coordinate system is zero, then the value of that variable at a point depends only on the coordinates of the point and not at all on the path by which we arrived at that point  Such a variable is often called a ‘state variable’, meaning that it has a value that is characteristic only of the state of the system, regardless of how that state was arrived at  Examples for state variables are:  Internal energy ‘EInter’,  Gravitational potential energy ‘Ug’,  Pressure ‘P’, and  Temperature ‘T’ 64  The change in entropy between any two states ‘i’ and ‘f’ is then f f S  S1  S2   dS   dQ reversible process  i i T Where the integral is evaluated over any reversible path connecting those two states 65  EXAMPLE A lump of ice whose mass ‘m’ is 235 g melts (reversibly) to water, the temperature remaining at 0 C throughout the process. What is the entropy change for the ice? (The heat of fusion of ice is 333 kJ kg1) 66  ANSWER The requirement that we melt the ice reversibly means that we must put the ice in contact with a heat reservoir whose temperature exceeds 0 C by only a differential amount, thereby melting only a small bit of ice (If we then lower the reservoir temperature by the same differential amount, the melted ice would freeze, thus the process is reversible). For such a reversible process, water water dQ S  S water  Sice   dS ice   ice T 67 In the limit of infinitesimal temperature differences between the isotherms, the temperature is constant, water 1 Q S  S water  Sice  T  ice dQ  T Here dQ means the small elements of heat energy that enter the ice from the heat reservoir, and the total of all these elements is just the total heat absorbed by the ice, or   Q  m L  0.235 kg  333 kJ kg 1  7.83 x 104 J Thus, Q 7.83 x 10 4 J S  S water  Sice    287 J K 1 T 273 K 68 END OF THE CHAPTER - 7 69