TThermodynamics 647
Chapter 14 Thermodynamics
Thermodynamics is a branch of science which deals with exchange of (2) Thermodynamic variables and equation of state : A thermodynamic heat energy between bodies and conversion of the heat energy into system can be described by specifying its pressure, volume, temperature, mechanical energy and vice-versa. internal energy and the number of moles. These parameters are called thermodynamic variables. The relation between the thermodynamic Some Definitions variables (P, V, T) of the system is called equation of state. For moles of an ideal gas, equation of state is PV = RT and for 1 (1) Thermodynamic system mole of an it ideal gas is PV = RT (i) It is a collection of an extremely large number of atoms or (3) Thermodynamic equilibrium : In steady state thermodynamic molecules variables are independent of time and the system is said to be in the state (ii) It is confined with in certain boundaries. of thermodynamic equilibrium. For a system to be in thermodynamic (iii) Anything outside the thermodynamic system to which energy or equilibrium, the following conditions must be fulfilled. matter is exchanged is called its surroundings. (i) Mechanical equilibrium : There is no unbalanced force between the system and its surroundings. Surrounding (ii) Thermal equilibrium : There is a uniform temperature in all parts of the system and is same as that of surrounding. (iii) Chemical equilibrium : There is a uniform chemical composition through out the system and the surrounding. (4) Thermodynamic process : The process of change of state of a Gas System system involves change of thermodynamic variables such as pressure P, volume V and temperature T of the system. The process is known as thermodynamic process. Some important processes are (i) Isothermal process : Temperature remain constant Fig. 14.1 (iv) Thermodynamic system may be of three types (ii) Adiabatic process : No transfer of heat (a) Open system : It exchange both energy and matter with the (iii) Isobaric process : Pressure remains constant surrounding. (b) Closed system : It exchange only energy (not matter) with the (iv) Isochoric (isovolumic process) : Volume remains constant surroundings. (v) Cyclic and non-cyclic process : Incyclic process Initial and final states (c) Isolated system : It exchange neither energy nor matter with the are same while in non-cyclic process these states are different. surrounding. (vi) Reversible and irreversible process :
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(5) Indicator diagram : Whenever the state of a gas (P, V, T) is (iii) Change in internal energy does not depend on the path of the changed, we say the gaseous system is undergone a thermodynamic process. process. So it is called a point function i.e. it depends only on the initial and The graphical representation of the change in state of a gas by a final states of the system, i.e. U U U thermodynamic process is called indicator diagram. Indicator diagram is f i plotted generally in pressure and volume of gas. (3) Work (W) : Suppose a gas is confined in a cylinder that has a Zeroth Law of Thermodynamics movable piston at one end. If P be the pressure of the gas in the cylinder, then force exerted by the gas on the piston of the cylinder F = PA (A = Area If systems A and B are each in thermal equilibrium with a third system of cross-section of piston) C, then A and B are in thermal equilibrium with each other.
Piston Insulating Conducting
System System System System A B A B dx
System System F C C Gas
Conducting Insulating Fig. 14.4 (A) (B) When the piston is pushed outward an infinitesimal distance dx, the (1) The zeroth law leads to theFig. 14.3concept of temperature. All bodies in work done by the gas dW F.dx P(A dx) P dV thermal equilibrium must have a common property which has the same For a finite change in volume from V to V value for all of them. This property is called the temperature. i f Vf (2) The zeroth law came to light long after the first and seconds laws Total amount of work done W P dV P(Vf Vi) Vi of thermodynamics had been discovered and numbered. It is so named because it logically precedes the first and second laws of thermodynamics. (i) If we draw indicator diagram, the area bounded by PV-graph and volume axis represents the work done Heat, Internal Energy and Work in Thermodynamics P (1) Heat (Q) : It is the energy that is transferred between a system P and its environment because of the temperature difference between them. Work = Area = P(V – V) 2 1 (i) Heat is a path dependent quantity e.g. Heat required to change the A temperature of a given gas at a constant pressure is different from that required to change the temperature of same gas through same amount at constant volume. V V2 V P 1 (ii) For gases when heat is absorbed and temperature changes Fig. 14.5
Q CT V2 P Work PdV P(V2 V1) V1 At constant pressure (Q)P CP T
At constant volume (Q) C T V V V dV V V (2) Internal energy (U) : Internal energy of a system is the energy 1 2 possessed by the system due to molecular motion and molecular Fig. 14.6 P configuration. The energy due to molecular motion is called internal kinetic energy U K and that due to molecular configuration is called internal potential energy U P Work = 0 i.e. Total internal energy U UK UP
(i) For an ideal gas, as there is no molecular attraction U 0 p V i.e. internal energy of an ideal gas is totally kinetic and is given by P Fig. 14.7
3 3 P2 U UK RT and change in internal energy U R T 2 2 Work = Area of the shown trapezium (ii) In case of gases whatever be the process 1 P1 (P1 P2)(V2 V1) f R R(T T ) 2 U RT C T T f i 2 V ( 1) 1 V V2 V 1 Fig. 14.8 RT RT (P V P V ) (ii) From W PV P(V V ) f i f f i i f i 1 1 If system expands against some external force then Vf Vi W = positive
TThermodynamics 649
If system contracts because of external force then Vf Vi Q = U + W W = negative (3) It makes no distinction between work and heat as according to it the internal energy (and hence temperature) of a system may be increased P B P B Negative Positive either by adding heat to it or doing work on it or both. work work (4) Q and W are the path functions but U is the point function. A A (5) In the above equation all three quantities Q, U and W must V V be expressed either in Joule or in calorie. (A) Expansion (B) Compression (6) The first law introduces the concept of internal energy. (iii) Like heat, work done is alsoFig. depends14.9 upon initial and final state of the system and path adopted for the process (7) Limitation : First law of thermodynamics does not indicate the P direction of heat transfer. It does not tell anything about the conditions, under which heat can be transformed into work and also it does not A 1 indicate as to why the whole of heat energy cannot be converted into P mechanical work continuously. A 2 B A1 V Table 14.1 : Useful sign convention in thermodynamics (A) Less area 1 B Quantity Sign Condition P V 2 + When heat is supplied to a system A Q – When heat is drawn from the system A2 B + When work done by the gas (expansion) W V – When work done on the gas (compression) A < A W < W (B) More area 1 2 1 2 (iv) In cyclic process, work doneFig is. 14.10equal to the area of closed curve. It + With temperature rise, internal energy increases U is positive if the cycle is clockwise and it is negative if the cycle is – With temperature fall, internal energy decreases anticlockwise.
P C P2 Isobaric Process Work = Area of triangle ABC When a thermodynamic system undergoes a physical change in such a 1 way that its pressure remains constant, then the change is known as (V V )(P P ) A 2 2 1 2 1 isobaric process. P1 B (1) Equation of state : In this process V and T changes but P remains
V1 V2 V constant. Hence Charle’s law is obeyed in this process. Fig. 14.11 P C V1 V2 P Hence if pressure remains constant V T 2 T T Work = Area of rectangle ABCD 1 2 (2) Indicator diagram : Graph 1 represent isobaric expansion, graph 2 = AB AD represent isobaric compression. A = (V – V) (P – P) P 2 1 2 1 1 B P P
V1 V2 V 2 1 P Fig. 14.12 dP P2 Slope = 0 Slope = dV V V Work = (P2 P1)(V2 V1) (A) Expansion 4 (B) Compression P 1 Fig. 14.14 (i) In isobaric expansion (Heating)
V1 V2 V Temperature increases so U is positive Fig. 14.13 First Law of Thermodynamics (FLOT) Volume increases so W is positive (1) It is a statement of conservation of energy in thermodynamical process. Heat flows into the system so Q is positive (ii) In isobaric compression (Cooling) (2) According to it heat given to a system (Q) is equal to the sum of increase in its internal energy (U) and the work done (W) by the system Temperature decreases so U is negative against the surroundings.
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Volume decreases so W is negative (2) Indicator diagram : Graph 1 and 2 represent isometric increase in pressure at volume V1 and isometric decrease in pressure at volume V2 Heat flows out from the system so Q is negative dP respectively and slope of indicator diagram (3) Specific heat : Specific heat of gas during isobaric process dV f (i) Isometric heating C P 1R P 2 (a) Pressure increases P 1 (4) Bulk modulus of elasticity : K 0 [As P = 0] (b) Temperature increases V V (c) Q positive (5) Work done in isobaric process V (d) U positive 1 V (A) V V f f (ii) Isometric cooling P W P dV P dV P[Vf Vi ] [As P = constant] V V i i (a) Pressure decreases 2 W P(V V ) R[T T ] R T f i f i (b) Temperature decreases
(6) FLOT in isobaric process : From Q U W (c) Q negative V V R 2 (d) U negative (B) U CV T T and W R T ( 1) Fig. 14.17 (3) Specific heat : Specific heat of gas during isochoric process R f (Q)P T R T R T CP T C R ( 1) 1 V 2 P P (7) Examples of isobaric process : All state changes occurs at constant (4) Bulk modulus of elasticity : K temperature and pressure. V 0 V Boiling of water Water Vapours (5) Work done in isochoric process (i) Water vapours W PV P[Vf Vi ] = 0 [As V = constant] (ii) Temperature constant (6) FLOT in isochoric process : From Fig. 14.15 (iii) Volume increases R P V P V W = 0 (Q) U C T T f f i i (iv) A part of heat supplied is used to change volume (expansion) V V 1 1 against external pressure and remaining part is used to increase it's potential energy (kinetic energy remains constant) Isothermal Process (v) From FLOT Q = U + W mL = U + P(V – V) When a thermodynamic system undergoes a physical change in such a f i way that its temperature remains constant, then the change is known as Freezing of water isothermal changes. Ice (i) Water ice Water (1) Essential condition for isothermal process (i) The walls of the container must be perfectly conducting to allow (ii) Temperature constant free exchange of heat between the gas and its surrounding. (iii) Volume increases Fig. 14.16 (ii) The process of compression or expansion should be so slow so as to provide time for the exchange of heat. (iv) Heat is given by water it self. It is used to do work against external atmospheric pressure and to decreases the internal potential Since these two conditions are not fully realised in practice, therefore, energy. no process is perfectly isothermal. (v) From FLOT Q = U + W – mL = U + P(V – V) (2) Equation of state : In this process, P and V change but T = f i constant i.e. change in temperature T = 0. Isochoric or Isometric Process Boyle’s law is obeyed i.e. PV= constant PV = PV When a thermodynamic process undergoes a physical change in such a 1 1 2 2 way that its volume remains constant, then the change is known as (3) Example of isothermal process : Melting of ice (at 0°C) and boiling isochoric process. of water (at 100°C) are common example of this process. (1) Equation of state : In this process P and T changes but V = (4) Indicator diagram : According to PV = constant, graph between P constant. Hence Gay-Lussac’s law is obeyed in this process i.e. P T and V is a part of rectangular hyperbola. The graphs at different temperature P P are parallel to each other are called isotherms. 1 2 constant P T1 T2
T1 T3 T2 T1 V Fig. 14.18 TThermodynamics 651 (ii) The system should be compressed or allowed to expand suddenly so that there is no time for the exchange of heat between the system and its surroundings. Since, these two conditions are not fully realised in practice, so no (i) Slope of isothermal curve : By differentiating PV = constant. We get process is perfectly adiabatic. (2) Some examples of adiabatic process P dV V dP 0 P (i) Sudden compression or expansion of a gas in a container with PdV VdP perfectly non-conducting walls. dP P (ii) Sudden bursting of the tube of bicycle tyre. Slope = tan dV V (iii) Propagation of sound waves in air and other gases. V (iv) Expansion of steam in the cylinder of steam engine. (ii) Area between the isotherm and volume axisFig. represents 14.19 the work done in isothermal process. (3) FLOT in adiabatic process : From If volume increases W = + (Area under curve) and if volume For adiabatic process Q 0 U W decreases W = – (Area under curve) If W = positive then U = negative so temperature decreases i.e. (5) Specific heat : Specific heat of gas during isothermal change is adiabatic expansion produce cooling. Q Q If W = negative then U = positive so temperature increases i.e. infinite. As C [As T = 0] mT m 0 adiabatic compression produce heating. (4) Equation of state : In adiabatic change ideal gases do not obeys (6) Isothermal elasticity (E ) : For this process PV = constant. Boyle's law but obeys Poisson's law. According to it dP Stress P dV V dP P E CP dV / V Strain PV = constant; where CV i.e. isothermal elasticity is equal to pressure E P (i) For temperature and volume 5 2 1 1 1 At N.T.P., E = Atmospheric pressure = 1.0110 N / m – 1 TV = constant T1V1 T2V2 or T V (7) Work done in isothermal process (ii) For temperature and pressure V V f f RT 1 W P dV dV [As PV = RT] V V T 1 1 1 i i V = const. T P T P or T P or P T P 1 1 1 2 2 V V f f W RT loge 2.303 RT log10 Table 14.2 : Special cases of adiabatic process Vi Vi 1 1 Type of gas P T 1 1 Pi Pi V P T V or W RT loge 2.303RT log10 P P f f Monoatomic 1 1 P 5 / 2 T 5 / 3 P T 2 / 3 (8) FLOT in isothermal process : From Q U W = 5/3 V V Diatomic 1 1 U 0 [As T = 0] Q W P P T7 / 2 T = 7/5 V 7 / 5 V 2 / 5 i.e. heat supplied in an isothermal change is used to do work against Polyatomic 1 4 1 external surrounding. P P T T = 4/3 V 4 / 3 V1 / 3 or if the work is done on the system than equal amount of heat energy will be liberated by the system. (5) Indicator diagram Adiabatic Process (i) Curve obtained on PV graph are called adiabatic curve. When a thermodynamic system undergoes a change in such a way that (ii) Slope of adiabatic curve : From PV constant no exchange of heat takes place between System and surroundings, the process is known as adiabatic process. By differentiating, we get P In this process P, V and T changes but Q = 0. dP V PV 1 dV 0 (1) Essential conditions for adiabatic process dP PV 1 P (i) There should not be any exchange of heat between the system and dV V its surroundings. All walls of the container and the piston must be perfectly V V insulating. Fig. 14.20 652 Thermodynamics P Slope of adiabatic curve tan Stop cock V Gas P Vacuum (iii) But we also know that slope of isothermal curve tan V (Slope) Hence (Slope) = (Slope) or Adi 1 Adi Iso (Slope) Insulated Iso W = 0 (Because walls are rigid)Fig. 14.2 3 (6) Specific heat : Specific heat of a gas during adiabatic change is Q = 0 (Because walls are insulated) Q 0 zero As C 0 [As Q = 0] U = U – U = 0 (Because Q and W are zero. Thus the final and mT mT f i initial energies are equal in free expansion. (7) Adiabatic elasticity (E) : PV constant Cyclic and Non-cyclic Process Differentiating both sides d PV PV 1dV 0 A cyclic process consists of a series of changes which return the dP Stress P E E P system back to its initial state. dV / V Strain In non-cyclic process the series of changes involved do not return the i.e. adiabatic elasticity is times that of pressure system back to its initial state. E C (1) In case of cyclic process as U U U U U 0 Also isothermal elasticity E P P f i f i E C V i.e. change in internal energy for cyclic process is zero and also i.e. the ratio of two elasticity of gases is equal to the ratio of two U T T = 0 i.e. temperature of system remains constant. specific heats. (2) From FLOT Q U W Q W (8) Work done in adiabatic process i.e. heat supplied is equal to the work done by the system. Vf Vf K [PV P V ] R(T T ) W P dV dV W i i f f i f (3) For cyclic process P-V graph is a closed curve and area enclosed by the V V V ( 1) ( 1) i i closed path represents the work done. (As PV = K, PV = RT and PV = RT) f f f i i i If the cycle is clockwise work done is positive and if the cycle is (i) W quantity of gas (either M or ) anticlockwise work done is negative. (ii) W temperature difference (T – T) i f P P 1 D (iii) W W < W < W A B A mono di tri mono dia tri 1 Positive work Negative work (9) Comparison between isothermal and adiabatic indicator diagrams : D C B C Always remember that adiabatic curves are more steeper than isothermal curves V V (i) Equal expansion : Graph 1 represent isothermal process and 2 (A) (B) Fig. 14.24 represent adiabatic process P (4) Work done in non cyclic process depends upon the path chosen or W > W the series of changes involved and can be calculated by the area covered isothermal adiabatic P > P between the curve and volume axis on PV diagram. isothermal adiabatic 1 T > T P P isothermal adiabatic 2 C B A (Slope) < (Slope) Isothermal Adiabatic V Vi Vf D B C (ii) Compression : Graph 1 represent adiabatic processFig. 14.2and1 2 represent isothermal process P A V W > W V Adiabatic Isothermal WABC= + Shaded area W = – Shaded area P > P ABCD Adiabatic Isothermal 1 (A) (B) T > T Adiabatic Isotherm al 2 Fig. 14.25 (Slope) < (Slope) Isothermal Adiabatic V Quasi Static Process V V f i When we perform a process on a given system, its state is, in general, (10) Free expansion : Free expansion is adiabatic processFig. 14.2 2in which no changed. Suppose the initial state of the system is described by the values work is performed on or by the system. Consider two vessels placed in a P1, V1, T1 and the final state by P2, V2, T2 . If the process is performed in system which is enclosed with thermal insulation (asbestos-covered). One such a way that at any instant during the process, the system is very nearly vessel contains a gas and the other is evacuated. When suddenly the in thermodynamic equilibrium, the process is called quasi-static. This means, stopcock is opened, the gas rushes into the evacuated vessel and expands we can specify the parameters P, V, T uniquely at any instant during such a freely. process. TThermodynamics 653 Actual processes are not quasi-static. To change the pressure of a gas, (iv) Produced by the passage of an electric current through a we can move a piston inside the enclosure. The gas near the piston is acted resistance is irreversible. upon by piston. The pressure of the gas may not be uniform everywhere (v) Heat transfer between bodies at different temperatures is also while the piston is moving. However, we can move the piston very slowly to irreversible. make the process as close to quasi-static as we wish. Thus, a quasi-static process is an idealised process in which all changes take place infinitely (vi) Joule-Thomson effect is irreversible because on reversing the flow slowly. of gas a similar cooling or heating effect is not observed. Reversible and Irreversible Process Mixed Graphical Representation (1) Reversible process : A reversible process is one which can be (1) PV-graphs reversed in such a way that all changes occurring in the direct process are P 4 exactly repeated in the opposite order and inverse sense and no change is 3 1 Isobaric (P-constant) 2 left in any of the bodies taking part in the process or in the surroundings. 1 For example if heat is absorbed in the direct process, the same amount of 2 Isothermal (Because P ) heat should be given out in the reverse process, if work is done on the V working substance in the direct process then the same amount of work 1 1 should be done by the working substance in the reverse process. The 3 Adiabatic (Because ) P conditions for reversibility are V V O (i) There must be complete absence of dissipative forces such as Fig. 14.26 4 Isochoric (V-constant) friction, viscosity, electric resistance etc. (2) PT-graphs (ii) The direct and reverse processes must take place infinitely slowly. P 2 3 (iii) The temperature of the system must not differ appreciably from 4 1 Isobaric (P-constant) its surroundings. 1 2 Isothermal (T-constant) Some examples of reversible process are (a) All isothermal and adiabatic changes are reversible if they are 3 Adiabatic (Because P T 1 ) performed very slowly. T (b) When a certain amount of heat is absorbed by ice, it melts. If the O 4 Isochoric (In isochoric P T) Fig. 14.27 same amount of heat is removed from it, the water formed in the direct process will be converted into ice. (3) VT-graphs (c) An extremely slow extension or contraction of a spring without V 3 setting up oscillations. 2 1 Isochoric (V-constant) 4 (d) When a perfectly elastic ball falls from some height on a perfectly 1 1 elastic horizontal plane, the ball rises to the initial height. 1 2 Adiabatic (Because V T ) (e) If the resistance of a thermocouple is negligible there will be no 3 Isothermal (T-constant) heat produced due to Joule’s heating effect. In such a case heating or cooling is reversible. At a junction where a cooling effect is produced due to T O 4 Isobaric (In isobaric V T ) Peltier effect when current flows in one direction and equal heating effect is Fig. 14.28 produced when the current is reversed. (f) Very slow evaporation or condensation. Heat Engine It should be remembered that the conditions mentioned for a Heat engine is a device which converts heat into work continuously through a cyclic process. reversible process can never be realised in practice. Hence, a reversible process is only an ideal concept. In actual process, there is always loss of The essential parts of a heat engine are heat due to friction, conduction, radiation etc. (1) Source : It is a reservoir of heat at high temperature and infinite (2) Irreversible process : Any process which is not reversible exactly is an thermal capacity. Any amount of heat can be extracted from it. irreversible process. All natural processes such as conduction, radiation, (2) Working substance : Steam, petrol etc. radioactive decay etc. are irreversible. All practical processes such as free (3) Sink : It is a reservoir of heat at low temperature and infinite expansion, Joule-Thomson expansion, electrical heating of a wire are also thermal capacity. Any amount of heat can be given to the sink. irreversible. Some examples of irreversible processes are given below (i) When a steel ball is allowed to fall on an inelastic lead sheet, its kinetic energy changes into heat energy by friction. The heat energy raises Source (T1) the temperature of lead sheet. No reverse transformation of heat energy occurs. Q1 Heat (ii) The sudden and fast stretching of a spring may produce vibrations W = Q1 – Q2 Engine in it. Now a part of the energy is dissipated. This is the case of irreversible Q process. 2 (iii) Sudden expansion or contraction and rapid evaporation or Sink (T2) The working substance absorbs heat Q from the source, does an condensation are examples of irreversible processes. 1 amount of work W, returns theFig. remaining 14.29 amount of heat to the sink and 654 Thermodynamics comes back to its original state and there occurs no change in its internal Q1 Q2 T1 T2 Q2 T2 energy. or Q2 T2 Q1 Q2 T1 T2 By repeating the same cycle over and over again, work is continuously T obtained. So coefficient of performance 2 T T The performance of heat engine is expressed by means of “efficiency” 1 2 where T = temperature of surrounding, T = temperature of cold body. which is defined as the ratio of useful work obtained from the engine to 1 2 It is clear that = 0 when T = 0 the heat supplied to it. 2 i.e. the coefficient of performance will be zero if the cold body is at the Work done W temperature equal to absolute zero. Heat input Q 1 (2) Relation between coefficient of performance and efficiency of For cyclic process U = 0 hence from FLOT Q = W refrigerator Q2 Q / Q Q1 Q2 Q2 We know 2 1 ….. (i) So W Q1 Q2 1 Q1 Q2 1 Q2 / Q1 Q1 Q1 Q Q A perfect heat engine is one which converts all heat into work i.e. But the efficiency 1 2 or 2 1 …..(ii) Q Q W Q1 so that Q2 0 and hence 1 . 1 1 But practically efficiency of an engine is always less than 1. 1 From (i) and (ii) we get, Refrigerator or Heat Pump Second Law of Thermodynamics A refrigerator or heat pump is basically a heat engine run in reverse direction. First law of thermodynamics merely explains the equivalence of work and heat. It does not explain why heat flows from bodies at higher It essentially consists of three parts temperatures to those at lower temperatures. It cannot tell us why the (1) Source : At higher temperature T. 1 converse is possible. It cannot explain why the efficiency of a heat engine is (2) Working substance : It is called refrigerant liquid ammonia and always less than unity. It is also unable to explain why cool water on stirring freon works as a working substance. gets hotter whereas there is no such effect on stirring warm water in a beaker. Second law of thermodynamics provides answers to these questions. (3) Sink : At lower temperature T. 2 Statement of this law is as follows Source (1) Clausius statement : It is impossible for a self acting machine to (Atmosphere) (T1) transfer heat from a colder body to a hotter one without the aid of an Q 1 external agency. Heat W = Q1 – Q2 From Clausius statement it is clear that heat cannot flow from a Engine body at low temperature to one at higher temperature unless work is done Q2 by an external agency. This statement is in fair agreement with our Sink (T2) experiences in different branches of physics. For example, electrical current (Contents of refrigerator) cannot flow from a conductor at lower electrostatic potential to that at Fig. 14.30 The working substance takes heat Q from a sink (contents of higher potential unless an external work is done. Similarly, a body at a lower 2 refrigerator) at lower temperature, has a net amount of work done W on it gravitational potential level cannot move up to higher level without work by an external agent (usually compressor of refrigerator) and gives out a done by an external agency. larger amount of heat Q to a hot body at temperature T (usually 1 1 (2) Kelvin’s statement : It is impossible for a body or system to atmosphere). Thus, it transfers heat from a cold to a hot body at the perform continuous work by cooling it to a temperature lower than the expense of mechanical energy supplied to it by an external agent. The cold temperature of the coldest one of its surroundings. A Carnot engine cannot body is thus cooled more and more. work if the source and sink are at the same temperature because work done The performance of a refrigerator is expressed by means of “coefficient by the engine will result into cooling the source and heating the of performance” which is defined as the ratio of the heat extracted from surroundings more and more. the cold body to the work needed to transfer it to the hot body. (3) Kelvin-Planck’s statement : It is impossible to design an engine Heat extracted Q Q that extracts heat and fully utilises into work without producing any other i.e. 2 2 Work done W Q1 Q2 effect. A perfect refrigerator is one which transfers heat from cold to hot From this statement it is clear that any amount of heat can never be body without doing work converted completely into work. It is essential for an engine to return some amount of heat to the sink. An engine essentially requires a source as well i.e. W = 0 so that Q1 Q2 and hence as sink. The efficiency of an engine is always less than unity because heat Q T cannot be fully converted into work. (1) Carnot refrigerator : For Carnot refrigerator 1 1 Q T 2 2 Carnot Engine TThermodynamics 655 (1) Carnot designed a theoretical engine which is free from all the V 4 V4 defects of a practical engine. This engine cannot be realised in actual W3 Q2 P dV RT2 loge V3 V practice, however, this can be taken as a standard against which the 3 performance of an actual engine can be judged. V3 RT2 loge Area CDFH V4 (iv) Fourth stroke (adiabatic compression) (curve DA) : Finally the cylinder is again placed on non-conducting stand and the compression is continued so that gas returns to its initial stage. Ideal gas V 1 R R W4 P dV (T2 T1) (T1 T2) Area ADFE V4 1 1 Source Insulating Sink (3) Efficiency of Carnot cycle : The efficiency of engine is defined as the ratio of work done to the heat supplied i.e. T1 K stand T2 K Work done W It consists of the following parts Fig. 14.31 Heat input Q1 (i) A cylinder with perfectly non-conducting walls and a perfectly Net work done during the complete cycle conducting base containing a perfect gas as working substance and fitted with a non-conducting frictionless piston W W1 W2 (W3 )(W4 ) W1 W3 Area ABCD (ii) A source of infinite thermal capacity maintained at constant higher [As W W ] temperature T 2 4 1. W W W Q Q W Q (iii) A sink of infinite thermal capacity maintained at constant lower 1 3 1 2 1 3 1 2 temperature T. Q W Q W Q 2 1 1 1 1 1 (iv) A perfectly non-conducting stand for the cylinder. RT log (V / V ) or 1 2 e 3 4 (2) Carnot cycle : As the engine works, the working substance of the RT1 loge (V2 / V1 ) engine undergoes a cycle known as Carnot cycle. The Carnot cycle consists Since points B and C lie on same adiabatic curve of the following four strokes 1 P 1 1 T1 V3 T1V2 T2V3 or …..(i) Q1 A(P1, V1) T2 V2 Also point D and A lie on the same adiabatic curve T1 1 W = Q – Q T V 1 2 1 1 or 1 4 …..(ii) T1V1 T2 V4 T2 V1 (P4, V4)D V V V V V V 3 4 3 2 3 2 T From (i) and (ii), or loge loge C(P3, V3) 2 V V V V V V Q2 2 1 4 1 4 1 V E F G H T So efficiency of Carnot engine 1 2 Fig. 14.32 (i) First stroke (Isothermal expansion) (curve AB) : T1 The cylinder containing ideal gas as working substance allowed to (i) Efficiency of a heat engine depends only on temperatures of source and expand slowly at this constant temperature T. sink and is independent of all other factors. 1 Work done = Heat absorbed by the system (ii) All reversible heat engines working between same temperatures are equally efficient and no heat engine can be more efficient than Carnot V2 V 2 Area ABGE engine (as it is ideal). W1 Q1 P dV RT1 loge V1 V 1 (iii) As on Kelvin scale, temperature can never be negative (as 0 K is (ii) Second stroke (Adiabatic expansion) (curve BC) : defined as the lowest possible temperature) and T and T are finite, l 2 The cylinder is then placed on the non conducting stand and the gas is efficiency of a heat engine is always lesser than unity, i.e., whole of heat can allowed to expand adiabatically till the temperature falls from T to T. never be converted into work which is in accordance with second law. 1 2 (4) Carnot theorem : The efficiency of Carnot’s heat engine depends V3 R W2 P dV = [T1 T2 ] Area BCHG only on the temperature of source (T) and temperature of sink (T), i.e., 1 2 V2 ( 1) (iii) Third stroke (Isothermal compression) (curve CD) : . The cylinder is placed on the sink and the gas is compressed at constant temperature T. Carnot stated that no heat engine working between two given 2 temperatures of source and sink can be more efficient than a perfectly Work done = Heat released by the system reversible engine (Carnot engine) working between the same two 656 Thermodynamics temperatures. Carnot's reversible engine working between two given temperatures is considered to be the most efficient engine. Table 14.3 : Difference Between Petrol Engine and Diesel Engine When a thermos bottle is vigorously shaken : Petrol engine Diesel engine Heat transferred to the coffee Q = 0 Working substance is a mixture of Working substance in this engine is a [As thermos flask is insulated from the surrounding] petrol vapour and air. mixture of diesel vapour and air. Work is done on the coffee against viscous force Efficiency is smaller (~47%). Efficiency is larger (~55%). W = (–) It works with a spark plug. It works with an oil plug. Internal energy of the coffee increases U = (+) It is associated with the risk of No risk of explosion, because only air explosion, because petrol vapour and is compressed. Hence compression and temperature of the coffee also increases T = (+) air is compressed. So, low compression ratio is kept large. Work done without the volume limits ratio is kept. Vf Petrol vapour and air is created with Spray of diesel is obtained through W P dV spark plug. the jet. Vi Entropy From this equation it seems as if work done can be calculated only when P-V equation is known and limits Vi and V f are known to us. Entropy is a measure of disorder of molecular motion of a system. But it is not so. We can calculate work done if we know the limits of Greater is the disorder, greater is the entropy. temperature. The change in entropy i.e. For moles of an ideal gas if P with temperature limits T Heat absorbed by system dQ T 0 dS or dS and 2T. Absolute temperatu re T 0 The relation is called the mathematical form of Second Law of RT RT2 2RT From PV = RT V dV dT Thermodynamics. P (1) For solids and liquids Vf 2T0 2RT (i) When heat given to a substance changes its state at constant W P dV dT 2RT0 Vi T0 T dQ mL temperature, then change in entropy dS T T Work done with spring : If mass less piston is attached to a spring of where positive sign refers to heat absorption and negative sign to heat force constant K and a mass m is placed over the piston. If the external evolution. pressure is P0 and due to expansion of gas the piston moves up through (ii) When heat given to a substance raises its temperature from T to a distance x then 1 T, then change in entropy 2 Total work done by the gas T2 dQ dT T2 W W1 W2 W3 dS mc mc loge M T T 1 T T1 where W = Work done against 1 M x external pressure (P ) T2 0 S 2.303mc log10 . T W = Work done against spring 1 2 P (2) For a perfect gas : Perfect gas equation for n moles is PV = nRT force (Kx) dQ C dT P dV W = Work done against gravitational force (mg) S V [As dQ = dU + dW] 3 T T 1 W P V Kx 2 mgx RT 0 2 CVdT dV S V The efficiency of an actual engine is much lesser than that of an T ideal engine. Actually the practical efficiency of a steam engine is about T2 dT V2 dV (8-15)% while that of a petrol engine is 40%. The efficiency of a diesel CV R [As PV = RT] T1 T V1 V engine is maximum and is about (50-55)%. T V 2 2 x When P and V bear the relation PV = constant, where x 1 or S CV loge R loge T1 V1 the process is called a polytropic one. In this process the molar heat R R R T2 P2 In terms of T and P, S C log R log capacity is, C CV P e e 1 x 1 1 x T1 P1 P V Enthalpy : Four quantities called “thermodynamic potentials” are and in terms of P and V S C log 2 C log 2 V e P e useful in the chemical thermodynamics of reactions and non-cyclic P1 V1 processes. They are internal energy, the enthalpy, the Helmoltz free energy and the Gibbs free energy. Enthalpy is defined by TThermodynamics 657 H = U + PV where P and V are the pressure and volume, and U is internal energy, Enthalpy is somewhat parallel to the first law of thermodynamics for a constant pressure system Q = U + PV since in this case Q = H. Confusion about FLOT It is typical for chemistry texts to write the first law as U = Q + W It is the same law, of course the thermodynamic expression of the conservation of energy principle. It is just that W is defined as the work done on the system instead of work done by the system. In the context of physics, the common scenario is one of adding heat to a volume of gas and using the expansion of that gas to do work, as in the pushing down of a piston in an internal combustion engine. In the context of chemical reactions and process, it may be more common to deal with situations where work is done on the system rather than by it. Possibilities P a b V If a Isothermal then b Must be adiabatic But If b adiabatic then it is not compulsory that a must be isothermal, it may be adiabatic also.