UNIT 6

Structure 6.1 Introduction i Objectives 6.2 Overview Thc Zeroth Law of Thermodynam~cs I Thc First Law of Thermodynamics Second Law of Thermodynamics Reversible and Irreversible Processes 6.3 The Real Heat Eng~nes Refr~gerators 6.4 Entropy and Its Significance Entropy and the Second Law of Thermodynamics Physical Concept of Entropy Combined Forms of the First and Second Law of Thermodynamics 6.5 Summary 6.6 Terminal Questions 6.7 Solutions and Answers

6.1 INTRODUCTION

How would you like to introduce this subject to your students? You may like to start by explaining that matter (solid, liquid or gas) is composed of a very large number of interacting atoms or molecules that can be regarded as systems of particles. Many processes involving the exchange of between matter and its surroundings can be studied without considering the atomic or molecular structure of matter. The study of such processes is the subject of thermodynamics which was developed during the eighteenth and nineteenth centuries as a rather formal and elegant empirical theory.

Thermodynamics demands greater understanding of concepts on the part of learners. The contents of this unit have been selected keeping in view the problem areas that need to be given greater focus by us in our classroom deliberations. In this context we assume that you are already familiar with the various basic terms and definitions used in the study of thermodynamics. You would know, for example, about (i) the thermodynamic system, (ii) classification of systems, i.e., open and closed systems, (iii) thermodynamic state of a system, (iv) thermodynamic variables, and (v) thermodynafiic equilibrium.

, In this unit we will briefly review the teaching-learning of the three laws of thermodynamics, and the sign conventions used in the development of various thermodynamical relations. We will also discuss different statements of the second law of thermodynamics, and the concept of reversible and irreversible processes, Carnot cycle, real heat engines and refrigerators. Finally, we will discuss the concept of entropy and its significance, and try to link entropy and the second law of thermodynamics.

Objectives

After studying this unit, you should be able to:

help your students learn better the laws of thermodynamics and their applications to various processes; acquire skills in explaining the different statements of second law of thermodynamics and their equivalence; Thermodynamics, Vibrations,Waves and explain with confidence the various steps involved in the derivation of expression Optics for the efficiency of Carnot cycle and extend its concept to real heat engines and refrigerators; assess the learning difficulties of your students vis-a-vis the concepts of thermodynamics and devise strategies to help them overcome these difficulties.

6.2 OVERVIEW

You may like to explain to your students in the very beginning that like classical mechanics, thermodynamics is also based on empirical laws -there is no way to prove them - and is therefore pheromenological. Nevertheless, it is very exact and powerful. Each of these laws introduces a new concept (viz. temperature, internal energy, entropy) which gives a definite meaning to physically measurable quantities and provides useful correlation between observable properties of matter. As mentioned in the introduction, we assume that you are already familiar with the various terms and definitions frequently used in the development and formulation of thermodynamics. These are

i) A thermodynamic system, ii) Concept of surroundings and boundary of a system, iii) State of a system and thermodynamic variables, iv) Thermodynamic equilibrium, and v) Thermodynamic process and its various forms.

We will use these concepts and various terms in the ensuing sections. You must have been using the term temperature quite freely, but you need to emphasise that its basis lies in a law of thermodynamics. Let us study this law which is known as Zeroth law.

6.2.1 The Zeroth Law of Thermodynamics

You know the statement of the Zeroth law of thermodynamics: Iftwo systems are separately in equilibrium with a third system then they must be in thermal equilibrium with one another.

You may have explained how this statement forms the basis of the concept of temperature: All the three systems can be said to possess a property that ensures their being in thermal equilibrium with each other. This property is called temperature. Thus temperature of a system may be defined as the property that determines whether or not the system is in thermal equilibrium with the neighbouring systems. Your students may like to know why it is called the Zeroth law.

Why do we call it the Zeroth Law?

The phenomenon that the two systems in contact tend towards a common temperature is so common that historically its importance had been overlooked. When physicists finally did appreciate its significance and its fundamental nature, it was decided to have it elevated to the status of a "law of thermodynamics". By that time the first and second laws of thermodynamics had already been developea. So in order to place it ahead of these laws, it was termed as 'Zeroth law'.

Equation pv = nrT is quite well known to you. Can such relations exist for other thermodynamic systems? In fact from the Zeroth law, it can be established mathematically that a relation exists between temperature and other thermodynamic variables associated with the system. Such a relation, as you know is called an equation of state. The general equation of state for any system is represented by Thermodynamics where f is a single valued function of pressurep, volume/V and the absolute temperature T. Charles7 and Boyle's laws give such relations for gaseous systems.

6.2.2 The First Law of Thermodynamics

You know that the first law of thermodynamics is the statement of the principle of conservation of energy and applies to every process in nature.

According to the first law of thermodynamics "when some quantity of heat (6Q) is supplied to a system capable of doing external , then the quantity of heat absorbed by the system (6Q) is equal to the sum of the increase in the internal energy of the system (dU) due to rise in temperature, and the external work done by the system (6W) in expansion", i.e.

This equation represents the differential form of the first law of thermodynamics. The first law of thermodynamics thus establishes an exact relationship between heat and work. You have to point out that the statement of this law involves the assumption that the internal energy is a function of the system only, i.e., dUdoes not depend on the path whereas 6Q and 6 Ware dependent on the path of the system followed in going from the initial to the final state. Mathematically, the differentials of such quantities are said to be inexact. It is to emphasize this difference between exact and inexact differentials that we have used different symbols - d for exact and 6 for inexact differentials.

You may like to consider giving simple examples of exadin-exact differentials. For a given function df = (3x2 + 3y) dx + (ex + 2y) dy, you could find c for which df is an exact differential. Suppose you want to integrate df from point (0,O) to (2, 1) along two different paths, one, along a straight line connecting the two points, and two, first along the x-axis from (0, 0) to (0, 1) and then along the y-axis from (0, 1) to (2, 1). You could show that the integrals differ unless c takes the value which made df exact, and which you have already found.

Significance

From the first law of thermodynamics, we learn that it is impossible to get work continuously from any machine without giving an equivalent amount of energy back to the machine. Just as the zeroth law of thermodynamics introduces the concept of temperature, the first law of thermodynamics introduces the concept of internal energy. Thus according to this law, the internal energy (and hence temperature) of a system can be increased by supplying heat to it or by doing work on the system or both.

Sign conventions

The system of using plus/minus sign in dealing with the three quantities 6Q, dUand 6 W involved in the first law of thermodynamics is found to be difficult by our students. You may like to explain it as follows:

(i) When heat is supplied to a system, 6Q is taken as positive. When it is drawn from the system, 6Q is taken as negative. (ii) When the temperature of the gas increases, its internal energy dU increases, and is taken as positive. When temperature of a gas decreases, its internal energy dU decreases, and is taken as negative. , (iii) When a gas expands, work done by the gas GW is taken as positive. When the gas is compressed, work is done on the gas, and hence G W is taken as negative. Thermodynamics, Vibrations, Waves and Optics In summary, the audit of energy or work done is done with respect to the system.

6.2.3 Second Law of Thermodynamics

The first law of thermodynamics gives us a statement concerning conservation of energy in thermal processes. However, it gives no information about the way a thermodynamic system evolves or the direction of flow of heat. Further, the first law does not give us any idea about the conditions under which conversion of heat takes place into work or vice versa. The second law takes into account both these conditions and can be stated in any of the following ways:

Kelvin-Planck statement

"It is impossible to construct a device, which operating in a cycle has the sole effect of extracting heat from a single reservoir and performing an equivalent amount of work". In other words "No process is possible whose sole result is complete conversion of heat into work".

This implies that one cannot devise a machine which just absorbs heat and produces 100% work. You may like to further explain this statement.

Explanation of Kelvin-Planck Statement

According to this statement, a single reservoir at a single temperature cannot continuously transfer heat into work. This form of the law is applicable to heat engines. The working substance of a , operating in a cycle, cannot convert all the extracted heat into work. It must reject a part of the heat to the sink at a lower temperature. So, in order to convert heat into work, it is necessary to have both.source and sink. Since all the heat extracted frdm the source can never be converted into work, therefore the efficiency of the engine is never 100%. For example, whatever is being taught is never fully understood in the first lecture!

Clausius statement

"No process is possible whose sole result is the transfer of heat from a body at a lower temperature to a body at a higher temperature".

Explanation of Clausius Statement

In simple words, according to Clausius statement, heat cannot by itself flow from a body at a lower temperature to a body at a higher temperature.

This form of the law is applicable to the performance of refrigerators, which are heat engines working in the reverse direction. A refrigerator transfers heat from a cold body to a hot body with the aid of an outside agency. That agency is the compressor, which is found to be hot (You may like to visit the site www.howstuffworks.com for material on this topic).

It is important to emphasize here that the two statements of the second law seem to be different yet their consequences are identical. In fact, each statement implies the other and the two are equivalent. If either of these statements is proved false, the other will also necessarily not be true. You may like to devise an activity to help your students appreciate this point better. Thermodynamics SAQ 1

Devise an activity to prove that the Kelvin-Planck statement and Clausius statements of the second law of thermodynamics are equivalent. Write a report on the activity including details of how successful your teaching strategy was.

6.2.4 Reversible and Irreversible Processes

Before you discuss Carnot cycle and the concept of entropy in the class, you would be explaining the concept of reversible and irreversible processes to your students.

Reversible process

"It is that process which can be retraced in the reverse direction so that the system and the surroundings pass through exactly the same states at each stage as in the direct process".

If some heat is absorbed from the surroundings in the direct process, then some amount of heat is given out to the surroundings in the reverse process. Similarly if sorne work is done by the system in the direct process, the same amount of work is done on the system in the reverse process. At the end of a reversible process, the system and the surroundings should get restored to their initial state.

You must explain that for a process to be reversible, the following conditions must be satisfied:

1. The process should proceed at an extremely slow rate so that the system remains in thermodynamic equilibrium at all times. 2. The system should be free from dissipative forces like friction, inelasticity, viscosity, electrical resistance, magnetic hysteresis etc.

You must help your students understand that a completely reversible process is only an idealized concept meant only for theoretical considerations. It can at best only be approximated.

You may like to give the following examples of reversible processes:

i) All isothermal and adiabatic changes performed slowly are reversible processes. ii) Electrclysis can be taken as reversible process provided resistance offered by the electrolyte is zero. iii) Slow compression, or extension, of a spring is a reversible process.

Irreversible process

"It is a process which is not exactly reversed, i.e., the system does not pass through the same intermediate states as in the direct process".

Every process in nature is an irreversible process, because there always occurs some loss of energy to the surroundings.

Examples of Irreversible process:

i) Most of the chemical reactions are irreversible because they involve changes in the internal structure of the constituents, e.g., rusting of iron is an irreversible process. ii) Joule heating in a resistance wire is irreversible. iii) Rapid isothermal and adiabatic charges are irreversible. Thermodyn :s, We now discuss the Carnot Cycle and its applications. Vibrations, les and Optics 6.3 THE CARNOT CYCLE 1

Before we start discussing Carnot cycle, it would be worthwhile to list the form of the first law, equations of state and the work done in various thermodynamic processes. These are listed in Table 6.1. You may like to prepare a chart based on Table 6.1 and use it as a teaching aid.

Table 6.1: Some useful relations in thermodynamics

Equation of State Work done during the First process w = Ip d~

SQ=S- W (Temperat AU=O W = nRT log, (F) ure remain (NO constants) internal energy = nRT log, (z)

-dU=S W i) p V=constant (No gain ii) TV-'=constant

or loss of ...\ I-YTY _ -- -.--A

I constant) 1 ( Isochoric I AV=O 1 6PdU I V 1 W=O, because A V=O

1 constant) 1

Representation of a thermodynamic process on an indicator diagram

We need to explain in detail the important aspect of representing a thermodynamic process on an indicator diagram as students generally fail to understand it. As a result they fail to understand the intricacies behind the Carnot cycle. With the help of Table 6.1 we are now in a position to represent the various thermodynamic processes through diagrams called the indicator diagrams. These diagrams are important as they help us in understanding the various steps involved in the derivation of the efficiency of the Carnot cycle.

You can begin by explaining that an indicator diagram is nothing but a graph between thermodynamic variables like p and V for a thermodynamic process. The graph can also be plotted between V and Tor between p and T. 1 1 Importance of indicator diagram @- V diagram)

You know that the work done by a system or on the system in going from one state to

another is p dV . This is just the area under the curve in ap-V diagram. You may like to use the following figures while discussing the three important Thermodynamics thermodynamic processes viz., isothermal, adiabatic and cyclic (Fig. 6.1). You could transfer them on an OHP transparency. a)

Isothermal Expansion Isothermal Compression

Work done by the gas = area .4BB 'A' Work done on the gas = area ABB 'A' b) Adiabatic Process

S W-tve, dl/ = -ve; :. Temperature decreases, SW=+ve, dU = -ve; :.Temperature decreases, i.e., i.e.,T2

s) Cyclic Process

'Traced clockwise Traced anti-clockwise (dU= 0, 6Q = 6W) (dU=O, SQ=SIY) A Expansion

t P t

curve curve v+ v+

Net work done by the gas = shaded area (AXBYA) Net work done on the gas = shaded area (RYBYA)

Fig.6.1: Representing isothermal, adiabatic and cyclic processes. The compression and expansion curves here are arbitrary, not necessarily isothermal or adiabatic

You should point out that every real process of transformation from p,, V,, TItopz, Vz,TZ is a combination of isothermal and adiabatic processes.

A thermal engine is a system that operates cyclically, absorbs heat and performs work. Carnot developed a theoretical ideal cycle for a heat engine in the year 1824. This Thermodynamics, cycle came to be known as the Carnot cycle. To create interest in the subject you may Vibrations, Waves and Optics like to tell your students about Carnot. A Carnot cycle is a cycle composed of two isothermal and two adiabatic transformations. It is represented by the figure ABCDA as shown on ap-V diagram (Fig. 6.2).

I Isothermal A (D,. VI) Exoansion

Adiabatic

.>.~.~-&:- AUlitDIClC

I : Isothermal ! I I I Compression I I I I I I I I I I I I I F H E G v --+ Fig.6.2: p-V diagram of an ideal gas Carnot cycle

You may explain the cycle as follows:

Step 1 (AB): Isothermal Expansion

The gas expands isothermally at source temp TI and absorbs heat Q1. Using first law, for one mole of a gas, we have

= area ABEF

Step 2 (BC): Adiabatic Expansion

The gas expands adiabatically until its temperature decreases to the sink temperature T2.Therefore, work done,

= area BCGE

Step 3 (CD): Isothermal Compression

The gas is compressed isothermally at T2,rejecting heat-QZ.Therefore,

(iii) = area DCGH - -

b Step 4 (DA): Adiabatic Compression Thermodynamics

The gas is compressed adiabatically until it returns back to the initial state A. Therefore, work done,

(iv) = area BCGE

Now work done by the engine per cycle = Ql-Q2. Also the net work done by the engine in one complete cycle = area ABEF + area BCGE

- area DCGH - area ADHF i = area ABCD L Thus the work done in one cycle is represented on ap-Vdiagram by the area of the cycle. You can then calculate the efficiency. We give these calculations here for ready reference.

Net work done by the engine per cycle

I Using Eqs. (i), (ii), (iii) and (iv) and simplifLing we get

Using equations for adiabatic processes for DA and BC,

and

(vii)

Comparing Eqs. (vi) and (vii), we have

Hence Eq. (v) takes the form

(viii)

Since W = Q - Q2

(ix) Thermodynamics, Efficiency of the engine Vibrations, Waves and Optics The efficiency of the engine is given by

Heat converted into work 'rl= Heat drawn in from the source

psing Eqs. (ix) & (i)]

Hence the efficiency of the Carnot cycle or heat engine is given by

A discussion along the following lines may be useful for helping your students understand the underlying physics.

Equation (x) reveals that the efficiency of Carnot's ideal heat engine depends on the temperature of the source TI and temperature of the sink T2 and is thus independent of the nature of the working substance. The greater the difference between the two temperatures, higher is the efficiency of the Carnot's engine. As right hand side of Eq. (x) is < 1, the efficiency of Camot's engine is less than 100%.

For q=1 (i.e., 100% efficiency), either TI = or T2 = 0 K. AS source at infinite temperature or sink at OK are not attainable, Carnot's heat engine cannot have 100% efficiency.

If T2 = TI,then from Eq. (x)

i.e., Carnot's engine will not work. Hence, it is not possible to convert heat energy into mechanical work unless source and sink are at different temperatures.

when T2 f OK, Q2# 0, i.e., some heat must be rejected to the sink. Hence efficiency of even an ideal Carnot's cycle or heat engine can never be 100%. You may like to think about why Carnot engine is an ideal heat engine? Here is an Thermodynamics exercise.

SAQ 2

Explain why Carnot engine is an ideal heat engine.

6.3.1 Real Heat Engines

The requirements for Carnot's heat engine are difficult, rather impossible, to be satisfied in actual practice. Therefore Carnot's heat engine can be designed only theoretically. That is why it is said to be an ideal engine. In this context we would like to discuss some real heat engines. We would restrict ourselves to a brief discussion of their cycles without going into the details of the derivation of their efficiencies. The main objective here is to see how the cycles of these real heat engines compare with each other and with the ideal Carnot cycle with respect to various operations, and to see how the four stages of the Carnot cycle are implemented in real heat engines.

Types of real heat engines

i) External combustion engine

In this categoj of heat engines, the source of heat lies outside the engine. is an example of this class of heat engines.

Rankine cycle: The ideal cycle for a simple steam engine is the Rankine cycle (Fig. 6.3). Steam engines work omthis principle. In this cycle the working substance is water and it changes its state during the operation. It performs the cycle in the boiler, cylinder and the condenser, instead of one cylinder.

Fig.6.3: The Rankine cycle

There are four stages of this cycle. First stage AB; second stage BC; third stage CD and fourth stage DA. Heat is absorbed during the operation AB.and BC while it is rejected during operation DA.

i i) Internal combustion engine

In this category of heat engines, the source of heat lies inside the engine. Petrol engine and diesel engine are examples of this class of heat engines.

a) Petrol Engine: The petrol engine was developed by Dr. Otto, a German engineer, in 1876. Unlike the steam engine, which has a double acting stroke, i I the petrol engine operates in four strokes. The working substance is a mixture of petrol and air. It consists of 2% petrol vapours and 98% air. Therefore, basically, one can assume air as the working substance, because its percentage is much larger as compared to petrol vapours. I Thermodynamics, Otto cycle: The air standard Otto cycle is an ideal cycle that approximates a spark Vibrations, Waves and Optics ignition internal combustion engine. In this cycle, the working substance air is supposed to be enclosed in a cylinder whose sides are perfectly non-conducting and the bottom is of a conducting material. As was the case in Carnot cycle here also there is a source and a sink in addition to an insulating cap.

(31 C @J, T3) Adiabatic

cooling

(1)

Fig.6.4: Theoretical Otto cycle

The working of ideal cycle (Fig. 6.4) consists of two constant volume lines and two adiabatic curves. It also has four stages. It starts from point A and goes like this: A--+B-+C--+D-+A.

Actual p-Vdiagram for Otto cycle: For an Otto engine running on petrol or gas the actual p-V diagram is somewhat different from the theoretical p- V diagram of the Otto-cycle. The cycle involves six processes out of which the movement of piston takes place in four. The engine is called four stroke engine (see Fig. 6.5).

Fig.6.5: Actual Otto cycle (Source: www.tpub.com/engine3/ en3-files/imageOl4.jpg)

You may like to explain why the four stroke engine is more efficient than a two stroke one.

b) Diesel Engine: The engine was developed in 1898 by Rudolph Diesel (1 858-1913), a German engineer. Like petrol engine, it is also a four stroke engine. The main difference lies in the fact that in the diesel engine there is no sparkplug. Also the air is sucked in only the suction stroke; diesel is not mixed with it. Air alone is compressed in the second stroke and it is Thermodynamics compressed to a much larger extent, than in the case of petrol engine. The pressure may rise to about 35 atmosphere. This raises the efficiency, but it requires a cylinder of greater strength.

Diesel cycle: The air standard Diesel cycle is the ideal cycle for the Diesel engine. This cycle was first introduced by Diesel and consists of two adiabates, one constant pressure and one constant volume part. This cycle of operation is used in big engines like heavy motor vehicles, ships and stationary power plants. It is also . known as constant pressure cycle since in this cycle heat is taken at constant pressure. The various stages of this cycle are shown in Fig. 6.6.

-: U +stroke volume -ln "' I

Fig.6.6: Theoretical Diesel cycle

Actual p- Vdiagram of Diesel cycle: A diesel cycle uses heavy oil as fuel and differs from other oil engines in the method of ignition. Air is compressed to a very high pressure raising its temperature, which is sufficient to ignite the fuel sprayed in the cylinder. Thep-V diagram for the actual diesel engine is shown below:

AB = Suction Stroke BC = Compression Stroke Combustion CD = Combustion Stroke DE = Working or Power Stroke E = Exhaust Stroke BA = Exhaust Stroke Adiabatic

A Exhaust

Fig.6.7: Actual Diesel cycle

In Table 6.2, we summarise the information about the efficiency of heat engines. Thermodynamics, Vibrations, Waves and Table 6.2: Efiiciency of Real Heat Engines Optics External Combustion Engine Internal Combustion Engine Rankine cycle Otto cycle Diesel cycle (Steam Engine) (Petrol Engine) (Diesel Engine) 7-1 Work done kY -1 ' = Heat Input where HI-H2 - p = adiabatic compression ratio where HI -Hw2 p - adiabatic compression ratio =-VI k=-- where v2 r adiabatic expansion ratio y = CJC,, = 1.4 Hw, = Enthalpy at point A. Vl v4 In a petrol engine the value of p p = -, r = - Theoretically the efficiency of cannot be increased beyond lo. v2 v3 engine for the value of It is because if we compress the pressure and temperature as 20 air petrol mixture beyond this atmospheres and 250°C comes value then the ignition out to be 32%. However, the temperature of the mixture is actual efficiency is only about reached much before the 15%. completion of the stroke. As a result, the four strokes will not occur with the proper timing. For this reason p is seldom greater than 5. Using p = 5, we find q = 52%. However, q is much below this value. 6.3.2 Refrigerators An ideal refrigerator can be regarded as Carnot's ideal heat engine working in the (Source) reverse direction. Thus in a refrigerator, heat Q2 is extracted from a cold body at T2, for which work W is done on the system, and a larger amount of heat Ql is transferred I+' to a hot body at TI as shown in Fig. 6.8. Refrigerators can thus reduce the temperature of the cooler body. In a real or actual refrigerator, vapours of freon act as the working substance. Water and other food stuff to be cooled in the refrigerator act as sink at lower temperature T2. Heat energy Q2 is absorbed from the sink. A certain amount of work W is performed by the compressor of the refrigerator on the working substance. The compressor is operated by an electric motor. Atmospheric or surrounding air at room temperature TI is the source to which heat Q, is rejected by the radiator fixed at the back of the refrigerator.

Having explained the guiding principle of a refrigerator, you may like to ask your Fig.6.8: Refrigerator students: How' is its effectiveness measured? It is usually expressed as the coefficient principle of performance, COP, 0.

The coefficient of performance of a refrigerator is defined as the ratio of total heat Q2 extracted at lower temperature T2 to the amount of input work done W. Mathematically, we can write it as

where Q2 is heat extracted at lower temperature (cooler body) at temperature T2 and Ql is heat rejected at higher temperature Tl. In a Carnot cycle Thermodynamics

Substituting for Q2 in Eq. (6.1), we get

(iii) If we merely substitute TI and T2, we would get 0 = 2 to 6. For a refrigerator, the value of is less than that calculated from Eqs. (6.1) or (6.2).

(iv) In a heat engine, the efficiency can never exceed 100%. But in the case of a refrigerator, the coefficient of performance may be much higher than 100%. This is because coefficient of performance is not the efficiency. In fact, we / could make P + a, as TI + T2. If the outside temperature is only slightly larger than the inside temperature, then the refrigerator performance is extremely high! P + w represents + 1.

(v) As the refrigerator works, T2 goes on decreasing due to formation of ice. There is practically no change in TI. This decreases the value of P. However, if the refrigerator is defrosted, T2 shall increase and consequently the value of P. So it is necessary to defrost the refrigerator from time to time. Hence, nowadays people prefer to use frost free refrigerators.

SAQ 3

You could administer a test to your students containing problems of the following kind. a) Calculate the coefficient of performance of a refrigerator working between -3OC and 27OC. b) Refrigerator A works between -1 O°C and 27"C, while refrigerator B works between -20°C and 17OC both removing heat equal to 2000 J from the freezer. Which of the two is a better refrigerator? c) An electric refrigerator transfers heat from the cold cooling coils to the warm surroundings. Is it against the second law of thermodynamics? Justify your answer.

You could add more problems to this list including problems on laws of thermodynamics. How did your students perform? Include your test items and analysis in your folder. Thermodynamics, Vibrations, Waves and 6.4 ENTROPY AND ITS SIGNIFICANCE Optics Before you develop and discuss the definition of entropy and its significance, -you could convince your students from the experiences of daily life, that all natural processes proceed in a particular direction, i.e., they are unidirectional. Now raise the question. What determines the direction of a natural process and what is the thermodynamic criterion which predicts the direction of this change? The answer to these and many other such questions was given by Rudolf Clausius in 1854 when he introduced another concept that is very useful in describing various thermal processes.

Later on in 1865, he called this concept entropy (a dord from the Greek words 'trepein', which means 'change' and 'en' which means 'in' or 'content'.) The letter S is used to designabe entropy. Clausius showed that for all processes, entropy of the universe can never decrease. That is, all natural processes proceed in the direction of increase of entropy. Now the question, what is entropy? It is a device used to represent the direction in which a natural process takes place. It is a very effective tool to study the thermodynamic behaviour of a system. To discover an answer to the question, what is entropy, let us consider a system that undergoes a reversible process from state 1 to 2 along the path A. Let us assume that the cycle is completed along B which is also reversible (Fig. 6.9).

Fig.6.9: Defining entropy using two reversible paths forming a closed cycle

Suppose we are at some point C and we add an infinitesimal amount of heat 6 QR. Suppose this takes us to the point D along the path 1A2. We do this at every step and go from State 1 to State 2 along 1A2. The net change in the value of ('p- ) is given

via A

The amount of heat 6 QRis infinitesimal and will not take us straight from 1 to 2; also Twill be different at each point.

If we characterize the initial and final states by thermodynamic variables (V,, TI)and (V2, T2), respectively, we can readily carry out the integration and the result tells us

that the change in the value of -"R depends upon TI, T2, VI and V2. That is, the T change in !@- is determined by the initial and final states only (the subscript R T denotes a reversible process). This means that for any system, -defines a new T property which is a characteristic of the state. Now 6QRis an inexact differential, i.e., Thermodynamics a function of path. But the value of -6QR , which is the ratio of a path function and a T state variable is determined by the initial and final states only. This can happen only when the ratio denotes a new function. This new function is called entropy of T the system and is denoted by the symbol S. It is a function of the state and is independent of the path followed, and we write it as

In other words, when we add an infinitesimal amount of heat 6QRreversibly to a

system at constant temperature T, the entropy of the system changes by -"R . The T above equation enables us to write an expression for the change of entropy for any system

(ii)

where the limits on the integral refer to two thermodynamic states of entropies S, and S2. We can draw the following inferences from the above equation. '

i) For a reversible cyclic process 1A2 Bl, there will be no net entropy change. Thus we can write

(iii)

ii) We can calculate only the entropy change rather than the absolute value of entropy.

iii) Therefore the entropy of a system can only be given with reference to that of another system. In fact, this is quite like energy in which case also the difference of energy is of any significance while its value for a system is meaningless.

6.4.1 Entropy and the Second Law of Thermodynamics

So far we have introduced 'entropy' and the calculation of entropy change in a reversible process. How does such a change influence the system, its surroundings and hence the universe? This study leads us to the linkage between entropy and the second law of thermodynamics .which is a universal law. It applies to processes taking part in our body, to combustion of fuel in an automobile, an aeroplane and a rocket as well as working of refrigerators. It enables us to specify the direction of evolution of all natural processes.

Let an infinitesimal amount of heat 6Q flow from the surrounding at temperature TAU, to the system under consideration at temperature T,,. The net change in the entropies of the system and surroundings is given by .Thermodynamics, If T,,, > T,,,, the heat transfer isfrom the system to the surroundings and both 6Q and Vibrations, Waves and Optics will be negative, yielding the same result.

We may therefore conclude that for all possible processes that a system in a given surroundings can undergo, the increase in entropy is

The equal and greater than signs hold for reversible and irreversible processes, res'pectively. Since 9atural processes are irreversible, the above equation means that the entropy of the universe is increasing. This result is known as the principle of increase of entropy; The significance of Eq. (vi) is that it dictates unidirectionality for evolution of a process. This allows us to make a quantitativ, general statement of the second law of thermodynamics as follows:

"The entropy of the universe can never decrease."

A more general statement of the second law of thermodynamics in terms of entropy reads as follows:

"When an isolated (closed) system undergoes a change, its entropy cannot decrease; it increases or remains constant."

6.4.2 Physical Concept of Entropy

Earlier you introduced the concept of entropy basically through mathematical relations. Next you could go deeper into the significance of entropy. Your students may wonder what this quantity 'entropy' is that is piling up in the universe and how long will entropy continue to increase? What is the future ofthe natural world? Ultimately, it will reach an upper limit and the principle of increase of entropy implies that the universe is marching towards a stagnant state to suffer thermal death. It is a state when all matter is at the same T. That will be the upper limit of S. Hence we wish to give an insight into the physical and philosophical aspects of the concept of entropy.

We recall that entropy increases during an irreversible process. So, the word entropy serves as an effective vehicle for communicating our observations in respect to evolution of irreversible processes, like heat conduction along a bar or cooling a cup of coffee. We also know that entropy of a system increases when heat is added to it. Since heat is chaos, from kinetic theory viewpoint, continuous withdrawal of heat makes the molecular arrangement increasingly ordered. That is the entropy decrease implies a transition from chaotic or more random state to an ordered state. Conversely, the more disordered the state is, the larger is its entropy. This implies that gases have much greater entropy than solids.

Another point you can make regarding entropy is that it can be related to the probability of occurrence of a state. For example, entropy increase during an irreversible process can be associated with a change of state from a less probable to a mare probable state, e.g., as the coffee cools, the entropy of the coffee plus surroundings increzses and the system changes from a less probable to a more probable state. In fact, when a system is left to itself, it passes from an ordered (least probable) state to a state of maximum disorder (most probable), i.e., a system has a tendency to change in the direction of increasing entropy. This argument led Boltzmann to relate entropy with probability. Mathematically, he expressed it by the Thermodynamics relation S = k In W , where Sand W denote the entropy and the thermodynamic probability of the state, k being the Boltzmann constant.

6.4.3 Combined Forms of the First and Second Law of Thermodynamics

You must have convinced your students that the first law of thermodynamics is a statement of the law of conservation of energy and the second law relates (available) energy to entropy. You can also relate the two laws as follows:

The differential form of the first law of thermodynamics is given by

and for an infinitesimal reversible process between two equilibrium states, the second law of thermodynamics tells us that

These'equations may be combined to give

TdS =dU+GW (iii)

I for ap-V-T system, 6 W = pdV. Thus the above equation takes the form

TdS = dU + pdV (iv)

You could point out that this is one of the most important thermodynamical relations connecting the first and second laws of thermodynamics.

We now end this unit with a summary of 'its contents.

6.5 SUMMARY

In this unit we have dealt with the teaching-learning of the following concepts: , . The Zeroth law of thermodynamics states that if two systems are separately in equilibrium with a third system, then they must be in thermal equilibrium with one another. Temperature of a system may be defined as the property that determines whether or not the system is in thermal equilibrium with the neighbouring system. The general equation of state for any gaseous system is represented by

where f is a single valued function of pressure p, volume V and the absolute . temperature T. The differential form of the first law of thermodynamics is given by , . -. 6Q=dU+6W

Kelvin-Planck statement of the second law'of thermodynamics.- No process .is. . . possible whose sole result is complete cohv.ersi?n of heat into'work. , , . . Thermodynamics, a Vibrations, Waves and Clausius statement of the second law of thermodynamics -No process is possiblc Optics whose sole result is the transfer of heat from a body at a lower temperature to a body at a higher temperature.

The efficiency of a Carnot's heat engine is maximum and is given by

The coefficient of performance (COP), P of a refrigerator is given by

Entropy is defined through the relation

Entropy is a state function.

Entropy is a measure of disorder of the system; the more chaotic the system, greater will be its entropy.

The second law of thermodynamics is essentially the principle of increase of entropy. It states that, when a closed system undergoes a change, its entropy cannot decrease; it either increases or remains constant. Mathematically, it may be expressed as

a TdS = dU + 6 W = dU+pdV is a mathematical relation which connects the first and second laws of thermodynamics.

6.6 TERMINAL QUESTIONS

1. You may like to give your students the following problems to solve:

a. 1 g of water at 1OO°C changes into steam occupying a volume of 1760 cm3 at atmospheric pressure. Calculate (a) the work done in joules, and (b) the increase in internal energy. Given latent heat of steam = 540 cal g-', 1 atm = l.Olxl~~~m-~.

b. A person consumes a diet of 1O~J per day and spends total energy of 1.2 x 1045per day. Determine the daily changes in the internal energy. If the net energy spent comes from sucrose at the rate of 1.6 x 1045 kg-' in how many days will the person reduce his mass by 1 kg?

c. What is the more effective way to increase q of a Carnot engine: to increase TI or lower T2? Comment. d. A Carnot engine is independent of the state and nature of the working substance. We expect that the real engines will similarly be independent of the nature of the working substance, at least to some extent. Why then are we so concerned about finding suitable fuels such as coal, gasoline or fissionable material for real engines? e. Can a kitchen be cooled by leaving the door of an electric refrigerator open? Explain. f. In the tropics, the temperatures at the surface of the ocean and at a depth of Thermodynamics 300 m are 25°C and 5"C, respectively. Will you recommend to tap this energy?

g. Calculate the increase in entropy of ice when 1 kg of it melts at 0°C. The latent heat of fusion of ice is 3.36 x 105Jkg-'. Assume that melting is an isothermal reversible process. h. A block of copper weighing 1.5 kg is heated from 300 K to 350 K. Calculate the entropy change of the block. The specific heat capacity of copper is 389 J kg-'. Assume that heat is added irreversibly. i. A huge copper block at 1000 K is joined to another huge copper block at 500 K by a copper rod. The rate of heat conduction is 1o45s-'. Calculate the increase in entropy of the universe due to this process.

j. I0 A current is passed through 25 R resistor for 1 second. Its temperature is maintained at 27°C. Calculate the entropy change of (a) the resistor and (b) the universe.

2. Assess your students' performance. What difficulties remained and how did you help your students overcome them?

6.8 SOLUTIONS AND ANSWERS

SAQs

To prove this, suppose that a refrigerator transfers an amount of heat Q2 from a colder to a hotter body without any work done on it by some external agency. Such a refrigerator violates the Clausius statement of the second law. Next, suppose that an engine works between the same hot and cold bodies and it extracts heat Ql from the hot body and gives out Q2 to the cold body. The engine as such does not violate any law. If the two - the engine and the refrigerator - are coupled together to work simultaneously, they will constitute a self acting device taking in heat (el-Q2) from the cold body and converting all of it into work without rejecting any heat to the cold body (because the cold body is left unchanged, an amount of heat Q2 is extracted from it by the refrigerator and the same amount Q2 is rejected to it by the engine).

This system therefore violates Kelvin's and Planck's statement of the second law. Similarly, it can be argued that a violation of Kelvin & Planck's statement will lead to the violation of Clausius statement. We, therefore conclude that both the statements are equivalent in meaning in all respects. Please bring the report of your activity to the ECP.

2. A Carnot engine is an ideal heat engine from the following points of view.

a) There is absolutely no friction between the walls of cylinder and the piston. b) The working substance is an ideal gas.

In a real engine, these conditions cannot be fulfilled and hence no heat engine working between the same two temperatures can have efficiency greater than that of Carnot engine. Thermodynamics, 263 263 Vibrations, Waves and b) --- = 7.1 Optics "=300-263 37

Therefore, refiigerator A is a better refiigerator as its COP is greater than that of B.

c) According to the second law of thermodynamics, the heat energy by itself cannot flow fiom a body at a lower temperature to a body at a higher temperature. However, it can flow if some external agent performs work to do so. In a refrigerator the external work performed by its compressor transfers heat from the cold cooling pipes (or coils) to the warm surroundings. It is therefore not against the second law of thermodynamics.

Terminal Questions

6Q = mL

= 1g x 540 cal g-' = 540 cal

f540x 4.18J - 2257 J . Change in internal energy dU =6Q-6W ~(2257-178) 5~2079J

b) The diet consumed is like heat supplied and energy spent is the work done by the system. Thus

Therefore change in the internal energy per day is given by

dU=-6W

This decrease corresponds to loss of sucrose. Therefore sucrose lost per day Thermodynamics

1 Hence, number of days required for the loss of 1 kg = -= 8 0.125 c) To increase the efficiency of a Carnot engine, the more effective way is to increase the temperature of the source (TI). This is because the lower temperature would correspond, in a real situation, to the temperature of the surroundings which is almost constant. d) It is true that a Carnot engine is independent of the state and nature of the working substance. But the choice of fuel is determined by the abundance, quality and economics. We always tend to develop engines suited to local requirements. e) A kitchen cannot be cooled by a refrigerator because the surroundings in this case act as sink.

This is highly inefficient system and it is not advisable to tap this source of energy.

h) Although the heat has been added irreversibly, we can calculate AS using

Now 6QR= rns AT

= (1.5 kg) (389 J kg-' K-') x AT

so that Thermodynamics, = x x Vibrations, Waves and 583.5 2.3 0.0671 Optics = 90.0 J K-'

By carrying out the heat transfer reversibly, we can calculate that

= (q/Tl)-(q/~2) = (.[104J/s (1/500~)- = ~oJK-~s-~

Thus the entropy of the universe increases by 10~~-'per second.

(a) The point to remember is that the resistor is in the same state throughout and its entro y change must be zero. Electrical energy is converted into heat at rate IP R and is passed on the universe.

(b) The entropy gain of the universe is

'Q 12' 25R a=---_-- - = 8.33JK-1 TT 300K

2. Please bring your report to the ECP.