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CHAPTER-20 Entropy and Second Law of Thermodynamics.

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1 CHAPTER-20 Entropy and Second Law of Thermodynamics

2 CHAPTER-20 Entropy and Second Law of Thermodynamics 1.Topics to be covered: 2.Reversible processes 3.Entropy 4.The Carnot engine 5.Refrigerators 6.Real engines

3 Ch 20-2 Irreversible Process and Entropy  Irreversible Process: One way process  If the irreversible process proceed in a closed system in a reverse way, we will wonder about it. A melted piece of ice freezes by itself. Although from energy conservation, the process can proceed in the reverse path i.e.heat released in in the environment in melting ice can be recovered back into ice piece to solidify it. Energy conservation does not prohibit the process to proceed one way or the reverse way.  Which parameter the one ay direction of the process?.  Entropy set the direction of irreversible process and irreversible process proceed through that path in which entropy of the system always increases  Entropy Postulate: If an irreversible process occurs in a closed system, the entropy S of the system always increases

4  Change in entropy: (The arrow of time)-time reversal impossible due to entropy violation  Entropy definition: Two methods: i) Using system’s temperature and heat loss/gain by the system ii) By counting the ways in which the atoms or molecules of the system can be arranged iii) Entropy S: state property like P,V,T, S iv) Change in entropy  S=  Q/T v)  S=S f -S i =  f i  Q/T Ch 20-2 Irreversible Process and Entropy

5 Free Expansion of an ideal gas- Irreversible Process: An ideal gas expands from i to f state in an adiabatic process such that no work is done on or by the gas and no change in the internal energy of the system i.e. T i =T f ; p i V i =p f V f ;W=  E int =0  S=S f -S i =  f i  Q/T Integral from i to f cannot be solved for an irreversible process but can be solved for a reversible process. Since S is state variable (  S if )rev= (  S if )irrev  S irrev =  S rev = S f -S i =  f i  Q/T (irreversible) To find the entropy change for an irreversible process in a closed system, replace that process with any reversible process that connects that connects same initial and final states and calculate its change in entropy Ch 20-3 Change in Entropy

6  Calculation of Entropy in Reversible Process  In free expansion T i =T f, i.e. isothermal process then reversible isothermal expansion process can be used to calculate change in entropy in free expansion. Then   S rev = S f -S i =  f i  Q/T= 1/T  f i  Q (isothermal)  An ideal gas confined in a insulated cylinder fitted with a piston. The cylinder rest on a thermal reservoir maintained at temp T.  Gas is expanded isothermally from I to f state and heat is absorbed by the gas from the reservoir.  S is positive because  Q is positive   S= S f -S i =  Q/T

7 Ch 20-3,4 Change in Entropy, Second Law of Thermodynamics  Entropy as a State function  S= S f -S i =  Q/T but  E int =  Q-W or  Q =  E int +W ; W=pdV;  E int = nCV  T Then  S=   Q/T=  i f (nC V dT/T) +  i f (pdV/T) =nC V  i f (dT/T) +  i f (nRdV/V)  S =nC V ln(T f /T i ) + nR ln(V f /V i )  Second Law of Thermodynamics In a closed system entropy remains constant for reversible process but it increases for irreversible process.  Entropy never decreases Change in Entropy of a reversible compression of an ideal gas in a closed system reservoir + gas   S sys =  S gas +  S res =-Q/T+Q/T=0

8  A Carnot engine is a device that extracts heat QH from a reservoir at temperature TH, does useful work W and rejects heat QL to a reservoir at temperature TL.  On a p-V diagram Carnot cycle ( executed by a cylinder fitted with a piston and in thermal contact with one of two reservoirs at temperatures T H and T L.) is bound by two isotherms and two adiabatic. Ch 20-5 Entropy in Real World: Engines

9 Heat Q H is absorbed at Temperature T H and heat Q L is rejected at T L.  Work: According to first law of thermodynamics,  E int =  Q-W,  E int =0 and W=  Q=  Q H  -  Q L   Entropy Change  S: In Carnot cycle  S=0  S=0= Q H /T H -Q L /T L Q H /T H =Q L /T L

10 Ch 20-5 Entropy in Real World: Engines  Thermal Efficiency of an Engine  :   = work we get /energy we pay for   =  W  /  Q H  For a Carnot engine W=  Q H  -  Q L  Then Carnot engine efficiency  C is   C :  Q H  -  Q L  /  Q H    C =1-  Q L  /  Q H  =1-  T L  /  T H  There is no perfect engine i.e. Q L =0

11 Ch 20-6 Entropy in Real World: Refrigerators

12 A Refrigerator: A device that uses work to transfer energy Q L from a low temperature reservoir at temperature T L to a high-temperature reservoir at temperature T H Coefficient of performance of a refrigerator K K= what we want /what we pay for K=  Q L  /  W  but W=  Q H  -  Q L  K =  Q L  /  W  =  Q L  /  Q H -  Q L  K =  Q L  /  Q H  =  T L  /  T H  -1 K=T L /(T H -T L ) There is no perfect refrigerator i.e. W=0

13 Suggested problems Chapter 20

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