1184 Adiabats and entropy (Hiroshi Matsuoka In this section, we will define the absolute temperature scale and entropy Quasi-static adiabatic processes and adiabats Suppose that we have two equilibrium states! and! of a macroscopic system and that these states can be connected by a quasi-static adiabatic process We then denote this relation by! "# We can readily show that for three equilibrium states of the system,!,!, and!, if a! " a # and! " a #, then! " # We can then classify all equilibrium states of this system into a set a of groups, in each of which any state can be connected to any other state by a quasi-static adiabatic process These groups are called adiabats Two general properties of adiabats Let us specify each equilibrium state of a macroscopic system by its empirical temperature, volume, and mole number,!,v,n 1 For the equilibrium states in a particular adiabat A, the volume of the system ranges over all possible finite values: 0 < V min! V < +", where V min is the minimum volume the system can take before the atoms in the system are fused together to form a nuclear matter To show this property, just pick an equilibrium state in this adiabat with volume V 0 Through an adiabatic expansion or compression, we can then change the volume of the system from V 0 to any value in the above range 2 For the equilibrium states in a particular adiabat A, the empirical temperature of the system ranges over all possible finite values: 0 <! < +" It is quite reasonable to assume this because of an experimental observation that through a quasi-static adiabatic expansion or compression of the volume of the system, we can adjust the internal energy of the system to almost any possible value since in the quasi-static adiabatic process the internal energy changes according to!u = W, where W is the work done on the system through an adiabatic expansion or compression of the volume Furthermore, for a fixed value of the volume, the internal energy is a monotonically increasing function of the empirical temperature so that we can always find an equilibrium state with a desired value of the temperature somewhere in the given adiabat From now on we will assume this property of adiabats Entropy as a label for adiabats Just as temperature was introduced as a label or index for each group of equilibrium states that are in thermal equilibrium with each other, we can introduce a label for adiabats We call
this label entropy Being the label for adiabats, the entropy of the system must take the same value for all the equilibrium states belonging to a given adiabat and take different values for different adiabats To define entropy, we must first find how to calculate an entropy difference between two adiabats For this purpose, we will use a quasi-static isothermal process connecting two equilibrium states each chosen from one of these two adiabats so that the system starts at an initial state in one of the adiabats and ends at a final state in the other adiabat Since in a quasistatic isothermal process the temperature of the system stays constant, we must choose a particular temperature scale to specify each quasi-static isothermal process To define the entropy difference between two adiabats, we will need a special temperature scale called the absolute temperature scale Before we define this absolute temperature scale, let us briefly sketch how we can define the entropy difference between two adiabats Suppose that along a quasi-static isothermal process connecting the two adiabats the temperature of the system, measured in the absolute temperature scale, remains at a constant value of T and that after this process some amount of heat Q has flowed into the system The entropy difference!s between the two adiabats is then defined by!s = Q T For this definition to make sense, we need to show: (i For any pair of two adiabats, we can always find a quasi-static isothermal process connecting them (ii For a given pair of two adiabats, any quasi-static isothermal process connecting them gives the same value for the entropy difference!s defined above
To show that the first statement holds, we only need to show that for any pair of two adiabats, there exists a common empirical temperature value for two equilibrium states chosen from these adiabats, respectively, because once we find these two equilibrium states at the same empirical temperature, we can attach the system to a heat reservoir at this common temperature and take the system isothermally from one of the states to the other by varying the volume of the system quasi-statically As we have stated in our discussion on the general property 2 of adiabats in the last sub-section, we assume that for equilibrium states in each adiabat, the empirical temperature ranges over all possible finite values: 0 <! < +" so that we can always find two states with the same empirical temperature each from two distinct adiabats We will thus assume that the first statement is always valid We will justify the second statement in the next sub-section Two isothermal processes each connecting two adiabats and Carnot cycles To show that for a given pair of two adiabats, A 1 and A 2, of a macroscopic system, any quasi-static isothermal process starting in one of the adiabats and ending in the other gives the same value for the entropy difference!s defined in the last sub-section, we must understand how two quasi-static isothermal processes each going from one of the adiabats to the other are related with each other Let us assume that in one of the two isothermal processes, the system absorbs heat Q ( from a heat reservoir at an empirical temperature of while in the other isothermal process, the system absorbs heat Q (! 2 from a heat reservoir at an empirical temperature of! 2 A A 1 2 Q (!! 2 Q (! 2 V The main tool we will use to study the relationship between these isothermal processes is a Carnot cycle, which is a quasi-static cyclic process consisting of one of the isothermal processes and the reverse process of the other as well as two adiabatic processes, each of which remains in one of the two adiabats, A 1 and A 2 More specifically, the system undergoes the following four quasi-static processes in succession:
(i The isothermal process at that takes the system from an initial equilibrium state in adiabat A 1 to an equilibrium state in adiabat A 2 In this process, the system absorbs heat Q ( from a heat reservoir at and receives work W 1 (ii An adiabatic process in adiabat A 2 from the equilibrium state at to an equilibrium state at! 2 In this adiabatic process, the system remains in adiabat A 2 and receives work W 2 (iii The reverse isothermal process at! 2 that takes the system from the equilibrium state in adiabat A 2 to an equilibrium state in adiabat A 1 In this process, the system absorbs heat!q(" 2 from a heat reservoir at! 2 and receives work W 3 (iii An adiabatic process in adiabat A 1 from the equilibrium state at! 2 to the initial equilibrium state at In this adiabatic process, the system remains in adiabat A 1 and receives work W 4 Q ( " W "Q(! 2 After the system goes through this cyclic process, it returns to its initial state so that there is no change in its internal energy:!u = 0 According to the first law of thermodynamics, we must have 0 =!U = Q (" 1 # Q (" 2 + W,! 2 where W = W 1 + W 2 + W 3 + W 4 is the total amount of work done on the system in the cycle, so that
Q ( "Q(! 2 = "W Two properties of Q ( and Q (! 2 1 Q ( Q(! 2 > 0 if "! 2 This implies that the signs of Q ( and Q (! 2 are the same, or equivalently, the signs of Q ( and! Q (" 2 are opposite In all the isothermal processes from adiabat A 1 to adiabat A 2, heat must always either flow into the system or flow out of the system In other words, in a Carnot cycle, some heat must flow into the system from one of the reservoirs while some heat must be discarded to the other reservoir We can show this property using Kelvin s statement of the second law Proof: If either Q ( or Q (! 2 were zero, then in either the Carnot cycle or its reverse cycle, the system would absorb heat and fully convert it to work, which contradicts with Kelvin s statement of the second law If both Q ( and! Q (" 2 were positive so that the signs of Q ( and Q (! 2 were opposite, then the work done by the system or W would be positive We could then fully convert the total amount of heat Q ( "Q(! 2 into work W while Q ( and! Q (" 2 each can be drawn from a third heat reservoir at temperature! 3 higher than both and! 2, which contradicts Kelvin s statement of the second law If both Q ( and! Q (" 2 were negative, we could run the Carnot cycle backward and fully convert the heat!q(" 1 + Q (" 2 = W into work W, which again contradicts Kelvin s statement of the second law! 3 Q ( "Q(! 2 Q ( " W "Q(! 2! 2
2 If >! 2, then Q ( > Q (! 2 Between a pair of isothermal processes from adiabat A 1 to adiabat A 2, the one at a higher temperature accepts or discards more heat than the other at a lower temperature We can show this property using Kelvin s statement of the second law " Q (! 2 Because of property 1 above, both Q ( and have the same sign so that without any loss of generality, we can assume that in the > 0 from the colder reservoir at! 2 < 0 from the reservoir at Because of the first law, after this! Q (" 1 > 0 We can absorbed by the hotter reservoir back to the colder reservoir Proof: Let s assume that Q Q! 2 cycle reverse Carnot cycle, the system absorbs heat Q! 2 while it absorbs heat! Q " 1 Carnot cycle the system does some work on the outside as!w = Q " 2 then transfer the heat Q through heat conduction between the reservoirs We therefore effectively extract the heat Q (! 2 " Q ( from the colder reservoir and fully convert it into the work!w, which contradicts Kelvin s statement of the second law " Q ( < 0 Q ( > 0 "W = Q (! 2 " Q ( > 0 Q (! 2 > 0! 2 The Carnot theorem: the universal heat ratio for two isothermal processes between two adiabats Consider two quasi-static isothermal processes going from adiabat A 1 to adiabat A 2 of a macroscopic system In one of the isothermal processes, the system absorbs heat Q ( at an empirical temperature, whereas in the other isothermal process, the system absorbs heat Q (! 2 at an empirical temperature! 2 < For any pair of two distinct adiabats of any macroscopic system, the ratio between Q ( and Q (! 2, Q ( Q (! 2 or Q (! 2 Q ( has the same universal value, which is positive according to Property 1 shown in the last sub-section and depends only on and! 2
Proof Consider two Carnot cycles C ( A and C ( B in two different systems A and B Each of the Carnot cycles is based on a quasi-static isothermal process at the temperature and the reverse process of a quasi-static isothermal process at the temperature! 2 In the Carnot cycle C ( A, system A absorbs heat Q A adiabat A 1 A ( A to A 2 ( A going from adiabat A 2 Q ( B during the isothermal process at going from (" 2 during the isothermal process at! 2 while it absorbs heat! Q A ( A to A 1 Similarly, in the Carnot cycle C ( B, system B absorbs heat ( B to A 2 while it absorbs B during the isothermal process at going from adiabat A 1 B during the isothermal process at! 2 going from adiabat A 2 heat! Q B " 2 Applying the first law to these cycles, we find to A 1 B and!w ( A = Q ( A " 1!W ( B = Q ( B " 1 { " 2 } +!Q A { " 2 }, +!Q B where!w ( A and!w ( B are the work done by the systems A and B Without any loss of generality, we can assume >! 2 so that and Q ( A ( > Q ( A (! 2 > 0 Q ( B ( > Q ( B (! 2 > 0, where we have taken into account both Property 1 and Property 2 in the last sub-section Let us suppose that " Q ( B ( Q B Q ( A Q ( A! 2! 2 and we will show that this will lead to a contradiction with Kelvin s statement of the second law, which implies that these two ratios must be equal Without any loss of generality, we can assume that > Q ( B ( Q B Q ( A Q ( A! 2! 2 By adjusting the mole number of the system for the cycle C ( B we can have a scaled cycle!c ( B of this cycle for which we have
Q ( A (! 2 = "Q ( B (! 2 = Q ("B (! 2 where! is a positive scale factor determined by! = Q ( A (" 2 Q ( B (" 2 We then have > Q ( B ( Q B = "Q ( B ( "Q B = Q ( "B ( Q "B = Q ( "B ( Q A Q ( A Q ( A! 2! 2! 2! 2! 2 so that Q ( A ( > Q ("B ( If we run the reverse cycle of the scaled cycle!c ( B, the system B absorbs the heat > 0 from the heat reservoir at temperature! 2 while it absorbs the heat < 0 from the heat reservoir at temperature If we run the cycle A and the Q (!B " 2! Q ("B # 1 reverse cycle of the scaled cycle!c ( B together, the combined cycle then effectively absorbs the heat Q A " Q #B > 0 from the heat reservoir at temperature while it absorbes no net heat from the heat reservoir at temperature! 2 This implies that the combined cycle fully converts the heat from the heat reservoir at temperature into work, which contradicts with Kelvin s statement Therefore, the heat ratio Q ( Q (! 2 is universal for two isothermal processes at the temperatures and! 2 as long as they take a system from one adiabat to another " Q (#B ( Q ( A ( W (#B B A "W ( A Q (#B (! 2 = Q ( A (! 2 " Q ( A (! 2! 2
The absolute temperature scale In this sub-section, we will define the absolute temperature scale For each value! of a particular empirical temperature, we need to assign a unique value T (! of the absolute temperature The Carnot theorem shown in the last sub-section implies that the positive ratio Q ( Q (! 2 must be a function of only and! 2 and equivalently of only T ( and T (! 2 : = f T Q Q! 2 (,T (! 2 According to the two properties of Q ( and Q (! 2, this function f must satisfy ( >1 f T (,T! 2 if >! 2 We also need to preserve the ordering of temperature: if >! 2, we should have T ( > T (! 2 so that if we assume that the absolute temperature is always positive, > 1, T T! 2 which suggests the following relation: = T ( Q Q! 2 T! 2 This relation in fact allows us to define and measure the absolute temperature T (! for an empirical temperature! as follows: (1 Have a system go through a quasi-static isothermal process that connects two adiabats and proceeds at the empirical temperature! tr of the triple point of water (ie, the equilibrium state of water where solid ice, liquid water, and gaseous steam can coexist (2 Measure the heat Q (! tr absorbed by the system in this isothermal process (3 Have the system go through another quasi-static isothermal process that connects the same two adiabats and proceeds at an empirical temperature! (4 Measure the heat Q (! absorbed by the system in this second isothermal process
(5 We then demand that the two temperatures in the absolute temperature scale satisfy Q! = T! Q! tr T! tr By setting T (! tr = 27316 K, we define the absolute temperature T (! by T (! " ( 27316 K Q (! Q! tr Because of the Carnot theorem, the absolute temperature scale thus defined is universal or depends only on the value of! The choice of T (! tr = 27316 K is made so that the absolute temperature scale thus defined will coincide with the ideal gas temperature scale as shown in the appendix at the end of this section For two arbitrary empirical temperatures and! 2, using a particular pair of adiabats, we can also show so that ( 27316 K Q ( = Q (! tr ( 27316 K Q (! 2 T T! 2 Q Q! tr = Q (! 2, T T! 2 = Q Q! 2 which will help us define entropy below The unattainability of the absolute zero temperature No macroscopic system can be found at the absolute zero temperature because according to the definition of the absolute temperature scale: T (! " ( 27316 K Q (!, Q! tr the absolute zero temperature T = 0 implies that the heat absorbed in a quasi-static isothermal process at T = 0 must be also zero, which contradicts Kelvin s statement of the second law since we can then use a Carnot cycle operating between a heat reservoir at T = 0 and another at T! > 0 to fully convert the heat absorbed from the reservoir at T! into work
A unique value of entropy assigned to each adiabat Entropy is a label for adiabats so that we can assign a unique value of entropy to each adiabat and distinguish adiabats by their values of entropy Let us finally define entropy Suppose that along a quasi-static isothermal process connecting two adiabats the temperature of the system, measured in the absolute temperature scale, remains at a constant value of T and that in this process some amount of heat Q flows into the system The entropy difference!s between the two adiabats is then defined by!s = Q T The entropy difference thus defined is uniquely determined for the two adiabats because as we have shown at the end of the sub-section, The absolute temperature scale, if in another quasistatic isothermal process that connects the same two adiabats, the system absorbs heat Q! at an absolute temperature of T!, then we find Q T = Q! T!, which implies that the value of the entropy difference!s between the two adiabats is uniquely determined or that for any quasi-static isothermal process between the two adiabats, the ratio Q T between the heat the system absorbs and the absolute temperature T has the same value Entropy as T! 0 and the third law of thermodynamics Experiments have shown that for any macroscopic system, the entropy difference between two equilibrium states both at the same absolute temperature T approaches zero as T approaches the absolute zero temperature:!s = S 2 " S 1 # 0 as T! 0, which suggests that the entropy of the system approaches a universal value as the absolute temperature of the system approaches zero The third law of thermodynamics simply set this universal value to be zero so that S! 0 as T! 0 As far as thermodynamics or macroscopic physics is concerned, this choice of zero for the universal entropy value at the absolute zero temperature limit is just an arbitrary choice, though it turns out that there is a microscopic meaning to this particular choice of the universal entropy value By thus choosing the universal entropy value at the absolute zero temperature limit to be zero, we can then assign a unique value of entropy to each adiabat as well as to all the equilibrium states in this adiabat
An infinitesimal change of entropy ds We can also define a small change ds of entropy when the system absorbs a small amount of heat!q T qs in an infinitesimal quasi-static isothermal process at absolute temperature T, which takes the system from an adiabat with entropy S to another adiabat with entropy S + ds : ds! "Q qs T T We can actually remove the restriction that the heat!q qs T flows into the system through an infinitesimal quasi-static isothermal process and replace!q qs T with a small amount of heat!q qs in a general infinitesimal quasi-static process so that ds =!Qqs T Proof Consider two adiabats whose values of entropy are S and S + ds An infinitesimal quasi-static isothermal process at absolute temperature T, takes the system from an equilibrium state at T,V,n to another at ( T,V +!V,n and has the system absorb the heat!q T qs, while a general infinitesimal quasi-static process takes the system from the state at T,V,n to ( T + dt, V + dv,n and has the system absorb the heat!q qs In general,!v " dv We want to show!q T qs "!Q qs Applying the first law of thermodynamics on the general infinitesimal quasi-static process, we have du =!Q qs +!W qs =!Q qs " P( T,V,ndV We can also take the system from the state at ( T,V,n to the state at ( T,V +!V,n through the infinitesimal quasi-static isothermal process and then from the state at T,V +!V,n to ( T + dt, V + dv,n through an infinitesimal quasi-static adiabatic process in the adiabat with entropy S + ds Applying the first law of thermodynamics on these two processes in succession, we find so that or du =!Q qs T " P( T, V,n!V " P( T, V +!V,n(!V " dv!q qs " P( T,V,ndV =!Q qs T " P( T, V,n!V " P( T,V +!V,n( dv "!V
!Q qs "!Q T qs = " P T,V +!V,n { " P( T, V,n }( dv "!V = "!P dv "!V # 0 where!p = P( T, V +!V, n " P( T,V,n and we neglect!p( dv "!V because and!p( dv "!V << P( T,V +!V,n!P( dv "!V << P( T,V,n We therefore find!q T qs "!Q qs so that ds =!Qqs T The fundamental equation or F = ma of thermodynamics Consider an infinitesimal quasi-static process where a small amount of heat!q qs flows into a system and a small amount of work!w qs is done on the system According to the first law, a small change du in the internal energy is then given by where and du =!Q qs +!W qs,!w qs = "PdV!Q qs = TdS, so that we have du = TdS! PdV, which is a special case (ie, the mole number n is kept constant of the fundamental equation of thermodynamics: du = TdS! PdV! µ dn Once we have the fundamental equation, we can derive all sorts of equations in thermodynamics including the following equation derived in Sec91:
ds = C V T dt +! " T dv Since "!S % $ ' #!T & V, n = C V T > 0, the entropy is a monotonically increasing function of the temperature if the volume and the mole number are kept constant We will use this property when we derive the modern statement of the second law in the next section Appendix to Sec1184: the ideal gas temperature = the absolute temperature In practice, the absolute temperature defined in this section is not all that practical It would be more practical if we could use a thermometer to measure temperatures in the absolute temperature scale Fortunately, the ideal gas temperature scale! actually coincides with the absolute temperature scale so that we can use an ideal gas thermometer to measure temperatures in the absolute temperature scale To see this coincidence, we must examine a Carnot cycle in a low-density gas: (a An isothermal expansion from an initial state (,V 1,n to a state (,V 2, n: 0 =!U 12 = Q (" 1 + W 12 # Q (" 1 = $W 12 = nr" 1 V 2 & % dv = nr" V 1 ln V 2 ( + ' *, V 1 V 1 where we have used an experimental result that the internal energy of low-density depends only on temperature so that the internal energy remains constant in a quasistatic isothermal process (b An adiabatic expansion from a state (,V 2, n to a state (! 2,V 3,n Along this adiabatic process, the first law becomes du =!Q qs +!W qs = "PdV = " nr# V dv,
where!q qs = 0 Experiments have also shown that for a low-density gas, we always find a temperature range where its heat capacity at constant volume is practically constant or as well as so that C V! (! C V "!U % $ ' #!V & # = "U & % ( $ "! ' (, n ( # d! = "U & % ( $ "! ' = 0 V,n V,n = const # d! + "U & % ( dv = du = nr! $ "V '!, n V dv We thus have (! C V d! = du = " nr! V dv By solving this differential equation, we find (! (! nr/ C! V nr/ C 2 V 3 =! V 1 V 2 (c An isothermal compression from a state (! 2,V 3,n to a state (! 2,V 4,n: 0 =!U 34 = Q (" 2 + W 34 # Q (" 2 = $W 34 = $ nr" 2 & % dv = $nr" V 2 ln V 4 ( + ' * (d An adiabatic compression from a state (! 2,V 4,n to a state (,V 1,n As in (b, we find V 4 V 3 V 3 (! (! nr/ C! V nr/ C 2 V 4 =! V 1 V 1 We therefore obtain where we have used = Q! 2 Q V 2 V 1 = V 3 V 4, # "nr! 2 ln V & 4 % $ V ( 3 ' # nr ln V & 2 % ( $ ' V 1 #! 2 ln V & 3 % $ V ( = 4 ' # ln V & 2 % ( $ ' V 1 =! 2,
which follows from (! nr/ C! V 2 V 4 (! nr/c! V 1 V 1 = 1 =! 2 (! nr/ C V V 3! nr/ C! V 1 V 2 Q ( =! 2 we have just derived is identical to Q ( = T (! 2 T ( so that the ideal gas temperature scale must coincide with the The equation Q! 2 Q! 2 absolute temperature scale 1185 The modern statement of the second law The modern statement of the second law distinguishes between reversible and irreversible processes using entropy as follows: For an adiabatic process from equilibrium state 1 to equilibrium state 2: (1 the process is reversible if and only if the entropy difference vanishes (ie,!s = S 2 " S 1 = 0 ; (2 the process is irreversible if and only if the entropy difference is positive (ie,!s > 0 Using Kelvin s statement of the second law, we will prove the following statement that is equivalent to the modern statement: For a pair of equilibrium states 1 and 2: (1 If S 2 = S 1, if and only if there exists a quasi-static (ie, reversible adiabatic process from state 1 to state 2; (2 If S 2 > S 1, then there exists an adiabatic process from state 1 to state 2; (3 If S 2 < S 1, then there is no adiabatic process from state 1 to state 2
Proof (1 If S 2 = S 1, then the states 1 and 2 are in the same adiabat and therefore there is a quasistatic adiabatic process from state 1 to state 2 Conversely, if there is a quasi-static adiabatic process from state 1 to state 2, then these states 1 and 2 are in the same adiabat so that S 2 = S 1 (2 Assume S 2 > S 1 Suppose that the equilibrium state 1 is at T 1,V 1,n equilibrium state 2 is at ( T 2,V 2,n while the We can then come up with a sequence of adiabatic processes from state 1 to state 2 as follows First, we follow the adiabat going through state 1 until the volume becomes V 2 (ie, ( T 1,V 1,n! ( T 3,V 2, n As the entropy remains constant along the adiabat, S 3 = S 1 so that S 2 > S 3 because S 2 > S 1 Entropy is a monotonically increasing function of the temperature and therefore T 2 > T 3 The internal energy is also a monotonically increasing function of the temperature so that U 2 > U 3 We can then perform adiabatic work with no net change of volume W = U 2! U 3 on the system to bring the system adiabatically from state 3 to state 2 Therefore, we have the sequence of two adiabatic processes from state 1 to state 3 to state 2 Adiabat 3 (3 Assume S 2 < S 1 Suppose that the equilibrium state 1 is at T 1,V 1,n equilibrium state 2 is at ( T 2,V 2,n while the Suppose that there exists an adiabatic process from state 1 to state 2 We will show that this assumption leads to a contradiction with Kelvin s statement of the second law, and we will thus conclude that this assumption is false Consider the following cyclic process starting and ending at state 1 First, we use the assumed adiabatic process from state 1 to state 2 The change in the internal energy after this adiabatic process is given by
!U 12 " U 2 # U 1 = W 12, where W 12 is the work done on the system during this process We then follow a quasi-static isothermal process at temperature T 2 until the system reaches state 3 that is in the same adiabat as state 1 (ie, ( T 2,V 2,n! ( T 2,V 3,n The change in the internal energy after this isothermal process is given by!u 23 " U 3 # U 2 = Q 23 + W 23, where Q 23 is the heat absorbed by the system during this isothermal process while W 23 is the work done on the system As states 3 and 1 are on the same adiabat, S 3 = S 1 so that S 2 < S 3 because S 2 < S 1 According to the definition of an entropy change, we find Q 23 T 2 = S 3! S 2 > 0, which leads to Q 23 > 0 since T > 0 The last leg of the cyclic process is an adiabatic process from state 3 back to state 1 in the adiabat that these two states share The change in the internal energy after this adiabatic process is then given by!u 31 " U 1 #U 3 = W 31, where W 31 is the work done on the system during this process For the entire cyclic process, we get so that 0 =!U 12 +!U 23 +!U 31 = W 12 + Q 23 + W 23 + W 31!W "!( W 12 + W 23 + W 31 = Q 23 > 0, which implies that through this cyclic process, the system absorbs the heat Q 23 from the heat reservoir at temperature T 2 and does work!w on the outside, which contradicts Kelvin s statement Therefore, our original assumption must be wrong and there is no adiabatic process from state 1 to state 2
SUMMARY FOR SEC1184 AND SEC1185 1 We classify all equilibrium states of a system into a set of groups called adiabats, in each of which any state can be connected to any other state by a quasi-static adiabatic process 2 Being the label for adiabats, the entropy of a system must take the same value for all the equilibrium states belonging to a given adiabat and take different values for different adiabats 3 The Carnot theorem Consider two quasi-static isothermal processes going from adiabat A 1 to adiabat A 2 of a system In one of the isothermal processes, the system absorbs heat Q ( at a higher empirical temperature, whereas in the other isothermal process, the system absorbs heat Q (! 2 at a lower temperature! 2 < For any pair of two distinct adiabats of any macroscopic system, the ratio between Q ( and Q (! 2, Q ( Q (! 2 or Q (! 2 Q ( has the same universal value, which is positive We then define the absolute temperature scale T (! for each value of empirical temperature! so that it satisfies = T ( Q Q! 2 T! 2 = 27316 K, where! tr, is the empirical temperature for the triple point of by 4 By setting T! tr water, we define the absolute temperature T! T (! " ( 27316 K Q (! Q! tr 5 We define a small change ds of entropy when a system absorbs a small amount of heat!q qs in an infinitesimal quasi-static process at absolute temperature T to be
ds =!Qqs T 6 Using Kelvin s statement of the second law, we can prove the following statement that is equivalent to the modern statement of the second law: For a pair of equilibrium states 1 and 2: (1 If S 2 = S 1, if and only if there exists a quasi-static (ie, reversible adiabatic process from state 1 to state 2; (2 If S 2 > S 1, then there exists an adiabatic process from state 1 to state 2; (3 If S 2 < S 1, then there is no adiabatic process from state 1 to state 2
Toward the modern statement of the second law The 0-th law! Empirical temperature! The law of conservation of energy + Existence of adiabatic processes with!v = 0! Ordering! : <! 2 if U! ( i,v,n < U (! ( f,v,n = U (! ( i,v,n + W adiabatic! #!U & % ( $!" ' V,n > 0 Kelvin s statement of the second law + All equilibrium states of a system are divided into adiabats! For two isothermal processes between two adiabats: (1 Q ( Q(! 2 > 0; (2 If >! 2, then Q ( > Q (! 2! The Carnot theorem: for "! 2, Q ( Q (! 2 is a function of and! 2 only! The definition of the absolute temperature scale: " Q ( T T! 2 Q! 2 and T (! " ( 27316 K Q (! Q! tr!the definition of entropy as a label for adiabats The entropy difference between 2 adiabats:!s " Qqs T + The third law: S! 0 as T! 0 K! The absolute value of entropy! ds =!Q qs T! du = TdS! PdV! "!S % $ ' #!T & V, n = C V T > 0 "!U % + Existence of adiabatic processes with!v = 0 + $ ' #!T &! The modern statement of the second law V,n "!S % > 0 + $ ' #!T & V, n > 0