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CHAPTER – 12, THERMODYNAMICS, Thermodynamics means heat, flow. Thermodynamics deals with the, conversion of heat into work and, work into heat., Thermal equilibrium, Two systems are said to be in, thermal equilibrium, if their, temperatures are the same. Then there, will not be any heat flow from one, system to another., Internal energy, It is the sum of all kinetic energies, and potential energies of the, molecules of the system., Zeroth Law of thermodynamics, This law was formulated by R.H., Fowler., The law states that “two systems, which are in thermal equilibrium with, a third system separately are in, thermal equilibrium with each other.”, , First Law of thermodynamics, The first law of thermodynamics is a, statement of law of conservation of, energy., , The law states that “If an amount of, heat is given to a system, a part of the, heat is used to increase the internal, energy and other part is used to do the, external work”, Q  U  W,  Q  H eat sup plied to the system.,  U  change in int ernal energy,  W  workdone by the system, But  W  P  V,  Q  U  P V, , Thermodynamic state variables, The physical quantities which, characterise a system are known as, state variables., Eg: - Pressure, volume, temperature,, mass, density, internal energy, heat, capacity, specific heat capacity etc., Thermodynamic state variables are of, two types:, i) Intensive state variables, They are the state variables which, do not depend on the size of the, system., Eg: - Pressure, temperature, density,, specific heat capacity etc., ii), Extensive sate variables, They are the state variables which, depend on the size of the system., Eg: -volume, mass, heat capacity,, internal energy etc., Note: - Heat and work are not state, variables., Thermodynamic Processes, (i) Quasi - static Process, A quasi-static process is an, infinitely slow process such that the, , SAJU K JOHN, M.Sc. Physics, NET, Doctoral Research Fellow at NIT Calicut, , 1

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system remains in thermal and, mechanical equilibrium with the, surroundings throughout the process., (ii) Isobaric Process, In an isobaric process, pressure is, constant throughout the process., If heat is applied, the piston moves up., A part of heat supplied is used, increase the internal energy and the, other part are used to do the work., Q   U   W., Q  nC P T, C P  Molar specific heat capacity, at cons tan t pressure., , (iii) Isochoric Process, In an isochoric process, volume is, constant throughout the process., V  0, , W  PV  0,  Q  U, The heat sup plied is completely, used to increase the int ernal energy., Q  nC V T, C V  molar specific heat capacity, at cons tan t volume., (iv) Isothermal process, It is a process taking place at, constant temperature., Equation for isothermal process, is PV = Constant [Boyle’s law], (Here the constant is µRT), Conditions for isothermal process, i. The process must be slow, ii. There should be a perfect conducting, wall (diathermic wall) between the, system and surroundings., , Eg: - The expansion of a gas in a, metallic cylinder placed in a large, reservoir of fixed temperature is an, example of isothermal process., Melting of ice at its normal melting, point, vaporization of a liquid at its, normal boiling point etc. are other, examples., Work done during an isothermal, process, Suppose a system of gas is, expanding from an initial volume V1, to a final volume V2 during an, isothermal process., The work done for the small change in, volume ‘dV’ is given by, , dW  PdV,  The total workdone,, W, , , , , V2, V1, , , , V2, , V1, , dW, PdV, ButPV  nRT, P, , nRT, V, , nRT, V1 V dV, V2 1,  nRT , dV, V1 V, , W, , V2, ,  nRT  log V V2, V, , 1, ,  nRT  log V2  log V1 , But we have, logA-logB=log, , A, B, , V ,  W  nRT log  2 ,  V1 , V , W  nRT log  2 ,  V1 , Adiabatic Process, , SAJU K JOHN, M.Sc. Physics, NET, Doctoral Research Fellow at NIT Calicut, , 2

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Problem4: A thermos flask contains, coffee. It is violently shaken., Considering the coffee as a system, answer the following:, a) Does the temperature rise?, b) Has heat been added to it?, c) Has internal energy changed?, Ans:, Problem2: Draw indicator diagrams, for isothermal and adiabatic, processes., Ans:, , Problem3: If a gas is compressed to, half its volume first rapidly and then, slowly, in which case the work done, will be greater?, , Problem5: Isothermal, isobaric,, isochoric and adiabatic processes are, some special thermodynamic, processes. In which of these, processes, the work done is maximum, when the gas expands from V1 to V2 ?, Ans:, , Ans:, , SAJU K JOHN, M.Sc. Physics, NET, Doctoral Research Fellow at NIT Calicut, , 4

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Problem6: A gas expands, adiabatically so that 50 J of work is, obtained. What is the change in, temperature in the above process if, the working substance is a, monoatomic gas? (R=8.314J/molK), Ans:, , Cyclic Process, In a cyclic process, the system, returns to its initial state., ,   U  0, for a cyclic process., Then  Q   W, , Heat Engines:, Heat engine is a device to convert, heat energy in to mechanical energy., , Problem7: A thermodynamic system, performs work without taking heat, from an external source., a) Which process is involved in this, case?, b) What is the source of energy for, this work?, c) By what factor does the pressure, of the system decrease if the, volume is doubled (γ=1.4)., , It consist of:, i. A very hot body of large specific heat, capacity called the source., ii., A working substance. Eg: a, mixture of fuel vapour and air in a, petrol or diesel engine or steam in a, steam engine., iii., An insulating stand, iv., A cold body of large specific, heat capacity called sink., Schematic representation of a heat, engine is given below., , Working substance absorbs heat of, amount Q 1 from the source; a part of, the heat is converted into useful work, SAJU K JOHN, M.Sc. Physics, NET, Doctoral Research Fellow at NIT Calicut, , 5

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W and other part Q2 is given to the, sink., , Efficiency of a heat engine (), , , W Q1  Q 2, Q, ,  1 2, Q1, Q1, Q1, ,   1, , Q2, Q1, , For Q 2  0, =1, i.e., 100% efficiency for heat engine,, which is never possible., Types of heat engines:, 1. External combustion engine – in, which heat is produced by burning, fuel outside the cylinder. Eg:, steam engine, 2. Internal combustion engine – in, such engines heat is produced by, burning fuel inside the cylinder., Eg. Petrol and diesel engines., Refrigerators and heat Pumps:, A refrigerator works in the reverse, order of a heat engine., The working substance for both, refrigerator and heat pump is Freon., , Here the working substance, absorbs heat Q 2 from the sink, some, work W is done on the working, substance by an external agency and, the working substance liberates a, large amount of heat Q1 to the source., Q1 = Q2 + W, , or W = Q1 - Q2, The coefficient of performance of a, refrigerator, ,  =, , Q2, Q2, , W, Q1  Q 2, , A heat pump is a device to pump heat, into a portion of space (room)., Coefficient of performance of a heat, pump, , =, , Q1, Q1, , W Q1 Q2, , [In a refrigerator, sink is the cooling, chamber. Source is the room in which, the refrigerator is placed. The work, (W) is the work done by the, compressor by consuming electricity.], [In a heat pump, sink is the, environment outside the room. Source, is the room which is to be heated.], Second Law of Thermodynamics, Kelvin – Planck Statement:, No process is possible whose sole, result is the absorption of heat from, a reservoir and the complete, conversion of heat into work., , Explanation: - This statement says, that the complete heat Q1 cannot be, converted into work. Thus the, efficiency of a heat engine cannot be, 100%., Clausius Statement: No process is, possible whose sole result is the, transfer of heat from a cold, reservoir to a hot reservoir., , SAJU K JOHN, M.Sc. Physics, NET, Doctoral Research Fellow at NIT Calicut, , 6

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The efficiency () of the carnot engine, Q,   1 2, Q1, V3, ), V4,  1,  (1), V2, nRT1 log ( ), V1, nRT2 log (, , Since step 2  3 is an adiabatic process,, , b) What are the sink, source and, working substance of a domestic, refrigerator?, Ans:, , T2 V2 1, , T1 V3 1, ,  T1V2 1  T2 V3 1 , T V ,  2  2 , T1  V3 , , Problem8: a) Which law of, thermodynamics is used to explain the, working of a heat engine?, , 1, , 1, , V, T,  2  ( 2 )  1  (2), V3, T1, Similarly step 4  1 is an adiabatic process,,  T2 V4 1  T1V11 , , T2 V11, , T1 V4  1, , T V ,  2  1 , T1  V4 , , 1, , 1, , V  T  1,  1   2   (3), V4  T1 , From (2) and (3), V1 V2, V, V, , or 3  2  (4), V4 V3, V4 V1, Substituting (4) in (1),   1, , T2, T1, , Problem9: (a) What is the working, substance of an ideal heat engine?, (b) Calculate the maximum, efficiency of a heat engine, working between steam point, and ice point. Can you design, an engine of 100% efficiency?, Ans:, , Carnot’s Theorem: i) Working between two given, temperatures T1 and T2 of the hot, and cold reservoirs respectively,, no engine can have efficiency, more than that of the Carnot, engine., ii) The efficiency of the Carnot, engine is independent of the nature, of the working substance., , SAJU K JOHN, M.Sc. Physics, NET, Doctoral Research Fellow at NIT Calicut, , 8

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Problem10: An ideal heat engine, utilizes a perfect gas. The source is at, 450 K and sink is at 320 K. If the, engine takes 3600J per cycle from the, source, calculate the efficiency of the, engine., Ans:, , Problem 11: a) Which, thermodynamic process is called an, iso-entropic process?, b) The efficiency of a Carnot, engine is 1/6. If on reducing the, temperature of the sink by 650C,, its efficiency becomes 1/3, find the, temperature of the sink and source., , SAJU K JOHN, M.Sc. Physics, NET, Doctoral Research Fellow at NIT Calicut, , 9

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Problem12: A Carnot engine working, between 5270C and 1270C has a work, output of 800J per cycle. How much, heat is supplied to the engine from the, source per cycle?, Ans:, , SAJU K JOHN, M.Sc. Physics, NET, Doctoral Research Fellow at NIT Calicut, , 10