The Second Law of Thermodynamics Entropy and The Second Law of Thermodynamics SMES1202 khkwek

A cup of hot coffee does not get hotter in a cooler room Direction of a Process A cup of hot coffee does not get hotter in a cooler room Transferring heat to a paddle wheel will not cause it to rotate Transferring heat to a wire will not generate electricity SMES1202 khkwek

The first law places no restriction on the direction of a process Processes occur in a certain direction, and not in the reverse direction The first law places no restriction on the direction of a process Satisfying the first law does not ensure that the process will actually occur There is a natural principle in addition to the first law which determines the direction in which a natural process will take place The second law of thermodynamics The reversed processes discusses earlier violates the second law This violation is easily detected with the help of a property called entropy (devised by Clausius) ONE WAY SMES1202 khkwek

The second law also tells us : that energy has quality as well as quantity how to determine the theoretical limits for the performance of commonly used engineering systems e.g. heat engines how to predict the degree of completion of chemical reactions SMES1202 khkwek

In terms of entropy, the second law can be stated: Processes in which the entropy of an isolated system would decrease do not occur. Or In every process taking place in an isolated system the entropy of the system either increases or remains constant. Note carefully that the statements above apply to isolated systems only. It is quite possible for the entropy of a nonisolated to decrease in an actual process, but it will be found that the entropy of other systems with which the first interacts increases by at least an equal amount. SMES1202 khkwek

Entropy Q2 is the heat flow into the system Q1 is the heat flow out of the system  the heat flows have opposite signs Hence, for a Carnot cycle, we should write: or Expressed in terms of the absolute values |Q1|and |Q2|. SMES1202 khkwek

P V isotherm adiabat P V An arbitrary reversible cyclic process can be approximated as closely as desired by performing a large number of small Carnot cycles, all traversed in the same sense. Those adiabatic portions of the cycles which coincide are traversed twice in the opposite directions and will cancel. The outstanding result consists of the heavy zigzag line. SMES1202 khkwek

or For Carnot cycle abcd: Similarly, for Carnot cycle efgh: If a similar equation is written for each pair of isothermal curves bounded by the same two adiabatic curves and if all the equations are added, then the result obtained is that: or SMES1202 khkwek

The sum can then be replaced by an integral: In the limit, as the cycles are made narrower, the zigzag processes correspond more and more closely to the original cyclic process. The sum can then be replaced by an integral: where d’Qris the infinitesimally small quantities of heat extracted and rejected, and the subscript “r” serves as a reminder that the result above applied to reversible cycles only. Recall that: dV or dU are exact differentials Therefore, the ratio d’Qr/T is also an exact differential. SMES1202 khkwek

Then in any cyclic process, Therefore, can define a property S of a system whose value depends only on the state of the system and whose differential dS is: Then in any cyclic process, For any path between states a and b : Unit: J K-1 Define specific entropy, s : Unit: J kmol-1 K-1 Unit: J kg-1 K-1 SMES1202 khkwek

< Clausius inequality Clausius inequality Recall that: Efficiency of an irreversible cycle, η’ Efficiency of an irreversible cycle, η < Low-temperature reservoir at TL  or Following the same reasoning as before, it is clear that for an arbitrary cycle that is partly or wholly irreversible, Thus: Reversible: Clausius inequality Irreversible: Sometimes taken as a statement of the second law SMES1202 khkwek

Entropy changes in reversible processes 1) Adiabatic process Any adiabatic process: Therefore, for any reversible adiabatic process: and i.e. S is constant  isentropic process 2) Reversible isothermal process T = constant Therefore, in a finite reversible isothermal process: SMES1202 khkwek

T1 T2 Q T1 T2 Q T2 > T1 and T2-T1 = ΔT = infinitesimal System Heat reservoir Q T2 > T1 and T2-T1 = ΔT = infinitesimal Heat flows into system, Qr is positive  Sb - Sa > 0 i.e. Entropy of system increases T1 T2 System Heat reservoir Q T2 < T1 and ΔT = infinitesimal Heat flows out of system, Qr is negative  Sb - Sa < 0 i.e. Entropy of system decreases SMES1202 khkwek

System undergoing phase change A common example of a reversible isothermal process: A change of phase at constant P and constant T System undergoing phase change Qr = heat of transformation, L In terms of per unit mass or per mole: Example: liquid water  water vapour At atmospheric P and T=373K : l23 = 22.6 x 105 J kg-1 The specific entropy of the vapour exceeds that of the liquid by: SMES1202 khkwek

Therefore need to evaluate to find In most processes, reversible heat flow is accompanied by a change in temperature. Therefore need to evaluate to find a) If process takes place at constant volume: Heat flow is:  If cv is constant: b) If process takes place at constant pressure: Heat flow is:  If cP is constant: SMES1202 khkwek

To raise the temperature from T1 to T2 reversibly, we would require a large number of reservoirs having temperatures T1+dT, T1+2dT,…,T2-dT, T2. T1 T1+dT System Heat reservoir dQ T1+dT T1+2dT System Heat reservoir dQ T2-2dT T2-dT System Heat reservoir dQ T2-dT T2 System Heat reservoir dQ SMES1202 khkwek

In any process where Q is reversible: Tsystem ≈ Tsurroundings T2 ≈ T1 = T Reversible heat flow into system: Qr = Qo  heat flow into surroundings:Qr = -Qo Qo T1 T2 system surroundings Isolated System + surrounding = Universe entropy change of the surroundings is equal and opposite in sign to that of the system for reversible process SMES1202 khkwek

How much does its entropy increase? (heat capacity of water =4.2 kJ kg-1 K-1) +heat 500g of water at 20oC 500g of water at 100oC 500g of water at 100oC +heat 500g of water vapour at 100oC Heat of transformation, l23 = 22.6 x 105 SMES1202 khkwek