11.4 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 rotational and vibrational motion of thedirection of flow of heat when two bodies are molecules (Fig. 11.3).placed in thermal contact. Heat flows from the 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. 11.3 (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. 11.4 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 energy transfer due to temperaturestate of the gas came about. Pressure, volume, 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 11.7). 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. Reprint 2025-26 230 PHYSICS What are the ways of changing internal 11.5 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. 11.4. Experience shows ∆Q = Heat supplied to the system by thethere are two ways of changing the state of the surroundingsgas (and hence its internal energy). One way is 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 (11.1) do work on the system, which again results in i.e. the energy (∆Q) supplied to the system goesincreasing the internal energy of the gas. Of in partly to increase the internal energy of thecourse, both these things could happen in the system (∆U) and the rest in work on thereverse direction. With surroundings at a lower environment (∆W). Equation (11.1) is known astemperature, heat would flow from the gas to the First Law of Thermodynamics. It is simplythe surroundings. Likewise, the gas could push 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. (11.1) in the alternative form and changing its internal energy. The notion of heat should be carefully ∆Q – ∆W = ∆U (11.2) 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, V1) to (P2, V2), we can first change the volume of the gas from V1 to V2, keeping itsinternal energy, not heat. A statement like ‘a pressure constant i.e. we can first go the stategas in a given state has a certain amount of 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, V2). Alternatively, we canof 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. (11.2), 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 ∆U = 0 (for example, isothermal expansion ofwhich, as already mentioned, is a state variable. an ideal gas, see section 11.8), 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. Reprint 2025-26 THERMODYNAMICS 231 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 = = (11.6) ∆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. (11.1) gives amount of substance. C depends on the nature ∆Q = ∆U + P ∆V (11.3) of the substance, its temperature and the conditions under which heat is supplied. The As an application of Eq. (11.3), 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 11.1 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 12 that predictions of specific heats of gases generally agree with Equation (11.3) then gives experiment. We can use the same law of equipartition of energy that we use there to ∆U = 2256 – 169.2 = 2086.8 J predict molar specific heat capacities of solids We see that most of the heat goes to increase (See Section 12.5 and 12.6). 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

11.6 SPECIFIC HEAT CAPACITY average energy of 2 × ½ kBT = kBT. In three dimensions, the average energy is 3 kBT.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 10) to be Now, at constant pressure, ∆Q = ∆U + P ∆V ≅ ∆ Q ∆U, since for a solid ∆V is negligible. Therefore, S = (11.4) ∆ T ∆ Q ∆U C = = = 3 R (11.7) We expect ∆Q and, therefore, heat capacity S ∆ T ∆ T to be proportional to the mass of the substance. Table 11.1 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 –v Speci"c heat Molar speci"c characteristic of the substance and Substance –1 –1 –1 –1 (J kg K ) heat (J mol K ) independent of its amount, we divide S by the mass of the substance m in kg : S  1  ∆Q s = (11.5) m =  m  ∆T s is known as the specific heat capacity of the substance. It depends on the nature of the As Table 11.1 shows, the experimentally substance and its temperature. The unit of measured values which generally agrees withspecific heat capacity is J kg–1 K–1. Reprint 2025-26 232 PHYSICS 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 (11.8) Specific heat capacity of water where Cp and Cv are molar specific heat capacities of an ideal gas at constant pressure The old unit of heat was calorie. One calorie and volume respectively and R is the universal was earlier defined to be the amount of heat gas constant. To prove the relation, we begin required to raise the temperature of 1g of water with Eq. (11.3) for 1 mole of the gas : by 1°C. With more precise measurements, it was found that the specific heat of water varies ∆Q = ∆U + P ∆V slightly with temperature. Figure 11.5 shows If ∆Q is absorbed at constant volume, ∆V = 0this variation in the temperature range 0 to 100 °C.  ∆Q   ∆ U   ∆ U  (11.9) C v =  ∆T  v =  ∆T  v =  ∆T  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  (11.10) C p =  ∆T  p =  ∆T  p + P  ∆ T  p The subscript p can be dropped from the Fig. 11.5 Variation of specific heat capacity of first term since U of an ideal gas depends only water with temperature. on T. Now, for a mole of an ideal gas For a precise definition of calorie, it was, PV = RTtherefore, necessary to specify the unit 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 (11.11)15.5 °C. Since heat is just a form of energy,  ∆ T  p it is preferable to use the unit joule, J. In SI units, the specific heat capacity of water Equations (11.9) to (11.11) give the desired is 4186 J kg–1 K–1 i.e. 4.186 J g–1 K–1. The so relation, Eq. (11.8). called mechanical equivalent of heat defined 11.7THERMODYNAMIC STATE VARIABLES as the amount of work needed to produce AND EQUATION OF STATE 1 cal of heat is in fact just a conversion factor 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. 11.6(a)]. During heat capacity at constant pressure. For an the rapid expansion, pressure of the gas may Reprint 2025-26 THERMODYNAMICS 233 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 that remain unchanged for each part areequilibrium state; again its temperature and intensive. The variables whose values get halvedpressure are not uniform [Fig. 11.6(b)]. in each part are extensive. It is easily seen, forEventually, the gas attains a uniform 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.) 11.8 THERMODYNAMIC PROCESSES 11.8.1 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. 11.6 (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 P V = µ R T 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 11.7 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. Reprint 2025-26 234 PHYSICS 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 large heat capacity.) In isobaric processes thesame is true of the temperature difference pressure is constant while in isochoricbetween the system and its surroundings processes the volume is constant. Finally, if the(Fig.11.7). To take a gas from the state (P, T ) to another state (P ′, T ′ ) via a quasi-static process, system is insulated from the surroundings and no heat flows between the system and thewe change the external pressure by a very small 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. 11.2 process infinitely slowly until the system achieves the pressure P ′. Similarly, to change Table 11.2 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 : 11.8.2 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. 11.7 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 d V temperature gradient, etc., are reasonably V1 approximation to an ideal quasi-static process. V2 d V V2We shall from now on deal with quasi-static = µ RT = µRT In ∫ (11.12) Vprocesses only, except when stated otherwise. V1 V 1 Reprint 2025-26 THERMODYNAMICS 235 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. V 2Thus, there is no change in the internal energy W = ∫ P d Vof an ideal gas in an isothermal process. The V1First Law of Thermodynamics then implies that heat supplied to the gas equals the work done by the gas : Q = W. Note from Eq. (11.12) 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 (11.15) compression, work is done on the gas by the environment and heat is released. From Eq. (11.14), the constant is P1V1γ or P2V2γ 11.8.3 Adiabatic process γ  γ  P2 V 2 P1V 1In an adiabatic process, the system is insulated W = 1 − 1   −1 γ γ − from the surroundings and heat absorbed or 1 − γ  V2 V1  released is zero. From Eq. (11.1), we see that work done by the gas results in decrease in its 1 µR(T1 − T2 ) = [ P2 V2 − P1V1 ] = (11.16)internal energy (and hence its temperature for 1 − γ γ − 1 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. (11.16), P V γ = const (11.13) T2 < T1. On the other hand, if work is done on the gas (W < 0), we get T2 > T1 i.e., thewhere γ is the ratio of specific heats (ordinary temperature of the gas rises. or molar) at constant pressure and at constant volume. 11.8.4 Isochoric process Cp In an isochoric process, V is constant. No work γ = Cv is done on or by the gas. From Eq. (11.1), 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 = P2 V2γ (11.14) heat is determined by the specific heat of the P1 V1 gas at constant volume. Figure11.8 shows the P-V curves of an ideal gas for two adiabatic processes connecting two 11.8.5 Isobaric process isotherms. In an isobaric process, P is fixed. Work done by the gas is W = P (V2 – V1) = µ R (T2 – T1) (11.17) 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. 11.8.6 Cyclic process In a cyclic process, the system returns to its initial state. Since internal energy is a stateFig. 11.8 P-V curves for isothermal and adiabatic variable, ∆U = 0 for a cyclic process. From processes of an ideal gas. Reprint 2025-26 236 PHYSICS Eq. (11.1), the total heat absorbed equals the 11.10 REVERSIBLE AND IRREVERSIBLE work done by the system. PROCESSES Imagine some process in which a thermodynamic11.9 SECOND LAW OF THERMODYNAMICS system goes from an initial state i to a final state The First Law of Thermodynamics is the f. During the process the system absorbs heat Q principle of conservation of energy. Common from the surroundings and performs work W on experience shows that there are many it. Can we reverse this process and bring both conceivable processes that are perfectly the system and surroundings to their initial allowed by the First Law and yet are never states with no other effect anywhere ? Experience observed. For example, nobody has ever seen suggests that for most processes in nature this a book lying on a table jumping to a height by is not possible. The spontaneous processes ofitself. But such a thing would be possible if nature are irreversible. Several examples can bethe principle of conservation of energy were cited. The base of a vessel on an oven is hotterthe only restriction. The table could cool than its other parts. When the vessel is removed,spontaneously, converting some of its internal heat is transferred from the base to the otherenergy into an equal amount of mechanical parts, bringing the vessel to a uniformenergy of the book, which would then hop to a height with potential energy equal to the temperature (which in due course cools to the mechanical energy it acquired. But this never temperature of the surroundings). The process happens. Clearly, some additional basic cannot be reversed; a part of the vessel will not principle of nature forbids the above, even get cooler spontaneously and warm up the base. though it satisfies the energy conservation It will violate the Second Law of Thermodynamics, principle. This principle, which disallows if it did. The free expansion of a gas is irreversible. many phenomena consistent with the First The combustion reaction of a mixture of petrol Law of Thermodynamics is known as the and air ignited by a spark cannot be reversed. Second Law of Thermodynamics. Cooking gas leaking from a gas cylinder in the The Second Law of Thermodynamics gives kitchen diffuses to the entire room. The diffusion a fundamental limitation to the efficiency of a process will not spontaneously reverse and bring heat engine and the co-efficient of the gas back to the cylinder. The stirring of a performance of a refrigerator. In simple terms, liquid in thermal contact with a reservoir will it says that efficiency of a heat engine can convert the work done into heat, increasing the never be unity. For a refrigerator, the Second internal energy of the reservoir. The process Law says that the co-efficient of performance cannot be reversed exactly; otherwise it would can never be infinite. The following two amount to conversion of heat entirely into work, statements, one due to Kelvin and Planck violating the Second Law of Thermodynamics. denying the possibility of a perfect heat engine, Irreversibility is a rule rather an exception and another due to Clausius denying the in nature. possibility of a perfect refrigerator or heat Irreversibility arises mainly from two causes: pump, are a concise summary of these one, many processes (like a free expansion, or observations. an explosive chemical reaction) take the system Kelvin-Planck statement to non-equilibrium states; two, most processes No process is possible whose sole result is the involve friction, viscosity and other dissipative absorption of heat from a reservoir and the effects (e.g., a moving body coming to a stop and complete conversion of the heat into work. losing its mechanical energy as heat to the floor and the body; a rotating blade in a liquid coming Clausius statement to a stop due to viscosity and losing its No process is possible whose sole result is the mechanical energy with corresponding gain in transfer of heat from a colder object to a the internal energy of the liquid). Since hotter object. dissipative effects are present everywhere and It can be proved that the two statements can be minimised but not fully eliminated, most above are completely equivalent. processes that we deal with are irreversible. Reprint 2025-26 THERMODYNAMICS 237 A thermodynamic process (state i → state f ) in a reversible heat engine operating between is reversible if the process can be turned back two temperatures, heat should be absorbed such that both the system and the surroundings (from the hot reservoir) isothermally and return to their original states, with no other released (to the cold reservoir) isothermally. We change anywhere else in the universe. From the thus have identified two steps of the reversible preceding discussion, a reversible process is an heat engine : isothermal process at temperature idealised notion. A process is reversible only if T1 absorbing heat Q1 from the hot reservoir, and another isothermal process at temperature T2it is quasi-static (system in equilibrium with the releasing heat Q2 to the cold reservoir. Tosurroundings at every stage) and there are no complete a cycle, we need to take the systemdissipative effects. For example, a quasi-static from temperature T1 to T2 and then back fromisothermal expansion of an ideal gas in a temperature T2 to T1. Which processes shouldcylinder fitted with a frictionless movable piston we employ for this purpose that are reversible?is a reversible process. A little reflection shows that we can only adopt Why is reversibility such a basic concept in reversible adiabatic processes for these thermodynamics ? As we have seen, one of the purposes, which involve no heat flow from any concerns of thermodynamics is the efficiency reservoir. If we employ any other process that is with which heat can be converted into work. not adiabatic, say an isochoric process, to take The Second Law of Thermodynamics rules out the system from one temperature to another, we the possibility of a perfect heat engine with 100% shall need a series of reservoirs in the efficiency. But what is the highest efficiency temperature range T2 to T1 to ensure that at each possible for a heat engine working between two stage the process is quasi-static. (Remember reservoirs at temperatures T1 and T2? It turns again that for a process to be quasi-static and out that a heat engine based on idealised reversible, there should be no finite temperature reversible processes achieves the highest difference between the system and the reservoir.) efficiency possible. All other engines involving But we are considering a reversible engine that irreversibility in any way (as would be the case operates between only two temperatures. Thus for practical engines) have lower than this adiabatic processes must bring about the limiting efficiency. temperature change in the system from T1 to T2 and T2 to T1 in this engine.11.11 CARNOT 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 Fig. 11.9 Carnot cycle for a heat engine with an effects, as remarked in the preceding section, ideal gas as the working substance. and lowers efficiency. A process is reversible if 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 Reprint 2025-26 238 PHYSICS cycle, called the Carnot cycle, shown in Fig. 11.9. We have taken the working substance In  V 3  of the Carnot engine to be an ideal gas. T2   V4  = 1 −  T1   V2  (11.23)(a) Step 1 → 2 Isothermal expansion of the gas In  V1  taking its state from (P1, V1, T1) to (P2, V2, T1). Now since step 2 → 3 is an adiabatic process, The heat absorbed by the gas (Q1) from the γ −1 γ −1 reservoir at temperature T1 is given by T1 V 2 = T2 V3 Eq. (11.12). This is also the work done (W1 → 2) by the gas on the environment. 1 /( γ −1 ) V2  T2   V2  i.e. = (11.24) 3  T1  W1 → 2 = Q1 = µ R T1 ln  V1  (11.18) V (b) Step 2 → 3 Adiabatic expansion of the gas Similarly, since step 4 → 1 is an adiabatic from (P2, V2, T1) to (P3, V3, T2) process Work done by the gas, using γ −1 γ −1 Eq. (11.16), is T2 V 4 = T1 V1 µR ( T1 − T2 ) 1 /γ −1 W2 →=3 (11.19) V1  T2  γ − 1 i.e. = (11.25) V4  T1  (c) Step 3 → 4 Isothermal compression of the gas from (P3, V3, T2) to (P4, V4, T2). From Eqs. (11.24) and (11.25), 2 Heat released (Q2) by the gas to the reservoir V3 = V (11.26) at temperature T2 is given by Eq. (11.12). This V4 V1 is also the work done (W3 → 4) on the gas by the environment. Using Eq. (11.26) in Eq. (11.23), we get  V 3  2 W 3 → 4 = Q2 = µRT2 ln η = 1 −T (Carnot engine) (11.27)  V 4  (11.20) T1 (d) Step 4 → 1 Adiabatic compression of the gas from (P4, V4, T2) to (P1,V1, T1). We have already seen that a Carnot engine is a reversible engine. Indeed it is the only Work done on the gas, [using Eq.(11.16), is reversible engine possible that works between two reservoirs at different temperatures. Each  T1 − T2  W4 → 1 = µR (11.21) step of the Carnot cycle given in Fig. 11.9 can  γ -1  be reversed. This will amount to taking heat Q2 from the cold reservoir at T2, doing work W on From Eqs. (11.18) to (11.21) total work done by the gas in one complete cycle is the system, and transferring heat Q1 to the hot reservoir. This will be a reversible refrigerator. W = W1 → 2 + W2 → 3 – W3 → 4 – W4 → 1 We next establish the important result  V2   V3  (sometimes called Carnot’s theorem) that = µ RT1 ln  V1  – µ RT2 ln  V4  (11.22) (a) working between two given temperatures T1 and T2 of the hot and cold reservoirs respectively, The efficiency η of the Carnot engine is no engine can have efficiency more than that of the Carnot engine and (b) the efficiency of the W Q 2 η = = 1 − Carnot engine is independent of the nature of Q1 Q1 the working substance. Reprint 2025-26 THERMODYNAMICS 239 To prove the result (a), imagine a reversible reservoir and delivers the same amount of work (Carnot) engine R and an irreversible engine I in one cycle, without any change in the source working between the same source (hot reservoir) or anywhere else. This is clearly against the and sink (cold reservoir). Let us couple the Kelvin-Planck statement of the Second Law of engines, I and R, in such a way so that I acts Thermodynamics. Hence the assertion ηI > ηR like a heat engine and R acts as a refrigerator. is wrong. No engine can have efficiency greater Let I absorb heat Q1 from the source, deliver than that of the Carnot engine. A similar work W ′ and release the heat Q1- W′ to the sink. argument can be constructed to show that a We arrange so that R returns the same heat Q1 reversible engine with one particular substance to the source, taking heat Q2 from the sink and cannot be more efficient than the one using requiring work W = Q1 – Q2 to be done on it. another substance. The maximum efficiency of Now suppose ηR < ηI i.e. if R were to act a Carnot engine given by Eq. (11.27) is as an engine it would give less work output independent of the nature of the system performing the Carnot cycle of operations. Thus we are justified in using an ideal gas as a system in the calculation of efficiency η of a Carnot engine. The ideal gas has a simple equation of I state, which allows us to readily calculate η, but the final result for η, [Eq. (11.27)], is true for any Carnot engine. R This final remark shows that in a Carnot cycle, Q1 T1 = W (11.28) T2 Q 2 is a universal relation independent of the natureFig. 11.10 An irreversible engine (I) coupled to a reversible refrigerator (R). If W ′ > W, this of the system. Here Q1 and Q2 are respectively, would amount to extraction of heat the heat absorbed and released isothermally W′ – W from the sink and its full (from the hot and to the cold reservoirs) in a conversion to work, in contradiction with Carnot engine. Equation (11.28), can, therefore, the Second Law of Thermodynamics. be used as a relation to define a truly universal thermodynamic temperature scale that is than that of I i.e. W < W ′ for a given Q1. With R independent of any particular properties of the acting like a refrigerator, this would mean system used in the Carnot cycle. Of course, for Q2 = Q1 – W > Q1 – W ′. Thus, on the whole, an ideal gas as a working substance, this the coupled I-R system extracts heat universal temperature is the same as the ideal (Q1 – W) – (Q1 – W ′) = (W ′ – W ) from the cold gas temperature introduced in section 11.9. 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. Reprint 2025-26 240 PHYSICS 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. 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 = 3 R 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  V 2   Q = W = µ R T ln   V1  9. In an adiabatic process of an ideal gas γ PV = constant C p where γ = C v Work done by an ideal gas in an adiabatic change of state from (P1, V1, T1) to (P2, V2, T2) is µR ( T1 − T2 ) W = γ – 1 Reprint 2025-26 THERMODYNAMICS 241 10. 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. 11. 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. 12. 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 T 2 η= 1 − (Carnot engine) T1 No engine operating between two temperatures can have efficiency greater than that of the Carnot engine. 13. 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 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. Reprint 2025-26 242 PHYSICS 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 11.1 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 ? 11.2 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.) 11.3 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. 11.4 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 ? 11.5 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) 11.6 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 ? 11.7 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? Reprint 2025-26 THERMODYNAMICS 243 11.8 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13) Fig. 11.11 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 Reprint 2025-26 CHAPTER TWELVE KINETIC THEORY 12.1 INTRODUCTION Boyle discovered the law named after him in 1661. Boyle, Newton and several others tried to explain the behaviour of 12.1 Introduction gases by considering that gases are made up of tiny atomic 12.2 Molecular nature of matter particles. The actual atomic theory got established more than 12.3 Behaviour of gases 150 years later. Kinetic theory explains the behaviour of gases 12.4 Kinetic theory of an ideal gas based on the idea that the gas consists of rapidly moving 12.5 Law of equipartition of energy atoms or molecules. This is possible as the inter-atomic forces, 12.6 Specific heat capacity which are short range forces that are important for solids and liquids, can be neglected for gases. The kinetic theory12.7 Mean free path was developed in the nineteenth century by Maxwell, Summary Boltzmann and others. It has been remarkably successful. It Points to ponder gives a molecular interpretation of pressure and temperature Exercises of a gas, and is consistent with gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It also relates measurable properties of gases such as viscosity, conduction and diffusion with molecular parameters, yielding estimates of molecular sizes and masses. This chapter gives an introduction to kinetic theory. 12.2 MOLECULAR NATURE OF MATTER Richard Feynman, one of the great physicists of 20th century considers the discovery that “Matter is made up of atoms” to be a very significant one. Humanity may suffer annihilation (due to nuclear catastrophe) or extinction (due to environmental disasters) if we do not act wisely. If that happens, and all of scientific knowledge were to be destroyed then Feynman would like the ‘Atomic Hypothesis’ to be communicated to the next generation of creatures in the universe. Atomic Hypothesis: All things are made of atoms - little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. Speculation that matter may not be continuous, existed in many places and cultures. Kanada in India and Democritus Reprint 2025-26 KINETIC THEORY 245 Atomic Hypothesis in Ancient India and Greece Though John Dalton is credited with the introduction of atomic viewpoint in modern science, scholars in ancient India and Greece conjectured long before the existence of atoms and molecules. In the Vaiseshika school of thought in India founded by Kanada (Sixth century B.C.) the atomic picture was developed in considerable detail. Atoms were thought to be eternal, indivisible, infinitesimal and ultimate parts of matter. It was argued that if matter could be subdivided without an end, there would be no difference between a mustard seed and the Meru mountain. The four kinds of atoms (Paramanu — Sanskrit word for the smallest particle) postulated were Bhoomi (Earth), Ap (water), Tejas (fire) and Vayu (air) that have characteristic mass and other attributes, were propounded. Akasa (space) was thought to have no atomic structure and was continuous and inert. Atoms combine to form different molecules (e.g. two atoms combine to form a diatomic molecule dvyanuka, three atoms form a tryanuka or a triatomic molecule), their properties depending upon the nature and ratio of the constituent atoms. The size of the atoms was also estimated, by conjecture or by methods that are not known to us. The estimates vary. In Lalitavistara, a famous biography of the Buddha written mainly in the second century B.C., the estimate is close to the modern estimate of atomic size, of the order of 10 –10 m. In ancient Greece, Democritus (Fourth century B.C.) is best known for his atomic hypothesis. The word ‘atom’ means ‘indivisible’ in Greek. According to him, atoms differ from each other physically, in shape, size and other properties and this resulted in the different properties of the substances formed by their combination. The atoms of water were smooth and round and unable to ‘hook’ on to each other, which is why liquid /water flows easily. The atoms of earth were rough and jagged, so they held together to form hard substances. The atoms of fire were thorny which is why it caused painful burns. These fascinating ideas, despite their ingenuity, could not evolve much further, perhaps because they were intuitive conjectures and speculations not tested and modified by quantitative experiments - the hallmark of modern science. in Greece had suggested that matter may consist of matter. The theory is now well accepted by of indivisible constituents. The scientific ‘Atomic scientists. However even at the end of the Theory’ is usually credited to John Dalton. He nineteenth century there were famous scientists proposed the atomic theory to explain the laws who did not believe in atomic theory ! of definite and multiple proportions obeyed by From many observations, in recent times we elements when they combine into compounds. now know that molecules (made up of one or The first law says that any given compound has, more atoms) constitute matter. Electron a fixed proportion by mass of its constituents. microscopes and scanning tunnelling The second law says that when two elements microscopes enable us to even see them. The form more than one compound, for a fixed mass size of an atom is about an angstrom (10 -10 m). of one element, the masses of the other elements In solids, which are tightly packed, atoms are are in ratio of small integers. spaced about a few angstroms (2 Å) apart. In To explain the laws Dalton suggested, about liquids the separation between atoms is also 200 years ago, that the smallest constituents about the same. In liquids the atoms are not of an element are atoms. Atoms of one element as rigidly fixed as in solids, and can move are identical but differ from those of other around. This enables a liquid to flow. In gases elements. A small number of atoms of each the interatomic distances are in tens of element combine to form a molecule of the angstroms. The average distance a molecule compound. Gay Lussac’s law, also given in early can travel without colliding is called the mean 19th century, states: When gases combine free path. The mean free path, in gases, is of chemically to yield another gas, their volumes the order of thousands of angstroms. The atoms are in the ratios of small integers. Avogadro’s are much freer in gases and can travel long law (or hypothesis) says: Equal volumes of all distances without colliding. If they are not gases at equal temperature and pressure have enclosed, gases disperse away. In solids and the same number of molecules. Avogadro’s law, liquids the closeness makes the interatomic force when combined with Dalton’s theory explains important. The force has a long range attraction Gay Lussac’s law. Since the elements are often and a short range repulsion. The atoms attract in the form of molecules, Dalton’s atomic theory when they are at a few angstroms but repel when can also be referred to as the molecular theory they come closer. The static appearance of a gas Reprint 2025-26 246 PHYSICS is misleading. The gas is full of activity and the is 6.02 × 1023. This is known as Avogadro number equilibrium is a dynamic one. In dynamic and is denoted by NA. The mass of 22.4 litres of equilibrium, molecules collide and change their any gas is equal to its molecular weight in grams speeds during the collision. Only the average at S.T.P (standard temperature 273 K and properties are constant. pressure 1 atm). This amount of substance is Atomic theory is not the end of our quest, but called a mole (see Chapter 1 for a more precise the beginning. We now know that atoms are not definition). Avogadro had guessed the equality of indivisible or elementary. They consist of a numbers in equal volumes of gas at a fixed nucleus and electrons. The nucleus itself is made temperature and pressure from chemical up of protons and neutrons. The protons and reactions. Kinetic theory justifies this hypothesis. neutrons are again made up of quarks. Even The perfect gas equation can be written as quarks may not be the end of the story. There PV = µ RT (12.3)may be string like elementary entities. Nature always has surprises for us, but the search for where µ is the number of moles and R = NA truth is often enjoyable and the discoveries kB is a universal constant. The temperature T is absolute temperature. Choosing kelvin scale forbeautiful. In this chapter, we shall limit ourselves absolute temperature, R = 8.314 J mol–1K–1.to understanding the behaviour of gases (and a Herelittle bit of solids), as a collection of moving molecules in incessant motion. M N µ = = (12.4) M 0 N A

1.20 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere? 43 Reprint 2025-26 Physics 1.21 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 mC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere? 1.22 An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density. 1.23 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10–22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates? Reprint 2025-26 Chapter Two ELECTROSTATIC POTENTIAL AND CAPACITANCE 2.12.12.12.12.1 IIINTRODUCTIONIINTRODUCTIONNTRODUCTIONNTRODUCTIONNTRODUCTION In Chapters 5 and 7 (Class XI), the notion of potential energy was introduced. When an external force does work in taking a body from a point to another against a force like spring force or gravitational force, that work gets stored as potential energy of the body. When the external force is removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy. The sum of kinetic and potential energies is thus conserved. Forces of this kind are called conservative forces. Spring force and gravitational force are examples of conservative forces. Coulomb force between two (stationary) charges is also a conservative force. This is not surprising, since both have inverse-square dependence on distance and differ mainly in the proportionality constants – the masses in the gravitational law are replaced by charges in Coulomb’s law. Thus, like the potential energy of a mass in a gravitational field, we can define electrostatic potential energy of a charge in an electrostatic field. Consider an electrostatic field EEEEE due to some charge configuration. First, for simplicity, consider the field E due to a charge Q placed at the origin. Now, imagine that we bring a test charge q from a point R to a point P against the repulsive force on it due to the charge Q. With reference Reprint 2025-26 Physics to Fig. 2.1, this will happen if Q and q are both positive or both negative. For definiteness, let us take Q, q > 0. Two remarks may be made here. First, we assume that the test charge q is so small that it does not disturb the original configuration, namely the charge Q at the origin (or else, we keep Q fixed at the origin by some unspecified force). Second, in bringing the charge q fromFIGURE 2.1 A test charge q (> 0) is moved from the point R to the R to P, we apply an external force Fext just enough to point P against the repulsive counter the repulsive electric force FE (i.e, Fext= –FE). force on it by the charge Q (> 0) This means there is no net force on or acceleration of placed at the origin. the charge q when it is brought from R to P, i.e., it is brought with infinitesimally slow constant speed. In this situation, work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q. If the external force is removed on reaching P, the electric force will take the charge away from Q – the stored energy (potential energy) at P is used to provide kinetic energy to the charge q in such a way that the sum of the kinetic and potential energies is conserved. Thus, work done by external forces in moving a charge q from R to P is WRP = – = (2.1) This work done is against electrostatic repulsive force and gets stored as potential energy. At every point in electric field, a particle with charge q possesses a certain electrostatic potential energy, this work done increases its potential energy by an amount equal to potential energy difference between points R and P. Thus, potential energy difference ∆U = U P − U R = W RP (2.2) (Note here that this displacement is in an opposite sense to the electric force and hence work done by electric field is negative, i.e., –WRP .) Therefore, we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating) charge q from one point to another for electric field of any arbitrary charge configuration. Two important comments may be made at this stage: (i) The right side of Eq. (2.2) depends only on the initial and final positions of the charge. It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force. The concept of the potential energy would not be meaningful if the work depended on the path. The path-independence of work done by an electrostatic field can be proved using the 46 Coulomb’s law. We omit this proof here. Reprint 2025-26 Electrostatic Potential and Capacitance (ii) Equation (2.2) defines potential energy difference in terms of the physically meaningful quantity work. Clearly, potential energy so defined is undetermined to within an additive constant.What this means is that the actual value of potential energy is not physically significant; it is only the difference of potential energy that is significant. We can always add an arbitrary constant a to potential energy at every point, since this will not change the potential energy difference: (U P + α) − (U R + α) = U P − U R Put it differently, there is a freedom in choosing the point where potential energy is zero. A convenient choice is to have electrostatic potential energy zero at infinity. With this choice, if we take the point R at infinity, we get from Eq. (2.2) Count Alessandro Volta (1745 – 1827) Italian W ∞ P = U P − U ∞ = U P (2.3) physicist, professor at Since the point P is arbitrary, Eq. (2.3) provides us with a Pavia. Volta established that the animal electri- COUNTdefinition of potential energy of a charge q at any point. city observed by LuigiPotential energy of charge q at a point (in the presence of field Galvani, 1737–1798, indue to any charge configuration) is the work done by the experiments with frog external force (equal and opposite to the electric force) in muscle tissue placed in bringing the charge q from infinity to that point. contact with dissimilar metals, was not due to 2.2 ELECTROSTATIC POTENTIAL any exceptional property of animal tissues but ALESSANDROConsider any general static charge configuration. We define was also generated potential energy of a test charge q in terms of the work done whenever any wet body on the charge q. This work is obviously proportional to q, since was sandwiched between the force at any point is qE, where E is the electric field at that dissimilar metals. This VOLTA point due to the given charge configuration. It is, therefore, led him to develop the convenient to divide the work by the amount of charge q, so first voltaic pile, orthat the resulting quantity is independent of q. In other words, battery, consisting of a (1745 work done per unit test charge is characteristic of the electric large stack of moist disks of cardboard (electro-field associated with the charge configuration. This leads to lyte) sandwiched the idea of electrostatic potential V due to a given charge between disks of metal –1827) configuration. From Eq. (2.1), we get: (electrodes). Work done by external force in bringing a unit positive charge from point R to P  U P − U R  = VP – VR = (2.4)  q  where VP and VR are the electrostatic potentials at P and R, respectively. Note, as before, that it is not the actual value of potential but the potential difference that is physically significant. If, as before, we choose the potential to be zero at infinity, Eq. (2.4) implies: Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V ) at that point. 47 Reprint 2025-26 Physics In other words, the electrostatic potential (V ) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point. The qualifying remarks made earlier regarding potential energy also apply to the definition of potential. To obtain the work done per unit test charge, we should take an infinitesimal test charge FIGURE 2.2 Work done on a test charge q dq, obtain the work done dW in bringing it from by the electrostatic field due to any given infinity to the point and determine the ratio charge configuration is independent dW/dq. Also, the external force at every point of the of the path, and depends only on path is to be equal and opposite to the electrostatic its initial and final positions. force on the test charge at that point. 2.3 POTENTIAL DUE TO A POINT CHARGE Consider a point charge Q at the origin (Fig. 2.3). For definiteness, take Q to be positive. We wish to determine the potential at any point P with position vector r from the origin. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point P. For Q > 0, the work done against the repulsive force on the test charge is positive. Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to the point P. At some intermediate point P¢ on the path, the electrostatic force on a unit positive charge is FIGURE 2.3 Work done in bringing a unit positive test charge from infinity to the Q × 1 rˆ ′ (2.5) point P, against the repulsive force of 2 4 πε0r ' charge Q (Q > 0), is the potential at P due to the charge Q. where ˆ′r is the unit vector along OP¢. Work done against this force from r¢ to r¢ + Dr¢ is Q ∆W = − 2 ∆′r (2.6) 4 πε0r ' The negative sign appears because for Dr¢ < 0, DW is positive. Total work done (W) by the external force is obtained by integrating Eq. (2.6) from r¢ = ¥ to r¢ = r, r Q Q r Q = dr ′ = ε 0r ′ 2 4 πε0r ′ ∞ 4 πε0r (2.7) W = − ∫4∞ π This, by definition is the potential at P due to the charge Q Q V (r ) = (2.8) 48 4 πε0r Reprint 2025-26 Electrostatic Potential and Capacitance Equation (2.8) is true for any sign of the charge Q, though we considered Q > 0 in its derivation. For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. [This is as it should be, since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are FIGURE 2.4 Variation of potential V with r [in units of in the same direction.] Finally, we (Q/4pe0) m-1] (blue curve) and field with r [in units of (Q/4pe0) m-2] (black curve) for a point charge Q.note that Eq. (2.8) is consistent with the choice that potential at infinity be zero. Figure (2.4) shows how the electrostatic potential ( 1/r) and the electrostatic field (1/r 2 ) varies with r. Example 2.1 (a) Calculate the potential at a point P due to a charge of 4 × 10–7C located 9 cm away. (b) Hence obtain the work done in bringing a charge of 2 × 10–9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought? Solution (a) = 4 × 104 V (b) W = qV = 2 × 10–9C × 4 × 104V = 8 × 10–5 J No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along EXAMPLE r and another perpendicular to r. The work done corresponding to the later will be zero. 2.1