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  1. Entropy heat engine Th=T2 Tc=T1 Let’s consider the efficiency of a Carnot engine again Taking into account and Clausius’ Theorem Generalization of the above considerations

  2. WD T2 T1 TN CN C2 C1 TR Clausius’ Theorem: The heats Qi exchanged at absolute temperatures Ti by a cyclic process of N steps satisfy . Note: the Qi might be irreversibly exchanged Theorem applies to any real engine cycle Consider a general cyclic device in thermal contact with a single heat reservoir Proof of the Clausius theorem: General cyclic Device

  3. Ti Ci TR General cyclic Device WD T2 T1 TN CN C2 C1 TR With a particular sign convention for work and heat we obtain: Consider convention for the case Ci is a Carnot engine: Flowing from the Carnot engine into the device at the lower temperature Ti Heat Work done by the engine on the surrounding Heat Flowing from the hot reservoir to the engine Consider our combined system as a black box

  4. The net effect is: Exchange work Exchange heat TR Kelvin statement: It is impossible to take heat from a single reservoir and change it entirely into work (we are allowed to do nothing or to do work on the device and transfer it entirely into heat)

  5. Heat exchanged by Carnot cycles According to our above sign convention we have Proof of Clausius’ Theorem Holds in general also for irreversible devices

  6. What happens in the particular case of a reversible device Corollary (of Clausius’ theorem): For an arbitrary reversible process . Proof: Let device operate in reversed direction and apply the Clausius theorem while

  7. Apply Clausius theorem to the device in forward direction yields original result With and

  8. Let’s consider now a reversible cyclic process (represented as closed contour in state space) isotherms P adiabats V Reversible cyclic process can be represented by subsequent Carnot cycles

  9. inner isothermal expansion and compression processes cancel mutually out Approximation of the closed contour by N steps with In the limit N From the exact differential theorem we conclude is exact dS is the differential of a state function called entropy S.

  10. Consequences of Existence of Entropy Entropy determined up to an additive constant ( which will be specified later by the 3rd law) Since is extensive, S and dS are extensive In some cases an inexact differential can be changed into an exact one by multiplying with an integrating factor is the integrating factor which makes exact (for more info go to Exact differentials and theory of differential equations) Relationship between the internal energy and the equation of state

  11. = dS exact Calculation of the derivatives yields P=P(T,V) Obviously: U=U(T,V) and not independent

  12. Example: We show for the ideal gas: is a consequence of U=U(T) 1 U=U(T) From is derived from U=U(T) 2 U=U(T) U independent of V

  13. CV Heat capacity in terms of entropy and With 0 Let’s take advantage of the new notation using differential forms @ V=const. dV=0

  14. V=const. P=const. Q m Q Let’s summarize the various representations of heat capacities for PVT systems Constant volume Constant pressure where H=U+PV

  15. Explicit expression for the entropy of an ideal gas for the ideal gas in general for the ideal gas Comparison (in order to obtain dimensionless argument of the logarithm introduce a reference state (Tr,Vr) ) (T,P) Change from (T,V) (P,V) Easily done with PV=nRT

  16. Isentropic processes isentropic Reversible adiabatic processes are isentropic processes that is, processes in which the entropy is constant. S=const. Isentropic processes in ideal gases

  17. A x Adiabatic bulk modulus We know already the isothermal bulk modulus Reminder: x P2=P1+P 0 x Proportionality constant defines BT T=const. We now define the adiabatic bulk modulus and the adiabatic compressibility

  18. Let’s calculate Since derived for Summary of important measurable properties of PVT systems Thermal expansion volume coefficient Isothermal bulk modulus Adiabatic bulk modulus Constant pressure heat capacity Constant volume heat capacity Easier to measure than CV measured in experiment from speed of sound

  19. Remember: Existence of entropy provides relationships between material properties Consider with with derived from general considerations

  20. X S S Y P V Z T T Second relationship looking for CP/CV expressed in terms of BS and BT Consider again and let’s calculate Applying for and

  21. Entropy and the Second Law Entropy never decreases We are going to show the meaning of the sloppy phrase: Therefore: Reminder: Clausius´ theorem Heat exchanged In a cyclic process of N steps @ absolute temperatures How to calculate the entropy change involved in a real (irreversible) process Answer: Find a reversible equilibrium process that takes the system between the initial and the final states of the actual processes.

  22. (hence we can later take advantage of Clausius´theorem ) Consider a cyclic process State space spanned by (P,V) for instance but not necessarily P cannot be represented by a line In state space real process 1 2 reversible process V Now let´s apply Clausius´ theorem

  23. Clausius Consider the reversible equilibrium part of the total cyclic process start “0” of real final “f” of real P 1 2 reversible process final of rev start of rev reversible heat exchange at T V System with single temperature adiabatic: Qi=0 Consider the real process Restrict to an adiabatic process

  24. Entropy statement of the second law: The total entropy of an adiabatically isolated system never decreases. The idea that entropy will increase, if it can is emphasized in Clausius summary of the second law: The entropy of the universe* tends toward a maximum. *Universe is a synonym for a completely isolated system (in contrast to adiabatically isolated systems of the above statement )

  25. Clausius statement and the principle of increase of entropy adiabatically isolated -Entropy change of the high temperature bath. -Entropy change of the low temperature bath. Heat is leaving the low temperature bath Total entropy change because Existence of a non-Clausius device violates entropy statement of 2nd law

  26. Kelvin statement and the entropy statement of the 2nd law adiabatically isolated -Entropy change of the high temperature bath: -Device: only effect is to accept heat from the reservoir and transform it entirely into work Device in the same state at the start and finish of the process Total entropy change Existence of a non-Kelvin device violates entropy statement of 2nd law

  27. Time’s arrow S1 S2 Entropy and Irreversibility Consider a process in an adiabatically isolated system where process allow by 2nd law Reversed process: Start and final state become exchanged Reversed process forbidden by 2nd law Entropy concept allows to quantify irreversibility S1 < S2 t1 t2

  28. Two examples for irreversible processes with corresponding entropy increase Free expansion of an ideal gas 1 V0 ,T0 Vf ,Tf Thermal insulation gas no heat flow fixed walls no work done by the container W=0 U=U(Tf)-U(T0)=0 With and ln1=0 ln2

  29. Let’s calculate S via an integral in state space using a path, L, of reversible equilibrium processes taking the system from the initial to the final states of the actual process. Fast free expansion is certainly not a sequence of reversible equilibrium processes only initial and final states can be represented in state space We know that free expansion of the ideal gas takes place at T=const.=T0=Tf We can therefore bring the system from (T0,V0) to (Tf,Vf) via an isothermal expansion T=const.

  30. Cu 1kg 2kg Temperature equalization 2 In order to have a specific example consider: No heat exchange with surrounding T0Cu=80C =353.15K T0W=10C =283.15K Final equilibrium temperature no heat exchange with surrounding heat flow into the water>0 heat flow out of the Cu block<0

  31. with Total entropy change Entropy of the Cu block decreases Total entropy of the isolated system increases

  32. Stability of thermodynamic systems According to Clausius: entropy is at maximum in systems in equilibrium at Constant U and V. Consider two systems in equilibrium with and entropy of combined system (not in equilibrium) is + Internal energy: 2U volume: 2V Compare with entropy of combined system is Internal energy: 2U volume: 2V

  33. We know entropy is extensive would evolve into 2 systems with otherwise system with Combined system with internal energy 2U and volume 2V, has entropy in equilibrium which is at maximum Alternative formal approach: We know: Absolute temperature T>0 Moreover, we want S=S(U,V) be a concave function of U S S(U0,V) Equation of the linear function U U0+U U0 U0-U

  34. Consider the limit U->0 With and for U->0

  35. Let’s apply the same consideration for the volume Consider the limit V->0 With For V->0

  36. Finally let’s consider the small changes U and V together (both inequalities yield the same result) Consider the limit U->0,V->0 With e.g., For all U,V in the vicinity of (U,V)=(0,0) Remember from math: sufficient condition for a maximum of f(x,y) at (x0,y0)

  37. For more background about the Hesse-matrix click here & determinant of Hesse-matrix>0 Our problem: in order to make sure (U,V)=(0,0) has to be a maximum of since f(0,0)=0 with > 0

  38. > 0 > 0 While the necessary conditions read What are the physical implications

  39. > 0 see textbook 1 < 0 From 2

  40. A x Intuitive meaning of >0 Consider a volume fluctuation V>0 x with x Pressure decreases 0 External pressure drives system back to equilibrium otherwise T=const.