Thermodynamics: the Second Law 자연과학대학 화학과 박영동 교수 The Direction of Nature and spontaneity

Thermodynamics: the Second Law 4.1 Entropy The direction of spontaneous change Entropy and the Second Law The entropy change accompanying expansion The entropy change accompanying heating The entropy change accompanying a phase transition Entropy changes in the surroundings Absolute entropies and the Third Law of thermodynamics The statistical entropy Residual entropy The standard reaction entropy The spontaneity of chemical reactions 4.2 The Gibbs energy Focusing on the system Properties of the Gibbs energy

The direction of spontaneous change

Can we convert heat to work completely?

The Second Law 1. Reversible vs Irreversible processes Reversible: A process for which a system can be restored to its initial state, without leaving a net influence on the system or its environment. * idealized, frictionless; * proceeds slowly enough for the system to remain in thermodynamic equilibrium. Irreversible: Not reversible * natural; * proceeds freely, drives the system out of thermodynamic equilibrium; * interacts with environment, can not be exactly reversed

Example: Gas-piston system under a constant temperature * Slow expansion and compression * Rapid expansion and compression

The Carnot cycle and entropy 1) 1 → 2: Isothermal expansion 2) 2 → 3: Adiabatic expansion The cycle operates with one mole ideal gas.

3) 3 → 4: Isothermal compression 4) 4 → 1: Adiabatic compression

4. Work for adiabatic expansion For an adiabatic process for an ideal gas, from chap. 2

The net heat transfer and the net work over the Carnot cycle are: Then, So the system absorbs heat and performs net work in the Carnot cycle, which behaves as a heat engine. 2→3, 4→1,

The Carnot cycle -summary qw∆U∆U∆H∆H∆S∆S 1→2RT h ln(V 2 /V 1 )-RT h ln(V 2 /V 1 )00Rln(V 2 /V 1 ) 2→30cV(Tc-Th)cV(Tc-Th)cV(Tc-Th)cV(Tc-Th)cP(Tc-Th)cP(Tc-Th)0 3→4-RT c ln(V 2 /V 1 )RT c ln(V 2 /V 1 )00-Rln(V 2 /V 1 ) 4→10cV(Th-Tc)cV(Th-Tc)cV(Th-Tc)cV(Th-Tc)cP(Th-Tc)cP(Th-Tc)0 cycleR(T h -T c )ln(V 2 /V 1 )-R(T h -T c )ln(V 2 /V 1 )000

For the Carnot cycle, we can also have: This relationship also hold for the reversed Carnot cycle. This is called the Carnot’s Theorem: The change of S is independent of path under a reversible process. S is a state function, means a system property. It is called entropy. (4.1)

3. The Second Law and its Various Forms To get the second law, we use the Clausius Inequality, i.e., for a cyclic process, which indicates during the cycle, 1)heat must be rejected to the environment; 2)heat exchange is larger at high temperature than at low temperature under reversible conditions; 3)the net heat absorbed is smaller under the irreversible condition than under the reversible condition.

Now, consider two cycles as shown in the plot. For the cycle which contains one reversible process and one irreversible process, (4.2) For the cycle which have two reversible processes, (4.3)

The difference between (4.2) and (4.3) gives Because states 1 and 2 are arbitrary, we have the second law, (4.4) It indicates that the heat absorbed by the system during a process has an upper limit, which is the heat absorbed during a reversible process. Combine (4.1) and (4.4), we have The first law relates the state of a system to work done on it and heat it absorbs. The second law controls how the systems move to the thermodynamic equilibriums, i.e., the direction of processes. or

16 The Second Law of Thermodynamics The Second Law of Thermodynamics establishes that all spontaneous or natural processes increase the entropy of the universe  S total =  S univ =  S sys +  S surr In a process, if entropy increases in both the system and the surroundings, the process is surely spontaneous

Several simplified forms of the second law : 1) For an adiabatic process, (4.4) becomes If the adiabatic process is reversible, then It is also isentropic (S is constant). 2) For an isochoric process, (4.1) becomes Because only state variables are involved, it holds for either reversible or irreversible processes. (4.5) and (4.6) show that : Irreversible work can only increase entropy; heat transfer can either increase or decrease entropy. (4.5) (4.6)

5. Thermodynamic Equilibrium * Consider an adiabatic process, the second law becomes or For an irreversible condition, is the entropy at the initial state. When reaches the maximum, the state is in thermodynamic equilibrium because the entropy can not increase anymore.

Calculate the changes in entropy as a result of the transfer of 100 kJ of energy as heat to a large mass of water (a) at 0°C (273 K) and (b) at 100°C (373 K). (a) ΔS at 0°C (273 K) (b) ΔS at 100°C (373 K)

Heat engines The engine will not operate spontaneously if this change in entropy is negative, and just becomes spontaneous as ΔS total becomes positive. This change of sign occurs ΔS total = 0, which is achieved when the efficiency, η, of the engine, the ratio of the work produced to the heat absorbed, is

Refrigerators, and heat pumps

Refrigerator power No thermal insulation is perfect, so there is always a flow of energy as heat into the sample at a rate proportional to the temperature difference. The rate at which heat leaks in can be written as A(T h – T c ), where A is a constant. Calculate the minimum power, P, required to maintain the original temperature difference? Assume the refrigerator is operating at 100% of its theoretical efficiency. Express P in terms of A, T h, T c.

The entropy change with isothermal expansion

The entropy change with heating In case heat capacity is constant,

The entropy change with heating In case heat capacity is temperature dependent,

The entropy change with a phase transition At phase transition, the temperature stays constant, T = T tr.

Absolute entropies and the Third Law of thermodynamics S (0) = 0 for all perfectly ordered crystalline materials.

ch04f10 Standard Molar Entropies In general, the more atoms in its molecules, the greater is the entropy of a substance

The statistical entropy U = 4ε

U = 0U = εU = 2εU = 3ε W = 1 W = 4 W = 10 W = 20 U W A simple 4-particle system (in equal-spaced energy levels) Energy S = k B ln( W ) W = C(n+3, n) = (n+3)(n+2)(n+1)/6 U = n ε

A simple 4-particle system W = C(n+3, n) = (n+3)(n+2)(n+1)/6 U = n ε S = k B ln( W )

U = 0U = ε A two-level system Two–Level system a very simplified version Energy In case when ε >> kT, almost no particle can reach upper state. According to Boltzmann, the probability of a particle at level 1 at T is p 1 = exp(- ε/kT ). p 1 = exp(- ε/kT )

Two–Level system a very simplified version U(T) = N ε exp(- ε/kT ) C v,m = (∂U/ ∂T) = N ε exp(- ε/kT ) (- ε/k) (- 1/T 2 ) = N ( ε 2 /kT 2 ) exp(- ε/kT ) ≈ 0 1. This is why we do not consider heat capacity contributions from subatomic particles such as electrons, etc. 2. This is why heat capacity approaches 0 when T approaches 0 for every material. ε1.66E-19"=100kJ/mol" kBkB 1.36E-23 J/K T, KU(T), J/molC v (T), J/mol E E E E-03

W and Energy Level Spacing

Residual entropy

Ice at 0 K

The spontaneity of chemical reactions at constant p, q p =ΔH. q if we define G ≡ H – TS, Gibbs Free Energy, ΔG = ΔH –TΔS at constant T. So, at constant T and p, ΔG = ΔH –TΔS ≤ 0.

Gibbs Energy, G

Calculate the change in molar entropy when one mole of argon gas is compressed from 2.0 dm 3 to 500 cm 3 and simultaneously heated from 300 K to 400 K. Take C V,m = (3/2)R p V T = 300 K T = 400 K

Calculate the change in entropy when 100 g of water at 80°C is poured into 100 g of water at 10°C in an insulated vessel given that C p,m = 75.5 J K −1 mol −1.

The enthalpy of vaporization of chloroform (trichloro-methane), CHCl 3, is 29.4 kJ mol −1 at its normal boiling point of K. (a) Calculate the entropy of vaporization of chloroform at this temperature. (b) What is the entropy change in the surroundings?

Suppose that the weight of a configuration of N molecules in a gas of volume V is proportional to V N. Use Boltzmann’s formula to deduce the change in entropy when the gas expands isothermally.

Without performing a calculation, estimate whether the standard entropies of the following reactions are positive or negative: