But how, in the i rst place, are entropy functions produced? One way to pro-duce an entropy function S ( E , V ) is to work backwards from known equations of state by integrating ( ∂ S/ ∂ E) T = 1/ T and ( ∂ S/ ∂ V) E = P / T . Suppose, for instance, we start with the equations of state for equilibrium or blackbody radiation
P E
= V
3 (1.35)
and
E = aVT 4 , (1.36)
where a is a universal constant called the radiation constant . Blackbody radia-tion can be considered a simple l uid because it occupies a volume V and exerts a scalar pressure P . Solving ( 1.35 ) and ( 1.36 ) for P / T and T in terms of E and
1.9 Equations of state 25 where in this case the “integration constant” f ( V ) is an undetermined function of V . Similarly integrating ( 1.38 ) in V while holding E constant produces requiring that the two expressions for the entropy, ( 1.40 ) and ( 1.41 ), lead to the same result. Thus we set the right-hand sides of ( 1.40 ) and ( 1.41 ) equal to each
We summarize. The entropy is an additive state variable that is a function of the energy E and the other extensive variables of the system. The partial deriva-tives of the entropy function in terms of its extensive variables generate the equations of state of the system. In particular,
(
∂ ∂∂E)
V =1T . Alternatively, we can produce the entropy function of a system from its equations of state.Entropy could also be expressed in terms of a mixture of extensive variables and intensive variables. However, expression exclusively in terms of extensive variables E , V , and mole number n , or particle number N , is for a simple l uid the most fundamental. For this reason we will emphasize the functions S ( E , V ) and S ( E , V , n ) or S ( E , V , N ). In other texts expect to see entropy expressed in terms of a great variety of variables.
The next section introduces a requirement on the entropy function imposed by the third law of thermodynamics.
Example 1.6 Modifying the ideal gas
Problem : Suppose we modify the ideal gas pressure equation of state by intro-ducing attraction between molecules that reduces the pressure exerted by the l uid. We use the parameter a where a is independent of any thermodynamic variable to mediate this attraction. Our modii cation is
P nRT to make this pair of equations of state consistent with the i rst and second laws of thermodynamics?
Solution : There is no infallible algorithm for answering this question. We start by guessing that the modii cation takes the form
E C TC TVV f V
( )
,where we insist that f ( V ) is chosen so that the i rst and second laws of thermo-dynamics are satisi ed and that CV remains a constant. In particular, we solve the two displayed equations of state for 1/ T and P / T in terms of the independent
Completing these partial derivatives we i nd that f ( V ) must satisfy the differ-ential equation
d
dV f V an
( )
= V22.1.10 The third law of thermodynamics 27
A solution of this equation is f ( V ) = − an 2 / V . Therefore, the modii ed energy equation of state is, up to an arbitrary constant, given by E = C V T − an 2 /V .
1.10 The third law of thermodynamics
Energy and entropy are thermodynamically parallel concepts. Each is an addi-tive state variable. And the existence of each is implied by the laws of ther-modynamics: energy by the i rst law and entropy by the i rst and second laws.
Furthermore, these laws identify ways of measuring their increments. Recall that dE = δ Q + δ W and dS = δ Q/T . Thus, one can measure the energy and entropy increments ∆ E and ∆ S by carefully controlling and measuring the energy transferred to or from a system by heating and cooling or by having work done on or by a system.
Yet energy and entropy play different roles within the science of thermody-namics. The energy of an isolated system remains constant, while the entropy of an isolated system cannot decrease. The third law of thermodynamics identi-i es another behavidenti-ior of entropy that has no energetidenti-ic parallel.
The empirically supported, physical content of the third law is: The entropy of a system approaches a dei nite and i nite value that is independent of the other thermodynamic variables as its thermodynamic temperature approaches abso-lute zero . Max Planck (1858–1948) added to this empirical content an extremely useful convention: The entropy S → 0 as the thermodynamic temperature T → 0.
A careful analysis of Planck’s convention reveals that it actually consists of two statements: (1) the entropy of all systems approach the same constant as T→ 0 , and (2) the entropy approached is S= 0 . The empirical content of the third law plus Planck’s convention sum to the following version of the third law: The entropy of every thermodynamic system approaches zero as its thermodynamic temperature approaches absolute zero . Symbolically, S → 0 as T → 0.
The physical content of the third law constrains the low temperature behav-ior of thermodynamic systems. Furthermore, Planck’s convention makes it possible to dei ne the absolute entropy of a system and so to produce tables of absolute entropies. The third law also has an interesting quantum statistical interpretation – a subject we take up in Chapter 4 .
An illustration
Here we simply observe whether or not the entropy of various model systems obeys the third law of thermodynamics, that is, whether or not S → 0 as T → 0.
For example, the entropy of an ideal gas is, in its most general form,
S E
(
,V)
nR nnRl VVVV CCVlnEE+ c, (1.43) where c stands for a constant independent of the any thermodynamic variable.We use the energy equation of state E = C V T to eliminate E from the entropy function ( 1.43 ) in favor of T . Doing so produces
S T
(
,V)
nR nnRl VVVV CCVlnTT+ c, (1.44) where the constant c has been redei ned. As T → 0, this function diverges to negative ini nity. Thus, there can be no, conventionally chosen, i nite value of the constant c that makes S → 0 as T → 0. Therefore, the entropy of an ideal gas violates the third law of thermodynamics.However, the ideal gas and other non-third-law-compliant models remain useful in high temperature regimes. If a model system violates the i rst or sec-ond laws of thermodynamics, it is useless and we discard it. But if a model vio-lates the third law of thermodynamics, we simply recognize its limited validity.
For this reason the third law of thermodynamics is neither as foundational nor as consequential as the i rst and second laws. Even so, the empirical content of the third law is invariably observed and cannot be derived from the other laws of thermodynamics.
Example 1.7 The third law and blackbody radiation
Problem: Determine whether or not the blackbody radiation equations of state obey the third law of thermodynamics.
Solution: The blackbody radiation equations of state are P = E /3 V and E = aT 4. According to Eq. ( 1.42 ) these imply the entropy function S E V
(
,)
bb+4a V1 4V1 4E3 4 3 where b is a constant independent of E and V . Using the energy equation of state E aT4 to eliminate E in S ( E , V ) in favor of the temperature T produces S T V(
,)
bb+4aVV1 4T 3. Therefore, S b as T → 0 and the constant b can be chosen to be zero. The blackbody radiation equations of state obey the third law.Problems
1.1 Compressing a l uid
A weight falls on a piston and irreversibly and adiabatically compresses the l uid system contained in the piston chamber ( Figure 1.11 ). Does the entropy of the l uid system increase, decrease, or remain the same and why?
Problems 29
1.2 Derivation
Derive Eqs. ( 1.6 ) from ( 1.5 ). (See hint in text.)1.3 Heat capacity
An object with constant heat capacity C
[
=]
absorbs heat and its tem-perature rises from TTT to Ti T . What is its entropy increase? f1.4 Isothermal compression of ideal gas
Consider an ideal gas whose pressure equation of state is PV=nRT and whose internal energy E E T
( )
is an unknown function of temperature. What is the increase in its entropy ΔS [ Sf Si] as the gas volume is changed from an ini-tial value VVV to a i nal value Vi VV while keeping the temperature T constant? (Hint: f Integrate dS=dE TdE T+P dV T .)1.5 Entropy increment
An ideal gas whose equations of state are PV=nRT and E 5nRT 2 evolves quasistatically from state A to state B to state C as shown in Figure 1.12 . Find the net entropy increment ΔS SSC−SSA per mole of ideal gas. (Hint: Use the graph to integrate dS=dE TdE T+PdV T .)
1.6 Valid and invalid equations of state
In the following l uid equations of state the symbols a and b stand for positive constants that are independent of all thermodynamic variables.
Figure 1.11 A weight falls on piston that irreversibly compresses a gas. See Problem 1.1.
(1) PV=aT and E b T (2) P E V/ and E bT (3) P aT and E bTV
(4) P aTsin( / )E V and E aT (bEV) (5) P aE and E bT
(a) Identify which two of these i ve pairs of equations of state do not obey the i rst and second laws by identifying which do not observe the equality of mixed partials ∂
(b) Derive the entropy function of the three pairs of equations of state that do obey the i rst and second laws.
1.7 Entropy function
An entropy function of a simple l uid takes the form S E V
(
,)
aEV2 where a is a positive constant. Derive the two equations of state of this hypothetical simple l uid.1.8 Room-temperature solid
One version of the entropy function of a room-temperature solid is
S E V E the system. (a) Find the energy E and pressure P of the room-temperature solid
6
Problems 31
as functions of V and T . (b) Does this system observe the third law of thermo-dynamics? Why or why not?
1.9 Valid and invalid entropy functions
Some of the following entropy functions S E V
(
,)
for a simple l uid do not result in a positive thermodynamic temperature, some do not observe the third law of thermodynamics, some violate both requirements, and some violate neither requirement. Identify which of the following entropy functions are not valid and indicate the reason or reasons for their invalidity. The constants a > 0 and b > 0 and are independent of all thermodynamic variables. Also P > 0 and V > 0.(1) S ( E , V ) = b ln( EV ) (2) S ( E , V ) = b ln( EV 2 ) (3) S ( E , V ) = b ln( V/E 2 ) (4) S ( E , V ) = b ln( E/V ) (5) S ( E , V ) = a ( EV ) (6) S ( E,V ) = a exp( − EV ) (7) S( ,EV) aV E