Entropy. Entropy Changes for an Ideal Gas

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1 Entropy and Entropy Changes for an Ideal Gas Ron Reifenberger Birck Nanotechnology Center Purdue University March 28, 2012 Lecture 10 1

2 Recall that we discussed an idealized process called reversible A thermodynamic process is reversible if at any stage, a differential, infinitesimal decrease in the variables causing a change will result in a reversal of the process itself, both in direction and all its quantitative effects. After a reversible change, it is possible to restore both the system and surroundings to their original conditions. A transfer of heat could be reversible if the body accepting the heat were always infinitesimally cooler than the body losing the heat. o transfer a finite amount of energy or to do a finite amount of work by a reversible process would require an infinite time. A reversible process is a limiting case which may be approximated but never actually realized. Reversible processes are most useful for logical analysis from H.C. Weber, hermodynamics for Chemical 2 Engineers, John Wiley and Sons, 1939 (pg. 11).

3 defines entropy difference ΔS Recall that when analyzing the Carnot engine, we found for a reversible process Q out c Q in h 0 Notation: Q out is the heat that flows to the cold reservoir; Q in is the heat that flows from the hot reservoir. Recall Q out is a negative number (1 st Law). 3

4 he thermodynamic function called S entropy is a state function, just like P, V, and E int. P S V V KEY IDEAS Entropy measures the amount of microscopic disorder in a thermal system. Because entropy is a state function, the change in entropy depends ONLY on the initial and final states t of the system, not on the details of the process by which the change occurs. 4

5 Defining Entropy Change ds dq rev 0 dq rev is defined as the heat in a reversible process required to drive the system from an initial to a final state. In principle, the above equation only applies when transfers of energy occur as heat flow and when any volume change occurs quasistatically. Reversible processes do not occur in nature. For irreversible processes, the above equation becomes dq ds actual 0 0 5

6 ENROPY (from the Greek, to turn, as in to turn useful energy into useless energy ) New thermodynamic variable hermodynamic state of a system S (Entropy)? Allowed,? irreversible? Allowed, reversible?? time Forbidden Which processes are allowed? Which processes are forbidden? time New idea: he entropy of a closed system can never decrease. Once the entropy of a system increases, it cannot return to its original value 6

7 Calculating the entropy change of the universe in an isothermal process Replace Q out, Q in notation with dq ds reservoir =-dq/ res res reservoir he system could be a volume of gas system ds system =dq/ system dq system Absorbs heat but temperature tu of system st does not change Heat lost but temperature of reservoir does not change S ds ds u= Universe u system reservoir In a reversible process, the entropy change of the universe (ΔS u ) is zero 7

8 Entropy is NO a conserved quantity in a real process! S ds ds u system res dq system dq res 1 1 dq system res Since heat flows from hot to cold, res > system 8 0

An understanding of entropy gives new meaning to the heat

Assume the entropy of the system decreases.

cold reservoir Can all the heat released be turned into

9 An understanding of entropy gives new meaning to the heat engine diagram Assume some exothermic (heat releasing) reaction occurs inside a system, e.g. chemicals react, fuel ignites, etc. Assume the entropy of the system decreases. ΔS System Environment W ΔS=Q Q c / c Q c, heat lost to cold reservoir Can all the heat released be turned into work? No, because work does not create Entropy? What condition must be satisfied for the process to occur spontaneously? 9

m=1kg =0 o C dq=ml f ; L f =latent heat of ice = 333 kj/kg dq res =20 o C Q ice S ice ice 1kg

10 An example: Entropy Change when Ice Melts he two systems are not isolated, but the combined system is. m=1kg =0 o C dq=ml f ; L f =latent heat of ice = 333 kj/kg dq res =20 o C Q ice S ice ice 1kg 333kJ kg 273 K 1.22 kj / K 1 kg 333 kj kg Qres Sres res 293 K 1.14 kj / K In this case, the colder object gains more entropy than the hot object loses. S kj / K 0.08 kj / K 0 u 10

11 An example where dq/ must be integrated A block of aluminum of mass m at -20 o C is thrown into a lake whose water temperature is 25 o C. What is change in entropy of the aluminum block and of the lake? Entropy change for aluminum i =-20 o C=253K; f =25 o C=298K dq dq malcd and ds lake lake d d dsblock mal c malc block lake Al block block Entropy change for lake dq m cd Al dq d m lake Alc ds lake m Alc d lake lake lake m c Al lake block m c n ds ds ds system block lake lake malcn block lake m c (1 0.85) Al lake lake block 11 m c (0.150) m c Al Al block

12 What about an ideal gas? A process goes from (P 1,V 1, 1 ) to (P 2,V 2, 2 ) What is the change in Entropy? P (P 1, V 1, 1 ) 2 dq in (P 2, V 2, 2 ) V here is ONE equation that accurately describes all these processes de int = dq in + dw on 12 Work/ideal gas

13 General calculation for the Change in Entropy for an Ideal Gas de dq dw dq PdV 1 st Law int in on in de C d Ideal Gas int V dv Cd dq nr Ideal Gas V V in Cd V dqin dv nr V Divide by ds Cd V nr dv V wo contributions to entropy fin Vfin S CV n nrn init Vinit by integration 13

14 EXAMPLE: A weather balloon is filled with 5 moles of Helium at sunrise. Energy is added to the balloon by heat (the sun rises!) and the volume of the balloon increases by 50%.he temperature also increases from 20 o C to 25 o C. What is the change in entropy of the gas if the expansion and temperature rise occur very slowly? Recall, for a monotonic gas like He, C V = 3/2 nr V V fin fin S CV n nrn init Vinit init = = 293K 1.5 V init = V o 293 fin = = 298K 298 V S 1.5nR n nr n o 293 V o 1.5(5)(8.3) n(1.017) 5(8.3) n(1.5) V fin = 1.5 V J / K ( greater than zero) o Note that S does not depend on the process which takes the gas from the initial to final state. 14

15 How much heat did the gas absorb? b? ΔQ <>S = ½(293K + 298K) x J/K = 5280 J 15

16 Free Expansion hree (3) moles of helium gas are confined to the left side of a container. When the valve is open, the volume of the gas doubles in an isothermal, irreversible ibl process. What is the entropy change? 2 V 2 S CV n nrn V S=(3)(8.3) ( ) l ln(2) ( ) = J/K 16

17 Can his Happen? Gas Gas Gas Vacuum hree (3) moles of helium gas fill the inside of a container as shown above. With the valve open, the volume of the gas suddenly decreases by a factor of two by way of an isothermal, irreversible process. What is the entropy change? 0 2 V 2 S C V n nr n 1 V1 S=(3)(8.314) ln(1/2) = J/K he entropy of the system decreases! he proposed process will not occur. 17

18 Special Cases V2 Isothermal Expansion: S nrn V V Free Expansion: 2 S nrn V 1 1 Isobaric Expansion: 2 S CP n 1 Heat Conduction: Q S c Q h Carnot Engine: S 0 18

19 he idea of entropy is also useful because it allows you to make a quantitative judgment on the quality of stored energy. For example, if you have 1kJ of energy, is it better to store that energy in a hot reservoir at 100 o C or in a cold reservoir at 0 o C? S hot 1000 J J J Scold 3.66 J 373 K K 273 K K he amount of energy stored is the same, but what you can do with it depends on its entropy: High quality energy (capacity to do work) has a low entropy value Low quality energy (less capacity to do work) has a high entropy value 19

20 he Second Law of hermodynamics After many, many, many years of experiments, observation showed that heat never flows from cold to hot, but always from hot to cold. his could NO have been predicted in advance, since energy is conserved in either process. In the late 1850 s, there was a sentiment that this indisputable Fact should be elevated to a Law. So, the 2 nd Law of hermodynamics was formulated: Any process that solely transfers heat from a low temperature to a higher temperature is impossible. his statement is very restrictive! It is limited to the transfer of heat from an object with one temperature to an object with a different temperature. his statement implies there are no other changes in the system st other than the transfer r of heat. 20

21 From this simple statement, t t a number of involved deductions can be made that ultimately leads to the concept of entropy and culminates with Boltzmann s famous equation: S=k B ln(w) or w = e S/k B where w = number of microstates possible consistent with a specified total energy and k B =1.38 x J/K. Entropy is just a logarithmic measure of the number of microstates available to a system. 21

22 he concept of ENROPY sheds light on a number of broad conceptual issues associated with thermodynamic processes involving an ideal gas. For example, consider the kinetic theory for gases. 22

23 wo problems with the Kinetic heory of Gasses 1. Why are so many thermodynamic processes (for example, heat flow) irreversible? he predictions of motion using Newtonian Mechanics are symmetric under time reversal (i.e they are reversible for t -t) If gas properties p are controlled by individual gas atoms interacting according to the Laws of Newton, then the interaction between gas atoms should be reversible Yet many (virtually all!) thermodynamic processes involving a gas are irreversible? 2. How do randomness and unpredictability on the atomic scale give rise to order and predictability at the macroscopic scale of matter? NEW IDEA: Maybe we can define a new thermodynamic variable that can predict in advance whether a certain thermodynamic process is allowed? 23

24 Entropy and Multiplicity Entropy of a closed system increases or remains the same. From a microscopic i point of view, Boltzmann argued that t an increase in Entropy is associated with the number of ways you can arrange a microstate without affecting a given macrostate. From this point of view, it is more correct to associate entropy with the number of possible ways to arrange constituents of a system i.e, the multiplicity of equivalent states t of a system. Consider your dorm room why does it tend to get disorderly? Because there are many, many more ways for your room to be disorderly then orderly. he large multiplicity of disorderly states suggests that random things that occur in your room are FAR more likely to contribute to the state of disorder d than to the state of order. 24

25 Caution: Associating the word disorder with the word entropy can be misleading.. because disorder has all kinds of aesthetic, moral, and political connotations. Really, entropy just measures the number of microstates available. 25

26 KEY IDEA Irreversible processes (e.g. heat flowing from hot to cold) occur naturally because there are many, many, many ways for heat to flow from hot to cold. In contrast, there are only a very few ways that heat could flow from cold to hot. In principle, according to conservation of energy considerations, heat energy could flow either way. he multiplicity of ways for heat to flow from hot to cold strongly favors this one direction over its opposite. It is really that simple! 26

27 Further implications of entropy In any irreversible (one way) process, energy is conserved, but some of the energy becomes wasted i.e. it becomes unavailable to do work. he amount of energy in the world is NO decreasing, but its availability (and ability) to do useful work is! It is impossible to convert a given amount of heat energy completely into work. In any energy conversion process, energy is always degraded in quality, so that its ability to do work is reduced. 27

28 he Second Law of hermodynamics can be re-stated as: he entropy of an isolated system never decreases 28

29 Why is the Entropy Change of a Cyclical Process Zero? o find the entropy change for a complicated cyclical process occurring in a closed system, replace the process with a series of mini-reversible isothermal/adiabatic closed cycle processes as shown: mini Carnot cycle: dq 1 1 P (P, V, ) arbitrary closed path thermodynamic process Q=0 Q=0 2 Reversible Processes... dq dq 2 dq V 29

Class 22 - Second Law of Thermodynamics and Entropy

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