Section 4.5: Entropy

The 2nd Law of Thermodynamics: The Entropy Difference is: (Infinitesimal, quasi-static processes) đQ = TdS or dS = (đQ/T) If the system is brought quasistatically between macrostate a & macrostate b, The Entropy Difference is: Sb - Sa = ∫dS = ∫(đQ/T) (1)

The Entropy Difference Between Macrostate a & Macrostate b: ΔS = Sb - Sa = ∫dS = ∫(đQ/T) (1) Use the definition of heat capacity at constant y, Cy(T)  (đQ/dT)y we can write đQ = Cy(T)dT so that (1) becomes ΔS = Sb - Sa = ∫[(Cy(T)dT)/T] (2)

ΔS = Sb - Sa = ∫[(Cy(T)dT)/T] (2) For a system which goes quasi- statically from macrostate a to macrostate b, the entropy change is: ΔS = Sb - Sa = ∫[(Cy(T)dT)/T] (2) Because S is a state function, ΔS is independent of the process by which the system goes from macrostate a to macrostate b, so this can be very useful.

ΔS = Sb - Sa = ∫[(Cy(T)dT)/T] (2) ΔS is independent of the process which takes the system from macrostate a to macrostate b. For the very special case where the macrostate of the system is characterized only by it’s temperature T, with all other parameters y held fixed, (2) can be written: ΔS = S(Tb) – S(Ta)  ∫[(Cy(T)dT)/T]

Section 4.6: Some Potentially Practical Consequences of the 3rd Law of Thermodynamics!!

The Zero of Energy is Always Arbitrary As we’ve seen, for all applications, only differences in the internal energy ΔĒ matter. This is because, as we know, The Zero of Energy is Always Arbitrary Similarly, for many applications, only Entropy DIFFERENCES ΔS are usually important.

The ABSOLUTE 3rd Law of Thermodynamics However, unlike internal energy, The ABSOLUTE Entropy S of a system can be known in principle from the 3rd Law of Thermodynamics

3rd Law of Thermodynamics: The ABSOLUTE Entropy S of a system can be known in principle from the 3rd Law of Thermodynamics: As the Temperature T  0, the Entropy S  S0, S0  a number independent of all system parameters. Often, S0 = 0 This fact can sometimes be put to practical use, as the following two examples (Reif, pages 145 – 148) illustrate.

1 Mole of Solid Gray Tin (α-Sn) 1 Mole of Solid White Tin (β-Sn) Example 1: 2 Systems: (pages 145-147; overview only!) System A: 1 Mole of Solid Gray Tin (α-Sn) (at room temperature & pressure) System B: 1 Mole of Solid White Tin (β-Sn) Both systems have exactly the same number of tin atoms at the same temperature & pressure!

The Entropy Difference at Room Temperature & Pressure Between System System A: 1 Mole of Solid Gray Tin (α-Sn) (at room temperature & pressure) System B: 1 Mole of Solid White Tin (β-Sn) The physical process to convert System A into System B is horrendously complicated!! However, in this example, The Entropy Difference at Room Temperature & Pressure Between System A & System B can be Calculated!

3rd Law of Thermodynamics! Example 1 (pages 145-147) This problem is solved by using the result that, if the temperature dependence of the heat capacity Cy(T) is known, the entropy difference at two temperatures can be calculated from ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) & by using (1) along with the 3rd Law of Thermodynamics!

3rd Law of Thermodynamics Example 1 (pages 145-147) ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) Use (1) along with the 3rd Law of Thermodynamics

3rd Law of Thermodynamics Example 1 (pages 145-147) ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) Use (1) along with the 3rd Law of Thermodynamics That is, as the Temperature T  0, System A Entropy SA  SA0 = Sg0 & System B Entropy SB  SB0 = Sw0, where SA0  SB0  S0 or Sg0  Sw0  S0

3rd Law of Thermodynamics Example 1 (pages 145-147) ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) Use (1) along with the 3rd Law of Thermodynamics That is, as the Temperature T  0, System A Entropy SA  SA0 = Sg0 & System B Entropy SB  SB0 = Sw0, where SA0  SB0  S0 or Sg0  Sw0  S0 Sw0  Sg0  S0 must be true, since The 3rd Law says that S0 is independent of the system parameters (so it is independent of the crystal phase of Sn!).

ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) Example 1 (pages 145-147) ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) Use (1) & the 3rd Law of Thermodynamics Temperature T  0, System A Entropy SA  SA0 = Sg0 & System B Entropy SB  SB0 = Sw0, where SA0  SB0  S0 or Sg0  Sw0  S0

ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) Example 1 (pages 145-147) ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) Use (1) & the 3rd Law of Thermodynamics Temperature T  0, System A Entropy SA  SA0 = Sg0 & System B Entropy SB  SB0 = Sw0, where SA0  SB0  S0 or Sg0  Sw0  S0 So, for System A: (0 K  300 K) ΔSA = SA(300) – SA(0) = ∫[(CAy(T)dT)/T] For System B: (0 K  300 K) ΔSB = SB(300) – SB(0) = ∫[(CBy(T)dT)/T]

∫[(CBy(T)dT)/T] - ∫[(CAy(T)dT)/T] Example 1 (pages 145-147) For System A: (0 K  300 K) ΔSA = SA(300) – SA(0) = ∫[(CAy(T)dT)/T] For System B: (0 K  300 K) ΔSB = SB(300) – SB(0) = ∫[(CBy(T)dT)/T] But, by the 3rd Law, SA(0) = SB(0) So, for System A (300K) System B(300K) ΔS(300) = SB(300) – SA(300) = ∫[(CBy(T)dT)/T] - ∫[(CAy(T)dT)/T]

+ 1 Mole of Solid Sulfur (S) 1 Mole of Solid Lead Sulfide (PbS) Example 2: 2 Systems: (pages 145-147; overview only!) System A: 1 Mole of Solid Lead (Pb) + 1 Mole of Solid Sulfur (S) (at room temperature & pressure) System B: 1 Mole of Solid Lead Sulfide (PbS) Both systems have exactly the same number of lead & sulfur atoms at the same temperature & pressure!

The Entropy Difference at Room Temperature & Pressure Between System Example 2: 2 Systems: (pages 147-148). System A: 1 Mole of Solid Lead (Pb) + 1 Mole of Solid Sulfur (S) (at room temperature & pressure) System B: 1 Mole of Solid Lead Sulfide (PbS). The chemical reaction to convert System A into System B is horrendously complEX!! However in this example, The Entropy Difference at Room Temperature & Pressure Between System A & System B can be Calculated!

3rd Law of Thermodynamics Example 2 (pages 147-148) This problem is solved by using the result that, if the temperature dependence of the heat capacity Cy(T) is known, the entropy difference at two temperatures can be calculated from ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) & by using (1) along with the 3rd Law of Thermodynamics

3rd Law of Thermodynamics Example 2 (pages 147-148) ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) Use (1) along with the 3rd Law of Thermodynamics

3rd Law of Thermodynamics Example 2 (pages 147-148) ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) Use (1) along with the 3rd Law of Thermodynamics That is, as the Temperature T  0, System A Entropy SA  SA0 = SPb + SS & System B Entropy SB  SB0 = SPbS where SA0  SB0  S0 or SPb + SS  SPbS  S0

3rd Law of Thermodynamics Example 2 (pages 147-148) ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1) Use (1) along with the 3rd Law of Thermodynamics That is, as the Temperature T  0, System A Entropy SA  SA0 = SPb + SS & System B Entropy SB  SB0 = SPbS where SA0  SB0  S0 or SPb + SS  SPbS  S0 SPb + SS  SPbS  S0 must be true, since The 3rd Law says that S0 is independent of the system parameters (so it is independent of the chemical environment of the Pb & S atoms!)

Section 4.7: Extensive & Intensive Parameters

Extensive & Intensive Parameters There are two general types of macroscopic parameters which can specify the system macrostate. Lets illustrate as them as follows. Let y be such a parameter. Start with a single system for which y has a certain value, as in the Figure  y Divide the system into two parts with a partition. The values of the parameter y for the two parts are y1 & y2, as in the Figure : y2 y1

2 possibilities: Extensive Parameter.  y y1 y2 2 possibilities: 1. If y = y1 + y2, the parameter y is an Extensive Parameter. 2. If y = y1 = y2, the parameter y is an Intensive Parameter.

An Extensive Parameter is: Heat Capacity C, Entropy S, ... In other words: An Extensive Parameter is: A system parameter that depends on the extent or size of the system. Examples: Mass m, Volume V, Internal Energy Ē, Heat Capacity C, Entropy S, ...

An Intensive Parameter is: A system parameter that is independent of the extent or size of the system. Examples: Temperature T, Mass Density ρ, Specific Heat C, Pressure p, ... The ratio of any 2 extensive parameters is intensive.