Thermodynamics PDF 2021-2022

CHAPTER TWELVE THERMODYNAMICS 12.1 INTRODUCTION In previous chapter we have studied thermal properties of matter. In this chapter we shall s...

CHAPTER TWELVE THERMODYNAMICS 12.1 INTRODUCTION In previous chapter we have studied thermal properties of matter. In this chapter we shall study laws that govern thermal energy. We shall study the processes where work is 12.1 Introduction converted into heat and vice versa. In winter, when we rub 12.2 Thermal equilibrium our palms together, we feel warmer; here work done in rubbing 12.3 Zeroth law of produces the ‘heat’. Conversely, in a steam engine, the ‘heat’ Thermodynamics of the steam is used to do useful work in moving the pistons, 12.4 Heat, internal energy and which in turn rotate the wheels of the train. work In physics, we need to define the notions of heat, 12.5 First law of temperature, work, etc. more carefully. Historically, it took a thermodynamics long time to arrive at the proper concept of ‘heat’. Before the 12.6 Specific heat capacity modern picture, heat was regarded as a fine invisible fluid 12.7 Thermodynamic state filling in the pores of a substance. On contact between a hot variables and equation of body and a cold body, the fluid (called caloric) flowed from state the colder to the hotter body ! This is similar to what happens 12.8 Thermodynamic processes when a horizontal pipe connects two tanks containing water 12.9 Heat engines up to different heights. The flow continues until the levels of 12.10 Refrigerators and heat water in the two tanks are the same. Likewise, in the ‘caloric’ pumps picture of heat, heat flows until the ‘caloric levels’ (i.e., the 12.11 Second law of temperatures) equalise. thermodynamics In time, the picture of heat as a fluid was discarded in 12.12 Reversible and irreversible favour of the modern concept of heat as a form of energy. An processes important experiment in this connection was due to Benjamin 12.13 Carnot engine Thomson (also known as Count Rumford) in 1798. He observed that boring of a brass cannon generated a lot of Summary heat, indeed enough to boil water. More significantly, the Points to ponder amount of heat produced depended on the work done (by the Exercises horses employed for turning the drill) but not on the sharpness of the drill. In the caloric picture, a sharper drill would scoop out more heat fluid from the pores; but this was not observed. A most natural explanation of the observations was that heat was a form of energy and the experiment demonstrated conversion of energy from one form to another–from work to heat. 2021-22 304 PHYSICS Thermodynamics is the branch of physics that in a different context : we say the state of a system deals with the concepts of heat and temperature is an equilibrium state if the macroscopic and the inter-conversion of heat and other forms variables that characterise the system do not of energy. Thermodynamics is a macroscopic change in time. For example, a gas inside a closed science. It deals with bulk systems and does not rigid container, completely insulated from its go into the molecular constitution of matter. In surroundings, with fixed values of pressure, fact, its concepts and laws were formulated in the volume, temperature, mass and composition that nineteenth century before the molecular picture do not change with time, is in a state of of matter was firmly established. Thermodynamic thermodynamic equilibrium. description involves relatively few macroscopic variables of the system, which are suggested by common sense and can be usually measured directly. A microscopic description of a gas, for example, would involve specifying the co-ordinates and velocities of the huge number of molecules constituting the gas. The description in kinetic theory of gases is not so detailed but it does involve molecular distribution of velocities. Thermodynamic description of a gas, on the other (a) hand, avoids the molecular description altogether. Instead, the state of a gas in thermodynamics is specified by macroscopic variables such as pressure, volume, temperature, mass and composition that are felt by our sense perceptions and are measurable*. The distinction between mechanics and thermodynamics is worth bearing in mind. In mechanics, our interest is in the motion of particles (b) or bodies under the action of forces and torques. Fig. 12.1 (a) Systems A and B (two gases) separated Thermodynamics is not concerned with the by an adiabatic wall – an insulating wall motion of the system as a whole. It is concerned that does not allow flow of heat. (b) The with the internal macroscopic state of the body. same systems A and B separated by a When a bullet is fired from a gun, what changes diathermic wall – a conducting wall that is the mechanical state of the bullet (its kinetic allows heat to flow from one to another. In this case, thermal equilibrium is attained energy, in particular), not its temperature. When in due course. the bullet pierces a wood and stops, the kinetic energy of the bullet gets converted into heat, In general, whether or not a system is in a state changing the temperature of the bullet and the of equilibrium depends on the surroundings and surrounding layers of wood. Temperature is the nature of the wall that separates the system related to the energy of the internal (disordered) from the surroundings. Consider two gases A and motion of the bullet, not to the motion of the bullet B occupying two different containers. We know as a whole. experimentally that pressure and volume of a given mass of gas can be chosen to be its two 12.2 THERMAL EQUILIBRIUM independent variables. Let the pressure and Equilibrium in mechanics means that the net volume of the gases be (PA, VA) and (PB, VB ) external force and torque on a system are zero. respectively. Suppose first that the two systems The term ‘equilibrium’ in thermodynamics appears are put in proximity but are separated by an * Thermodynamics may also involve other variables that are not so obvious to our senses e.g. entropy, enthalpy, etc., and they are all macroscopic variables. However, a thermodynamic state is specified by five state variables viz., pressure, volume, temperature, internal energy and entropy. Entropy is a measure of disorderness in the system. Enthalpy is a measure of total heat content of the system. 2021-22 THERMODYNAMICS 305 adiabatic wall – an insulating wall (can be law in 1931 long after the first and second Laws movable) that does not allow flow of energy (heat) of thermodynamics were stated and so numbered. from one to another. The systems are insulated The Zeroth Law clearly suggests that when two from the rest of the surroundings also by similar systems A and B, are in thermal equilibrium, adiabatic walls. The situation is shown there must be a physical quantity that has the schematically in Fig. 12.1 (a). In this case, it is same value for both. This thermodynamic found that any possible pair of values (PA, VA) will variable whose value is equal for two systems in be in equilibrium with any possible pair of values thermal equilibrium is called temperature (T ). (PB, VB ). Next, suppose that the adiabatic wall is Thus, if A and B are separately in equilibrium replaced by a diathermic wall – a conducting wall with C, TA = TC and TB = TC. This implies that that allows energy flow (heat) from one to another. TA = TB i.e. the systems A and B are also in It is then found that the macroscopic variables of thermal equilibrium. the systems A and B change spontaneously until We have arrived at the concept of temperature both the systems attain equilibrium states. After formally via the Zeroth Law. The next question that there is no change in their states. The is : how to assign numerical values to situation is shown in Fig. 12.1(b). The pressure temperatures of different bodies ? In other words, and volume variables of the two gases change to how do we construct a scale of temperature ? (PB ′, VB ′) and (PA ′, VA ′) such that the new states Thermometry deals with this basic question to of A and B are in equilibrium with each other*. which we turn in the next section. There is no more energy flow from one to another. We then say that the system A is in thermal equilibrium with the system B. What characterises the situation of thermal equilibrium between two systems ? You can guess the answer from your experience. In thermal equilibrium, the temperatures of the two systems are equal. We shall see how does one arrive at the concept of temperature in thermodynamics? The Zeroth law of thermodynamics provides the clue. 12.3 ZEROTH LAW OF THERMODYNAMICS (a) Imagine two systems A and B, separated by an adiabatic wall, while each is in contact with a third system C, via a conducting wall [Fig. 12.2(a)]. The states of the systems (i.e., their macroscopic variables) will change until both A and B come to thermal equilibrium with C. After this is achieved, suppose that the adiabatic wall between A and B is replaced by a conducting wall and C is insulated from A and B by an adiabatic wall [Fig.12.2(b)]. It is found that the states of A and B change no (b) further i.e. they are found to be in thermal Fig. 12.2 (a) Systems A and B are separated by an equilibrium with each other. This observation adiabatic wall, while each is in contact forms the basis of the Zeroth Law of with a third system C via a conducting Thermodynamics, which states that ‘two wall. (b) The adiabatic wall between A systems in thermal equilibrium with a third and B is replaced by a conducting wall, system separately are in thermal equilibrium while C is insulated from A and B by an with each other’. R.H. Fowler formulated this adiabatic wall. * Both the variables need not change. It depends on the constraints. For instance, if the gases are in containers of fixed volume, only the pressures of the gases would change to achieve thermal equilibrium. 2021-22 306 PHYSICS 12.4 HEAT, INTERNAL ENERGY AND WORK associated with various random motions of its The Zeroth Law of Thermodynamics led us to molecules. We will see in the next chapter that the concept of temperature that agrees with our in a gas this motion is not only translational commonsense notion. Temperature is a marker (i.e. motion from one point to another in the of the ‘hotness’ of a body. It determines the volume of the container); it also includes direction of flow of heat when two bodies are rotational and vibrational motion of the placed in thermal contact. Heat flows from the molecules (Fig. 12.3). body at a higher temperature to the one at lower temperature. The flow stops when the temperatures equalise; the two bodies are then in thermal equilibrium. We saw in some detail how to construct temperature scales to assign temperatures to different bodies. We now describe the concepts of heat and other relevant quantities like internal energy and work. Fig. 12.3 (a) Internal energy U of a gas is the sum of the kinetic and potential energies of its The concept of internal energy of a system is molecules when the box is at rest. Kinetic not difficult to understand. We know that every energy due to various types of motion bulk system consists of a large number of (translational, rotational, vibrational) is to molecules. Internal energy is simply the sum of be included in U. (b) If the same box is the kinetic energies and potential energies of moving as a whole with some velocity, these molecules. We remarked earlier that in the kinetic energy of the box is not to be thermodynamics, the kinetic energy of the included in U. system, as a whole, is not relevant. Internal energy is thus, the sum of molecular kinetic and potential energies in the frame of reference relative to which the centre of mass of the system is at rest. Thus, it includes only the (disordered) energy associated with the random motion of molecules of the system. We denote the internal energy of a system by U. Though we have invoked the molecular picture to understand the meaning of internal energy, as far as thermodynamics is concerned, U is simply a macroscopic variable of the system. The important thing about internal energy is that it depends only on the state of the system, not on how that state was achieved. Internal energy U of a system is an example of a thermodynamic ‘state variable’ – its value depends only on the given state of the system, not on history i.e. not on the ‘path’ taken to arrive at that state. Thus, the internal energy of a given mass of gas depends on its state described by Fig. 12.4 Heat and work are two distinct modes of specific values of pressure, volume and energy transfer to a system that results in temperature. It does not depend on how this change in its internal energy. (a) Heat is state of the gas came about. Pressure, volume, energy transfer due to temperature difference between the system and the temperature, and internal energy are surroundings. (b) Work is energy transfer thermodynamic state variables of the system brought about by means (e.g. moving the (gas) (see section 12.7). If we neglect the small piston by raising or lowering some weight intermolecular forces in a gas, the internal connected to it) that do not involve such a energy of a gas is just the sum of kinetic energies temperature difference. 2021-22 THERMODYNAMICS 307 What are the ways of changing internal 12.5 FIRST LAW OF THERMODYNAMICS energy of a system ? Consider again, for We have seen that the internal energy U of a simplicity, the system to be a certain mass of system can change through two modes of energy gas contained in a cylinder with a movable transfer : heat and work. Let piston as shown in Fig. 12.4. Experience shows there are two ways of changing the state of the ∆Q = Heat supplied to the system by the gas (and hence its internal energy). One way is surroundings to put the cylinder in contact with a body at a ∆W = Work done by the system on the higher temperature than that of the gas. The surroundings temperature difference will cause a flow of ∆U = Change in internal energy of the system energy (heat) from the hotter body to the gas, The general principle of conservation of thus increasing the internal energy of the gas. energy then implies that The other way is to push the piston down i.e. to ∆Q = ∆U + ∆W (12.1) do work on the system, which again results in increasing the internal energy of the gas. Of i.e. the energy (∆Q) supplied to the system goes course, both these things could happen in the in partly to increase the internal energy of the reverse direction. With surroundings at a lower system (∆U) and the rest in work on the temperature, heat would flow from the gas to environment (∆W). Equation (12.1) is known as the surroundings. Likewise, the gas could push the First Law of Thermodynamics. It is simply the general law of conservation of energy applied the piston up and do work on the surroundings. to any system in which the energy transfer from In short, heat and work are two different modes or to the surroundings is taken into account. of altering the state of a thermodynamic system Let us put Eq. (12.1) in the alternative form and changing its internal energy. The notion of heat should be carefully ∆Q – ∆W = ∆U (12.2) distinguished from the notion of internal energy. Heat is certainly energy, but it is the energy in Now, the system may go from an initial state transit. This is not just a play of words. The to the final state in a number of ways. For distinction is of basic significance. The state of example, to change the state of a gas from a thermodynamic system is characterised by its (P1, V 1) to (P2, V 2), we can first change the internal energy, not heat. A statement like ‘a volume of the gas from V1 to V2, keeping its gas in a given state has a certain amount of pressure constant i.e. we can first go the state heat’ is as meaningless as the statement that (P1, V2) and then change the pressure of the gas from P1 to P2, keeping volume constant, to ‘a gas in a given state has a certain amount take the gas to (P2, V 2). Alternatively, we can of work’. In contrast, ‘a gas in a given state first keep the volume constant and then keep has a certain amount of internal energy’ is a the pressure constant. Since U is a state perfectly meaningful statement. Similarly, the variable, ∆U depends only on the initial and statements ‘a certain amount of heat is final states and not on the path taken by the supplied to the system’ or ‘a certain amount gas to go from one to the other. However, ∆Q of work was done by the system’ are perfectly and ∆W will, in general, depend on the path meaningful. taken to go from the initial to final states. From To summarise, heat and work in the First Law of Thermodynamics, Eq. (12.2), thermodynamics are not state variables. They it is clear that the combination ∆Q – ∆W, is are modes of energy transfer to a system however, path independent. This shows that resulting in change in its internal energy, if a system is taken through a process in which which, as already mentioned, is a state variable. ∆U = 0 (for example, isothermal expansion of an ideal gas, see section 12.8), In ordinary language, we often confuse heat with internal energy. The distinction between ∆Q = ∆W them is sometimes ignored in elementary physics books. For proper understanding of i.e., heat supplied to the system is used up thermodynamics, however, the distinction is entirely by the system in doing work on the crucial. environment. 2021-22 308 PHYSICS If the system is a gas in a cylinder with a If the amount of substance is specified in movable piston, the gas in moving the piston does terms of moles µ (instead of mass m in kg ), we work. Since force is pressure times area, and can define heat capacity per mole of the area times displacement is volume, work done substance by by the system against a constant pressure P is S 1 ∆Q C= = (12.6) ∆W = P ∆V µ µ ∆T C is known as molar specific heat capacity of where ∆V is the change in volume of the gas. the substance. Like s, C is independent of the Thus, for this case, Eq. (12.1) gives amount of substance. C depends on the nature ∆Q = ∆U + P ∆V (12.3) of the substance, its temperature and the conditions under which heat is supplied. The As an application of Eq. (12.3), consider the unit of C is J mo1–1 K–1. As we shall see later (in change in internal energy for 1 g of water when connection with specific heat capacity of gases), we go from its liquid to vapour phase. The additional conditions may be needed to define measured latent heat of water is 2256 J/g. i.e., C or s. The idea in defining C is that simple for 1 g of water ∆Q = 2256 J. At atmospheric predictions can be made in regard to molar pressure, 1 g of water has a volume 1 cm3 in specific heat capacities. liquid phase and 1671 cm3 in vapour phase. Table 12.1 lists measured specific and molar heat capacities of solids at atmospheric pressure Therefore, and ordinary room temperature. ∆W =P (Vg –Vl ) = 1.013 ×105 ×(1671×10–6) =169.2 J We will see in Chapter 13 that predictions of specific heats of gases generally agree with Equation (12.3) then gives experiment. We can use the same law of ∆U = 2256 – 169.2 = 2086.8 J equipartition of energy that we use there to predict molar specific heat capacities of solids We see that most of the heat goes to increase (See Section 13.5 and 13.6). Consider a solid of the internal energy of water in transition from N atoms, each vibrating about its mean the liquid to the vapour phase. position. An oscillator in one dimension has 12.6 SPECIFIC HEAT CAPACITY average energy of 2 × ½ kB T = kBT. In three dimensions, the average energy is 3 k B T. Suppose an amount of heat ∆Q supplied to a For a mole of a solid, the total energy is substance changes its temperature from T to T + ∆T. We define heat capacity of a substance U = 3 kBT × NA = 3 RT (∵ kBT × NA = R ) (see Chapter 11) to be Now, at constant pressure, ∆Q = ∆U + P ∆V ≅ ∆Q ∆U, since for a solid ∆V is negligible. Therefore, S= (12.4) ∆T ∆Q ∆U We expect ∆Q and, therefore, heat capacity S C= = = 3R (12.7) ∆T ∆T to be proportional to the mass of the substance. Table 12.1 Specific and molar heat capacities Further, it could also depend on the of some solids at room temperature, i.e., a different amount of heat may temperature and atmospheric be needed for a unit rise in temperature at pressure different temperatures. To define a constant Substance Speci"c–v heat Molar speci"c characteristic of the substance and –1 –1 (J kg K ) –1 –1 heat (J mol K ) independent of its amount, we divide S by the mass of the substance m in kg : S  1  ∆Q s = =   (12.5) m  m  ∆T s is known as the specific heat capacity of the substance. It depends on the nature of the As Table 12.1 shows, the experimentally substance and its temperature. The unit of measured values which generally agrees with specific heat capacity is J kg–1 K–1. 2021-22 THERMODYNAMICS 309 predicted value 3R at ordinary temperatures. ideal gas, we have a simple relation. (Carbon is an exception.) The agreement is known to break down at low temperatures. Cp – Cv = R (12.8) Specific heat capacity of water where C p and C v are molar specific heat capacities of an ideal gas at constant pressure The old unit of heat was calorie. One calorie was earlier defined to be the amount of heat and volume respectively and R is the universal required to raise the temperature of 1g of water gas constant. To prove the relation, we begin by 1°C. With more precise measurements, it was with Eq. (12.3) for 1 mole of the gas : found that the specific heat of water varies ∆Q = ∆U + P ∆V slightly with temperature. Figure 12.5 shows this variation in the temperature range 0 to If ∆Q is absorbed at constant volume, ∆V = 0 100 °C.  ∆Q   ∆U   ∆U  Cv =  = =  ∆T  v  ∆T  v  ∆T  (12.9) where the subscript v is dropped in the last step, since U of an ideal gas depends only on temperature. (The subscript denotes the quantity kept fixed.) If, on the other hand, ∆Q is absorbed at constant pressure,  ∆Q   ∆U   ∆V  Cp =   =   +P  ∆T  p  ∆T  p  ∆T  p (12.10) The subscript p can be dropped from the Fig. 12.5 Variation of specific heat capacity of water first term since U of an ideal gas depends only with temperature. on T. Now, for a mole of an ideal gas For a precise definition of calorie, it was, therefore, necessary to specify the unit PV = RT temperature interval. One calorie is defined which gives to be the amount of heat required to raise the temperature of 1g of water from 14.5 °C to  ∆V  P  = R 15.5 °C. Since heat is just a form of energy,  ∆T  p (12.11) it is preferable to use the unit joule, J. In SI units, the specific heat capacity of water Equations (12.9) to (12.11) give the desired is 4186 J kg–1 K–1 i.e. 4.186 J g–1 K–1. The so relation, Eq. (12.8). called mechanical equivalent of heat defined as the amount of work needed to produce 12.7 THERMODYNAMIC STATE VARIABLES 1 cal of heat is in fact just a conversion factor AND EQUATION OF STATE between two different units of energy : calorie Every equilibrium state of a thermodynamic to joule. Since in SI units, we use the unit joule system is completely described by specific for heat, work or any other form of energy, the values of some macroscopic variables, also term mechanical equivalent is now called state variables. For example, an superfluous and need not be used. equilibrium state of a gas is completely As already remarked, the specific heat specified by the values of pressure, volume, capacity depends on the process or the temperature, and mass (and composition if conditions under which heat capacity transfer there is a mixture of gases). A thermodynamic takes place. For gases, for example, we can system is not always in equilibrium. For example, define two specific heats : specific heat a gas allowed to expand freely against vacuum capacity at constant volume and specific is not an equilibrium state [Fig. 12.6(a)]. During heat capacity at constant pressure. For an the rapid expansion, pressure of the gas may 2021-22 310 PHYSICS not be uniform throughout. Similarly, a mixture temperature do not. To decide which variable is of gases undergoing an explosive chemical extensive and which intensive, think of a reaction (e.g. a mixture of petrol vapour and relevant system in equilibrium, and imagine that air when ignited by a spark) is not an it is divided into two equal parts. The variables equilibrium state; again its temperature and that remain unchanged for each part are pressure are not uniform [Fig. 12.6(b)]. intensive. The variables whose values get halved Eventually, the gas attains a uniform in each part are extensive. It is easily seen, for example, that internal energy U, volume V, total temperature and pressure and comes to mass M are extensive variables. Pressure P, thermal and mechanical equilibrium with its temperature T, and density ρ are intensive surroundings. variables. It is a good practice to check the consistency of thermodynamic equations using this classification of variables. For example, in the equation ∆Q = ∆U + P ∆V quantities on both sides are extensive*. (The product of an intensive variable like P and an extensive quantity ∆V is extensive.) 12.8 THERMODYNAMIC PROCESSES 12.8.1 Quasi-static process Consider a gas in thermal and mechanical equilibrium with its surroundings. The pressure of the gas in that case equals the external Fig. 12.6 (a) The partition in the box is suddenly pressure and its temperature is the same as removed leading to free expansion of the that of its surroundings. Suppose that the gas. (b) A mixture of gases undergoing an external pressure is suddenly reduced (say by explosive chemical reaction. In both lifting the weight on the movable piston in the situations, the gas is not in equilibrium and container). The piston will accelerate outward. cannot be described by state variables. During the process, the gas passes through In short, thermodynamic state variables states that are not equilibrium states. The non- describe equilibrium states of systems. The equilibrium states do not have well-defined various state variables are not necessarily pressure and temperature. In the same way, if independent. The connection between the state a finite temperature difference exists between variables is called the equation of state. For the gas and its surroundings, there will be a example, for an ideal gas, the equation of state rapid exchange of heat during which the gas is the ideal gas relation will pass through non-equilibrium states. In due course, the gas will settle to an equilibrium PV=µRT state with well-defined temperature and For a fixed amount of the gas i.e. given µ, there pressure equal to those of the surroundings. The are thus, only two independent variables, say P free expansion of a gas in vacuum and a mixture and V or T and V. The pressure-volume curve of gases undergoing an explosive chemical for a fixed temperature is called an isotherm. reaction, mentioned in section 12.7 are also Real gases may have more complicated examples where the system goes through non- equations of state. equilibrium states. The thermodynamic state variables are of two Non-equilibrium states of a system are difficult kinds: extensive and intensive. Extensive to deal with. It is, therefore, convenient to variables indicate the ‘size’ of the system. imagine an idealised process in which at every Intensive variables such as pressure and stage the system is an equilibrium state. Such a * As emphasised earlier, Q is not a state variable. However, ∆Q is clearly proportional to the total mass of system and hence is extensive. 2021-22 THERMODYNAMICS 311 process is, in principle, infinitely slow, hence the A process in which the temperature of the name quasi-static (meaning nearly static). The system is kept fixed throughout is called an system changes its variables (P, T, V ) so slowly isothermal process. The expansion of a gas in that it remains in thermal and mechanical a metallic cylinder placed in a large reservoir of equilibrium with its surroundings throughout. fixed temperature is an example of an isothermal In a quasi-static process, at every stage, the process. (Heat transferred from the reservoir to difference in the pressure of the system and the the system does not materially affect the external pressure is infinitesimally small. The temperature of the reservoir, because of its very same is true of the temperature difference large heat capacity.) In isobaric processes the between the system and its surroundings pressure is constant while in isochoric (Fig.12.7). To take a gas from the state (P, T ) to processes the volume is constant. Finally, if another state (P ′, T ′ ) via a quasi-static process, the system is insulated from the surroundings we change the external pressure by a very small and no heat flows between the system and the surroundings, the process is adiabatic. The amount, allow the system to equalise its pressure definitions of these special processes are with that of the surroundings and continue the summarised in Table. 12.2 process infinitely slowly until the system achieves the pressure P ′. Similarly, to change Table 12.2 Some special thermodynamic the temperature, we introduce an infinitesimal processes temperature difference between the system and the surrounding reservoirs and by choosing reservoirs of progressively different temperatures T to T ′, the system achieves the temperature T ′. We now consider these processes in some detail : 12.8.2 Isothermal process For an isothermal process (T fixed), the ideal gas equation gives PV = constant i.e., pressure of a given mass of gas varies inversely as its volume. This is nothing but Boyle’s Law. Suppose an ideal gas goes isothermally (at temperature T ) from its initial state (P1, V1) to Fig. 12.7 In a quasi-static process, the temperature the final state (P2, V 2). At any intermediate stage of the surrounding reservoir and the with pressure P and volume change from V to external pressure differ only infinitesimally V + ∆V (∆V small) from the temperature and pressure of the system. ∆W = P ∆ V A quasi-static process is obviously a Taking (∆V → 0) and summing the quantity hypothetical construct. In practice, processes ∆W over the entire process, that are sufficiently slow and do not involve V2 accelerated motion of the piston, large W = ∫ P dV temperature gradient, etc., are reasonably V1 approximation to an ideal quasi-static process. V2 We shall from now on deal with quasi-static dV V = µ RT ∫ = µRT In 2 (12.12) processes only, except when stated otherwise. V1 V V1 2021-22 312 PHYSICS where in the second step we have made use of We can calculate, as before, the work done in the ideal gas equation PV = µ RT and taken the an adiabatic change of an ideal gas from the constants out of the integral. For an ideal gas, state (P1, V1, T1) to the state (P2, V2, T2). internal energy depends only on temperature. V2 Thus, there is no change in the internal energy W = ∫ P dV of an ideal gas in an isothermal process. The V1 First Law of Thermodynamics then implies that heat supplied to the gas equals the work done by the gas : Q = W. Note from Eq. (12.12) that for V2 > V1, W > 0; and for V2 < V1, W < 0. That is, in an isothermal expansion, the gas absorbs heat and does work while in an isothermal (12.15) compression, work is done on the gas by the environment and heat is released. From Eq. (12.14), the constant is P1V1γ or P2V2γ 12.8.3 Adiabatic process In an adiabatic process, the system is insulated 1  P2V2γ P1V1γ  W =  − γ −1  from the surroundings and heat absorbed or 1 − γ  V2γ −1 V1  released is zero. From Eq. (12.1), we see that work done by the gas results in decrease in its 1 µR(T1 − T2 ) internal energy (and hence its temperature for = 1−γ [P2V2 − P1V1 ] = γ −1 (12.16) an ideal gas). We quote without proof (the result that you will learn in higher courses) that for As expected, if work is done by the gas in an an adiabatic process of an ideal gas. adiabatic process (W > 0), from Eq. (12.16), P V γ = const (12.13) T2 < T1. On the other hand, if work is done on the gas (W < 0), we get T 2 > T 1 i.e., the where γ is the ratio of specific heats (ordinary temperature of the gas rises. or molar) at constant pressure and at constant volume. 12.8.4 Isochoric process Cp In an isochoric process, V is constant. No work γ = Cv is done on or by the gas. From Eq. (12.1), the heat absorbed by the gas goes entirely to change Thus if an ideal gas undergoes a change in its internal energy and its temperature. The its state adiabatically from (P1, V1) to (P2, V2) : change in temperature for a given amount of γ γ P1 V1 = P2 V2 (12.14) heat is determined by the specific heat of the Figure12.8 shows the P-V curves of an ideal gas at constant volume. gas for two adiabatic processes connecting two 12.8.5 Isobaric process isotherms. In an isobaric process, P is fixed. Work done by the gas is W = P (V2 – V1) = µ R (T2 – T1) (12.17) Since temperature changes, so does internal energy. The heat absorbed goes partly to increase internal energy and partly to do work. The change in temperature for a given amount of heat is determined by the specific heat of the gas at constant pressure. 12.8.6 Cyclic process In a cyclic process, the system returns to its Fig. 12.8 P-V curves for isothermal and adiabatic initial state. Since internal energy is a state processes of an ideal gas. variable, ∆U = 0 for a cyclic process. From 2021-22 THERMODYNAMICS 313 Eq. (12.1), the total heat absorbed equals the and W is the work done on the environment in work done by the system. a cycle. In a cycle, a certain amount of heat (Q2) may also be rejected to the environment. Then, 12.9 HEAT ENGINES according to the First Law of Thermodynamics, Heat engine is a device by which a system is over one complete cycle, made to undergo a cyclic process that results in conversion of heat to work. W = Q1 – Q2 (12.19) (1) It consists of a working substance–the system. For example, a mixture of fuel i.e., vapour and air in a gasoline or diesel engine or steam in a steam engine are the working Q substances. η =1 − 2 (12.20) Q1 (2) The working substance goes through a cycle consisting of several processes. In some of For Q2 = 0, η = 1, i.e., the engine will have these processes, it absorbs a total amount 100% efficiency in converting heat into work. of heat Q1 from an external reservoir at some Note that the First Law of Thermodynamics i.e., high temperature T1. the energy conservation law does not rule out (3) In some other processes of the cycle, the such an engine. But experience shows that working substance releases a total amount such an ideal engine with η = 1 is never possible, of heat Q2 to an external reservoir at some even if we can eliminate various kinds of losses lower temperature T2. associated with actual heat engines. It turns (4) The work done (W ) by the system in a cycle out that there is a fundamental limit on the is transferred to the environment via some efficiency of a heat engine set by an independent arrangement (e.g. the working substance principle of nature, called the Second Law of may be in a cylinder with a moving piston Thermodynamics (section 12.11). that transfers mechanical energy to the The mechanism of conversion of heat into wheels of a vehicle via a shaft). work varies for different heat engines. Basically, The basic features of a heat engine are there are two ways : the system (say a gas or a schematically represented in Fig. 12.9. mixture of gases) is heated by an external furnace, as in a steam engine; or it is heated internally by an exothermic chemical reaction as in an internal combustion engine. The various steps involved in a cycle also differ from one engine to another. Fig. 12.9 Schematic representation of a heat engine. 12.10 REFRIGERATORS AND HEAT PUMPS The engine takes heat Q1 from a hot A refrigerator is the reverse of a heat engine. reservoir at temperature T1, releases heat Q2 to a cold reservoir at temperature T2 Here the working substance extracts heat Q2 and delivers work W to the surroundings. from the cold reservoir at temperature T2, some external work W is done on it and heat Q1 is The cycle is repeated again and again to get released to the hot reservoir at temperature T1 useful work for some purpose. The discipline of (Fig. 12.10). thermodynamics has its roots in the study of heat engines. A basic question relates to the efficiency of a heat engine. The efficiency (η) of a heat engine is defined by W η= (12.18) Q1 Fig. 12.10 Schematic representation of a refrigerator where Q 1 is the heat input i.e., the heat or a heat pump, the reverse of a heat absorbed by the system in one complete cycle engine. 2021-22 314 PHYSICS Pioneers of Thermodynamics Lord Kelvin (William Thomson) (1824-1907), born in Belfast, Ireland, is among the foremost British scientists of the nineteenth century. Thomson played a key role in the development of the law of conservation of energy suggested by the work of James Joule (1818-1889), Julius Mayer (1814- 1878) and Hermann Helmholtz (1821-1894). He collaborated with Joule on the so-called Joule-Thomson effect : cooling of a gas when it expands into vacuum. He introduced the notion of the absolute zero of temperature and proposed the absolute temperature scale, now called the Kelvin scale in his honour. From the work of Sadi Carnot (1796-1832), Thomson arrived at a form of the Second Law of Thermodynamics. Thomson was a versatile physicist, with notable contributions to electromagnetic theory and hydrodynamics. Rudolf Clausius (1822-1888), born in Poland, is generally regarded as the discoverer of the Second Law of Thermodynamics. Based on the work of Carnot and Thomson, Clausius arrived at the important notion of entropy that led him to a fundamental version of the Second Law of Thermodynamics that states that the entropy of an isolated system can never decrease. Clausius also worked on the kinetic theory of gases and obtained the first reliable estimates of molecular size, speed, mean free path, etc. A heat pump is the same as a refrigerator. where Q2 is the heat extracted from the cold What term we use depends on the purpose of reservoir and W is the work done on the the device. If the purpose is to cool a portion of system–the refrigerant. (α for heat pump is space, like the inside of a chamber, and higher defined as Q1/W) Note that while η by definition temperature reservoir is surrounding, we call can never exceed 1, α can be greater than 1. the device a refrigerator; if the idea is to pump By energy conservation, the heat released to the heat into a portion of space (the room in a hot reservoir is building when the outside environment is cold), Q1 = W + Q2 the device is called a heat pump. In a refrigerator the working substance Q2 i.e., α = (12.22) (usually, in gaseous form) goes through the Q1 – Q2 following steps : (a) sudden expansion of the gas In a heat engine, heat cannot be fully from high to low pressure which cools it and converted to work; likewise a refrigerator cannot converts it into a vapour-liquid mixture, (b) work without some external work done on the absorption by the cold fluid of heat from the system, i.e., the coefficient of performance in Eq. region to be cooled converting it into vapour, (c) (12.21) cannot be infinite. heating up of the vapour due to external work 12.11 SECOND LAW OF THERMODYNAMICS done on the system, and (d) release of heat by the vapour to the surroundings, bringing it to The First Law of Thermodynamics is the principle the initial state and completing the cycle. of conservation of energy. Common experience The coefficient of performance ( α ) of a shows that there are many conceivable refrigerator is given by processes that are perfectly allowed by the First Law and yet are never observed. For example, Q2 nobody has ever seen a book lying on a table α= (12.21) W jumping to a height by itself. But such a thing 2021-22 THERMODYNAMICS 315 would be possible if the principle of conservation Experience suggests that for most processes in of energy were the only restriction. The table nature this is not possible. The spontaneous could cool spontaneously, converting some of its processes of nature are irreversible. Several internal energy into an equal amount of examples can be cited. The base of a vessel on mechanical energy of the book, which would an oven is hotter than its other parts. When then hop to a height with potential energy equal the vessel is removed, heat is transferred from to the mechanical energy it acquired. But this the base to the other parts, bringing the vessel never happens. Clearly, some additional basic to a uniform temperature (which in due course principle of nature forbids the above, even cools to the temperature of the surroundings). though it satisfies the energy conservation The process cannot be reversed; a part of the principle. This principle, which disallows many vessel will not get cooler spontaneously and phenomena consistent with the First Law of warm up the base. It will violate the Second Law Thermodynamics is known as the Second Law of Thermodynamics, if it did. The free expansion of Thermodynamics. of a gas is irreversible. The combustion reaction The Second Law of Thermodynamics gives a of a mixture of petrol and air ignited by a spark fundamental limitation to the efficiency of a heat cannot be reversed. Cooking gas leaking from a engine and the co-efficient of performance of a gas cylinder in the kitchen diffuses to the refrigerator. In simple terms, it says that entire room. The diffusion process will not efficiency of a heat engine can never be unity. spontaneously reverse and bring the gas back According to Eq. (12.20), this implies that heat to the cylinder. The stirring of a liquid in thermal released to the cold reservoir can never be made contact with a reservoir will convert the work zero. For a refrigerator, the Second Law says that done into heat, increasing the internal energy the co-efficient of performance can never be of the reservoir. The process cannot be reversed infinite. According to Eq. (12.21), this implies exactly; otherwise it would amount to conversion that external work (W ) can never be zero. The of heat entirely into work, violating the Second following two statements, one due to Kelvin and Law of Thermodynamics. Irreversibility is a rule Planck denying the possibility of a perfect heat rather an exception in nature. engine, and another due to Clausius denying Irreversibility arises mainly from two causes: the possibility of a perfect refrigerator or heat one, many processes (like a free expansion, or pump, are a concise summary of these an explosive chemical reaction) take the system observations. to non-equilibrium states; two, most processes Kelvin-Planck statement involve friction, viscosity and other dissipative No process is possible whose sole result is the effects (e.g., a moving body coming to a stop and absorption of heat from a reservoir and the losing its mechanical energy as heat to the floor complete conversion of the heat into work. and the body; a rotating blade in a liquid coming to a stop due to viscosity and losing its Clausius statement mechanical energy with corresponding gain in No process is possible whose sole result is the the internal energy of the liquid). Since transfer of heat from a colder object to a hotter dissipative effects are present everywhere and object. can be minimised but not fully eliminated, most It can be proved that the two statements processes that we deal with are irreversible. above are completely equivalent. A thermodynamic process (state i → state f ) 12.12 REVERSIBLE AND IRREVERSIBLE is reversible if the process can be turned back PROCESSES such that both the system and the surroundings Imagine some process in which a thermodynamic return to their original states, with no other system goes from an initial state i to a final change anywhere else in the universe. From the state f. During the process the system absorbs preceding discussion, a reversible process is an heat Q from the surroundings and performs idealised notion. A process is reversible only if work W on it. Can we reverse this process and it is quasi-static (system in equilibrium with the bring both the system and surroundings to their surroundings at every stage) and there are no initial states with no other effect anywhere ? dissipative effects. For example, a quasi-static 2021-22 316 PHYSICS isothermal expansion of an ideal gas in a from temperature T1 to T2 and then back from cylinder fitted with a frictionless movable piston temperature T2 to T1. Which processes should is a reversible process. we employ for this purpose that are reversible? Why is reversibility such a basic concept in A little reflection shows that we can only adopt thermodynamics ? As we have seen, one of the reversible adiabatic processes for these concerns of thermodynamics is the efficiency purposes, which involve no heat flow from any with which heat can be converted into work. reservoir. If we employ any other process that is The Second Law of Thermodynamics rules out not adiabatic, say an isochoric process, to take the possibility of a perfect heat engine with 100% the system from one temperature to another, we efficiency. But what is the highest efficiency shall need a series of reservoirs in the possible for a heat engine working between two temperature range T2 to T1 to ensure that at each reservoirs at temperatures T1 and T2 ? It turns stage the process is quasi-static. (Remember out that a heat engine based on idealised again that for a process to be quasi-static and reversible processes achieves the highest reversible, there should be no finite temperature efficiency possible. All other engines involving difference between the system and the reservoir.) irreversibility in any way (as would be the case But we are considering a reversible engine that for practical engines) have lower than this operates between only two temperatures. Thus limiting efficiency. adiabatic processes must bring about the temperature change in the system from T1 to T2 12.13 CARNOT ENGINE and T2 to T1 in this engine. Suppose we have a hot reservoir at temperature T1 and a cold reservoir at temperature T2. What is the maximum efficiency possible for a heat engine operating between the two reservoirs and what cycle of processes should be adopted to achieve the maximum efficiency ? Sadi Carnot, a French engineer, first considered this question in 1824. Interestingly, Carnot arrived at the correct answer, even though the basic concepts of heat and thermodynamics had yet to be firmly established. We expect the ideal engine operating between two temperatures to be a reversible engine. Irreversibility is associated with dissipative effects, as remarked in the preceding section, Fig. 12.11 Carnot cycle for a heat engine with an and lowers efficiency. A process is reversible if ideal gas as the working substance. it is quasi-static and non-dissipative. We have A reversible heat engine operating between seen that a process is not quasi-static if it two temperatures is called a Carnot engine. We involves finite temperature difference between have just argued that such an engine must have the system and the reservoir. This implies that the following sequence of steps constituting one in a reversible heat engine operating between cycle, called the Carnot cycle, shown in Fig. two temperatures, heat should be absorbed 12.11. We have taken the working substance of (from the hot reservoir) isothermally and released (to the cold reservoir) isothermally. We the Carnot engine to be an ideal gas. thus have identified two steps of the reversible (a) Step 1 → 2 Isothermal expansion of the gas heat engine : isothermal process at temperature taking its state from (P1, V1, T1) to T1 absorbing heat Q1 from the hot reservoir, and (P2, V2, T1). another isothermal process at temperature T2 releasing heat Q 2 to the cold reservoir. To The heat absorbed by the gas (Q1) from the complete a cycle, we need to take the system reservoir at temperature T 1 is given by 2021-22 THERMODYNAMICS 317 1 /( γ −1) Eq. (12.12). This is also the work done (W1 → 2) V2 T  by the gas on the environment. i.e. =  2 (12.29) V3  T1   V2  W1 → 2 = Q1 = µ R T1 ln   (12.23) Similarly, since step 4 → 1 is an adiabatic  V1  process (b) Step 2 → 3 Adiabatic expansion of the gas from (P2, V2, T1) to (P3, V3, T2) γ −1 γ −1 T2 V4 = T1 V1 Work done by the gas, using Eq. (12.16), is 1 / γ −1 V1  T2  µR (T1 − T2 ) i.e. V4 =  T1  (12.30) W2 → 3 = (12.24) γ −1 From Eqs. (12.29) and (12.30), (c) Step 3 → 4 Isothermal compression of the gas from (P3, V3, T2) to (P4, V4, T2). V3 V2 = (12.31) V4 V1 Heat released (Q2) by the gas to the reservoir at temperature T2 is given by Eq. (12.12). This Using Eq. (12.31) in Eq. (12.28), we get is also the work done (W3 → 4) on the gas by the environment. T2 η =1 − (Carnot engine) (12.32) V  T1 W3 → 4 = Q2 = µRT2 ln  3  (12.25)  V4  We have already seen that a Carnot engine (d) Step 4 → 1 Adiabatic compression of the is a reversible engine. Indeed it is the only gas from (P4, V4, T2) to (P1,V1, T1). reversible engine possible that works between two reservoirs at different temperatures. Each Work done on the gas, [using Eq.(12.16), is step of the Carnot cycle given in Fig. 12.11 can be reversed. This will amount to taking heat Q2  T − T2  W4 → 1 = µ R  1 from the cold reservoir at T2, doing work W on  γ -1  (12.26) the system, and transferring heat Q1 to the hot From Eqs. (12.23) to (12.26) total work done reservoir. This will be a reversible refrigerator. by the gas in one complete cycle is We next establish the important result (sometimes called Carnot’s theorem) that W = W1 → 2 + W2 → 3 – W3 →4 – W4 → 1 (a) working between two given temperatures T1  V2   V3  and T2 of the hot and cold reservoirs respectively, = µ RT1 ln   – µ RT2 ln   (12.27) no engine can have efficiency more than that of  1 V  V4  the Carnot engine and (b) the efficiency of the The efficiency η of the Carnot engine is Carnot engine is independent of the nature of the working substance. W Q2 η = =1 − To prove the result (a), imagine a reversible Q1 Q1 (Carnot) engine R and an irreversible engine I In  V3  working between the same source (hot reservoir)   and sink (cold reservoir). Let us couple the  T   V4  =1 −  2  engines, I and R, in such a way so that I acts  T1   V2  (12.28) like a heat engine and R acts as a refrigerator. In  V  1 Let I absorb heat Q1 from the source, deliver work W ′ and release the heat Q1- W′ to the sink. Now since step 2 → 3 is an adiabatic process, We arrange so that R returns the same heat Q1 γ −1 γ −1 to the source, taking heat Q2 from the sink and T1 V2 = T2 V3 requiring work W = Q1 – Q2 to be done on it. 2021-22 318 PHYSICS Now suppose η R < η I i.e. if R were to act than that of the Carnot engine. A similar as an engine it would give less work output argument can be constructed to show that a than that of I i.e. W < W ′ for a given Q1. With R reversible engine with one particular substance acting like a refrigerator, this would mean cannot be more efficient than the one using Q2 = Q1 – W > Q1 – W ′. Thus, on the whole, another substance. The maximum efficiency of the coupled I-R system extracts heat a Carnot engine given by Eq. (12.32) is (Q1 – W) – (Q1 – W ′) = (W ′ – W ) from the cold independent of the nature of the system reservoir and delivers the same amount of work performing the Carnot cycle of operations. Thus in one cycle, without any change in the source we are justified in using an ideal gas as a system or anywhere else. This is clearly against the in the calculation of efficiency η of a Carnot Kelvin-Planck statement of the Second Law of engine. The ideal gas has a simple equation of Thermodynamics. Hence the assertion ηI > ηR state, which allows us to readily calculate η, but is wrong. No engine can have efficiency greater the final result for η, [Eq. (12.32)], is true for any Carnot engine. This final remark shows that in a Carnot cycle, I Q1 T1 = (12.33) Q 2 T2 is a universal relation independent of the nature R of the system. Here Q1 and Q2 are respectively, the heat absorbed and released isothermally (from the hot and to the cold reservoirs) in a W Carnot engine. Equation (12.33), can, therefore, be used as a relation to define a truly universal Fig. 12.12 An irreversible engine (I) coupled to a thermodynamic temperature scale that is reversible refrigerator (R). If W ′ > W, this independent of any particular properties of the would amount to extraction of heat system used in the Carnot cycle. Of course, for W ′ – W from the sink and its full an ideal gas as a working substance, this conversion to work, in contradiction with universal temperature is the same as the ideal the Second Law of Thermodynamics. gas temperature introduced in section 12.11. SUMMARY 1. The zeroth law of thermodynamics states that ‘two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other’. The Zeroth Law leads to the concept of temperature. 2. Internal energy of a system is the sum of kinetic energies and potential energies of the molecular constituents of the system. It does not include the over-all kinetic energy of the system. Heat and work are two modes of energy transfer to the system. Heat is the energy transfer arising due to temperature difference between the system and the surroundings. Work is energy transfer brought about by other means, such as moving the piston of a cylinder containing the gas, by raising or lowering some weight connected to it. 3. The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) is taken into account. It states that ∆Q = ∆U + ∆W where ∆Q is the heat supplied to the system, ∆W is the work done by the system and ∆U is the change in internal energy of the system. 2021-22 THERMODYNAMICS 319 4. The specific heat capacity of a substance is defined by 1 ∆Q s= m ∆T where m is the mass of the substance and ∆Q is the heat required to change its temperature by ∆T. The molar specific heat capacity of a substance is defined by 1 ∆Q C= µ ∆T where µ is the number of moles of the substance. For a solid, the law of equipartition of energy gives C = 3R which generally agrees with experiment at ordinary temperatures. Calorie is the old unit of heat. 1 calorie is the amount of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C. 1 cal = 4.186 J. 5. For an ideal gas, the molar specific heat capacities at constant pressure and volume satisfy the relation Cp – Cv = R where R is the universal gas constant. 6. Equilibrium states of a thermodynamic system are described by state variables. The value of a state variable depends only on the particular state, not on the path used to arrive at that state. Examples of state variables are pressure (P ), volume (V ), temperature (T ), and mass (m ). Heat and work are not state variables. An Equation of State (like the ideal gas equation PV = µ RT ) is a relation connecting different state variables. 7. A quasi-static process is an infinitely slow process such that the system remains in thermal and mechanical equilibrium with the surroundings throughout. In a quasi-static process, the pressure and temperature of the environment can differ from those of the system only infinitesimally. 8. In an isothermal expansion of an ideal gas from volume V1 to V2 at temperature T the heat absorbed (Q) equals the work done (W ) by the gas, each given by  V2  Q = W = µRT ln  V   1 9. In an adiabatic process of an ideal gas γ PV = constant Cp where γ = Cv Work done by an ideal gas in an adiabatic change of state from (P1, V1, T1) to (P2, V2, T2) is µ R ( T1 − T2 ) W = γ –1 10. Heat engine is a device in which a system undergoes a cyclic process resulting in conversion of heat into work. If Q1 is the heat absorbed from the source, Q2 is the heat released to the sink, and the work output in one cycle is W, the efficiency η of the engine is: W Q η= =1− 2 Q1 Q1 2021-22 320 PHYSICS 11. In a refrigerator or a heat pump, the system extracts heat Q2 from the cold reservoir and releases Q1 amount of heat to the hot reservoir, with work W done on the system. The co-efficient of performance of a refrigerator is given by Q2 Q2 α= = W Q1 − Q2 12. The second law of thermodynamics disallows some processes consistent with the First Law of Thermodynamics. It states Kelvin-Planck statement No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work. Clausius statement No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. Put simply, the Second Law implies that no heat engine can have efficiency η equal to 1 or no refrigerator can have co-efficient of performance α equal to infinity. 13. A process is reversible if it can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. Spontaneous processes of nature are irreversible. The idealised reversible process is a quasi-static process with no dissipative factors such as friction, viscosity, etc. 14. Carnot engine is a reversible engine operating between two temperatures T1 (source) and T2 (sink). The Carnot cycle consists of two isothermal processes connected by two adiabatic processes. The efficiency of a Carnot engine is given by T2 η =1 − (Carnot engine) T1 No engine operating between two temperatures can have efficiency greater than that of the Carnot engine. 15. If Q > 0, heat is added to the system If Q < 0, heat is removed to the system If W > 0, Work is done by the system If W < 0, Work is done on the system Quantity Symbol Dimensions Unit Remark Co-efficienty of volume αv [K–1] K–1 αv = 3 α1 expansion Heat supplied to a system ∆Q [ML2 T–2] J Q is not a state variable Specific heat capacity s [L2 T–2 K–1] J kg–1 K–1 dt Thermal Conductivity K [MLT–3 K–1] J s–1 K–1 H = – KA dx 2021-22 THERMODYNAMICS 321 POINTS TO PONDER 1. Temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass. A bullet fired from a gun is not at a higher temperature because of its high speed. 2. Equilibrium in thermodynamics refers to the situation when macroscopic variables describing the thermodynamic state of a system do not depend on time. Equilibrium of a system in mechanics means the net external force and torque on the system are zero. 3. In a state of thermodynamic equilibrium, the microscopic constituents of a system are not in equilibrium (in the sense of mechanics). 4. Heat capacity, in general, depends on the process the system goes through when heat is supplied. 5. In isothermal quasi-static processes, heat is absorbed or given out by the system even though at every stage the gas has the same temperature as that of the surrounding reservoir. This is possible because of the infinitesimal difference in temperature between the system and the reservoir. EXERCISES 12.1 A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g ? 12.2 What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.) 12.3 Explain why (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2 )/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. 12.4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ? 12.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J) 12.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (a) What is the final pressure of the gas in A and B ? (b) What is the change in internal energy of the gas ? (c) What is the change in the temperature of the gas ? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ? 2021-22 322 PHYSICS 12.7 A steam engine delivers 5.4×108J of work per minute and services 3.6 × 109J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute? 12.8 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing? 12.9 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13) Fig. 12.13 Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F 12.10 A refrigerator is to maintain eatables kept inside at 90C. If room temperature is 360C, calculate the coefficient of performance. 2021-22