Atkins / Paula Physical Chemistry, 8th Edition Chapter 3. The Second Law
The direction of spontaneous change 3.1 The dispersal of energy 3.2 Entropy 3.3 Entropy changes accompanying specific processes 3.4 The Third Law of thermodynamics Concentrating on the system 3.5 The Helmholtz and Gibbs energies 3.6 Standard reaction Gibbs energies P.76
Combining the First and Second Laws 3.7 The fundamental equation 3.8 Properties of the internal energy 3.9 Properties of the Gibbs energy P.76
Some aspect of the world determines the spontaneous direction of change, the direction of change that does not require work to be done to bring it about. 2nd Law of thermodynamics (Kelvin) No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. Chapter 3. The Second Law P.77
3.1 The dispersal of energy Chapter 3. The Second Law P.77
3.1 The dispersal of energy (cont d) Chapter 3. The Second Law P.78
3.2 Entropy 2nd Law of thermodynamics: The entropy of an isolated system increases in the course of a spontaneous change: S tot > 0 S tot - total entropy of the system and its surroundings (a) The thermodynamic definition of entropy
Ex 3.1 Calculating the entropy change for the isothermal expansion of a perfect gas Calculate the entropy change of a sample of perfect gas when it expands isothermally from a volume V i to a volume V f. Answer At const temperature From eqn 2.11 When the volume occupied by 1.00 mol of any perfect gas molecules is doubled at any const temperature, V f /V i = 2 and
Illustration 3.1 Calculating the entropy change in the surroundings To calculate the entropy change in the surroundings when 1.00 mol H 2 O(l) is formed from its elements under standard conditions at 298 K, use H = 286 kj from Table 2-7. The energy released as heat is supplied to the surroundings, now regarded as being at constant pressure, so q sur = +286 kj. Therefore, This strongly exothermic reaction results in an increase in the entropy of the surroundings as energy is released as heat into them. Test 3.2 Calculate the entropy change in the surroundings when 1.00 mol N 2 O 4 (g) is formed from 2.00 mol NO 2 (g) under standard conditions at 298 K. Answer: 192 J K 1
3.2 Entropy Boltzmann distribution: Boltzmann formula statistical entropy Chapter 3. The Second Law P.81
3.2(b) The entropy as a state function (eqn 3.6) 1. Show that eqn 3.6 is true for a special cycle (a Carnot cycle ) involving a perfect gas. 2. Show that the result is true whatever the working substance. 3. The result is true for any cycle. Chapter 3. The Second Law P.82
Carnot cycle 1. Reversible isothermal expansion from A to B at T h ; the entropy change is q h /T h, q h is the energy supplied to the system as heat from the hot source. 2. Reversible adiabatic expansion from B to C. No energy leaves the system as heat, change in entropy is 0. Temp. falls from T h to T c, temp. of cold sink. Chapter 3. The Second Law P.81
Carnot cycle (cont d) 3. Reversible isothermal compression from C to D at T c. Energy is released as heat to the cold sink; the change in entropy of the system is q c /T c ; in this expression q c is negative. 4. Reversible adiabatic compression from D to A. No energy enters the system as heat, so the change in entropy is zero. The temperature rises from T c to T h.
For perfect gas eqn 3.7: Heating accompanying reversible adiabatic expansion The two temperatures in eqn 3.7 lie on the same adiabat. For a perfect gas: From the relations between temperature and volume for reversible adiabatic processes (eqn 2.28): V A T hc = V D T c c V c T cc = V B T h c Multiplication of the first of these expressions by the second gives V A V c T hc T cc = V D V B T hc T c c
efficiency, ε (epsilon), of a heat engine: Chapter 3. The Second Law P.83
Chapter 3. The Second Law P.84
Chapter 3. The Second Law P.84
Chapter 3. The Second Law P.84
Refrigeration coefficient of performance minimum power Chapter 3. The Second Law P.85
3.2 Entropy (c)the thermodynamic temperature (d) The Clausius inequality More energy flows as work under reversible conditions than under irreversible conditions. i.e., dw rev dw, or dw dw rev 0. dq rev dq = dw dw rev 0, or dq rev dq, so dq rev /T dq/t