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HEAT & THERMODYNAMICS, The energy that is transferred from one system to another, by force moving its point of application in its own direction, is called work., , 2. THERMODYNAMICS, It is the study of interelations between heat and other forms, of energy, , dx, , Thermodynamic System : A collection of large number of, molecules of matter (solid, liquid or gas) which are so, arranged that these posses certain values of pressure,, volume and temperature forms a thermodynamic system., •, , System, Fs = PsA, , The parameters pressure, volume, temperature, internal, energy etc which determine the state or condition of system, are called thermodynamic state variables., Work done by the system, , In thermodynamics we deal with the thermodynamic systems, as a whole and study the interaction of heat & energy during, the change of one thermodynamic state to another., ,   Ps Adx, , 2.1 Thermal Equilibrium, ,   Ps dV, , The term ‘equilibrium’ in thermodynamics implies the state, when all the macroscopic variables characterising the, system (P, V, T, mass etc) do not change with time., •, , Two systems when in contact with each other come to, thermal equilibrium when their temperatures become same., , •, , Based on this is zeroth law of thermodynamics. According, to zeroth law, when the thermodynamics systems A and B, are separately in thermal equilibrium with a third, thermodynamic system C, then the systems A and B are in, thermal equilibrium with each other also., , Where Ps Pressure of system on the piston. This work done, by system is positive if the system expands & it is negative, if the system contracts., •, , Work and Heat are path functions whereas internal energy, is a state function., , •, , Heat & work are two different terms through they might, look same., 2.3 Important Thermodynamics Terms, State Variables : P, V, T, moles, They can be extensive or intestive., , 2.2 Heat, Work and Internal Energy, , Equation of State : The equation which connects the, pressure (P), the volume (V) and absolute temperature (T), of a gas is called the equation of state., , Internal Energy is the energy possessed by any system, due to its molecular K.E. and molecular P.E. Here K.E & P.E, are with respect to centre of mass frame. This internal, energy depends entirely on state and hence it is a state, variable. For 1 real gases internal energy is only by virtue, of its molecular motion., Units, , nf RT, for ideal gases where, 3, , n, , =, , number of moles, , f, , =, , Degree of fredom, , R, , =, , Universal Gas Constant, , T, , =, , Temperature in Kelvin, , , , PV = constant, , (Boyle’s Law), , V, = constant, T, , (Charle’s Law), , PV = NRT, Thermodynamic Process : A thermodynamic process is, said to take place when some changes occur in the state of, a thermodynamic system, i.e., the thermodynamic, parameters of the system change with some important time., Types of these thermodynamic process are Isothermal,, Adiabatic, Isobaric and Isocboric, , Internal Energy can be change either by giving heat energy, or by performing some work., Heat Energy is the energy transformed to or from the system, because of the difference in temperatures by conduction,, convection or radiation., ,   F dx, , Quasi Static Process : A thermodynamic process which is, infinitely slow is called as quasi-static process., •, , In quasi static process, system undergoes change so slowly,, that at every instant, system is in equilibrium, both thermal, and mechanical, with the surroundings.

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HEAT & THERMODYNAMICS, •, , Quasi-static process is an idealised processed. We generally, assume all the processes to be quasistatic unless stated., , 2.5.1 Isothermal Process, Description : A thermodynamic process in which, temperature remains constant, , Indicator P-V, Diagram : A graph between pressure and, volume of a gas under thermodynamic operation is called, P-V. diagram., , Condition : The walls of the container must be perfectly, conducting to allow free exchange of heat between gas, and its surroundings., , P, , The process of compression or expansion should be slow, so as the provide time for exchange of heat., , a, b, d, , •, , a, , Isobaric, , b, , Isothermal, , c, , Adiabatic, , d, , Isochoric, , c, , These both conditions are perfectly ideal., Equation of State : T = Constant or Pv = Constant, , v, , Indicator Diagram :, P, , P, , Area under P – V diagram gives us work done by a gas., v, , st, , 2.4 1 Law of Thermodynamics, , T, , Let Q = Heat supplied to the system by the surroundings, W = Work done by the system on the surroundings, , Slope of P  V, , U = Change in internal energy of the system., , U = 0, , First law of thermodynamics states that energy can neither, be created nor be destroyed. It can be only transformed, from the form to another., , W=, , Sign Conventions :, , v2, , When heat is supplied to the system, then Q is positive, and when heat is withdrawn from the system, Q is, negative., , •, , When a gas expands, work done by the gas is positive &, when a gas contracts then w is negative, , •, , U is positive, when temperature rises and U is negative,, when temperature falls., Remember here we always take work done by the system., In chemistry, work done on the system is considered. Hence, st, there is some different look of 1 law of Thermodynamics in, chemistry., Q + W = U, where Q, U have same meanings but W stands for work, done on the system, 2.5 Application of the First of Law of Thermodynamics, st, , Here we see how 1 Law of Thermodynamics is applied to, various thermodynamic processes., , (Temperature remains constant), v2, , Mathematically : Q = U + Q, •, , P, at any point., V, , =, , , , v2, ,  P dV, g, , v2, , nRT dV, V V, , [Using PV = nRT), , v2, = nRT l n v, 1, First Law of Thermodynamics, Q = U + W, , , v2, Q = nRT l n v, 1, Remarks : All the heat supplied is used entirely to do work, against external sorroundings. It heat is supplied then the, gas expands & if heat is withdrawn then the gas contracts., Practical Examples :, Melting of ice at 0°C, Boiling of H2O at 100°C

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HEAT & THERMODYNAMICS, 2.5.2 Adiabatic Process, Description : When there is not heat exchange with, surroundings, Conditions : The walls of the container must be perfectly, non-conducting in order to prevent any exchange of heat, between the gas and its surroundings., The process of compression or expansion should be rapid, so there is no time for the exchange of heat., These conditions are again ideal condition and are hard to, obtain, , 1  P2 V2 r P1 V1r , , , , 1  r  V2r 1 V1r 1 , , , , W =, , r, , or, , r –1, , PT, , r, 1 r, , Q = UT + W, Substituting the values, ,  cons tan t, , We get Q = 0, Q = 0 is as expected, Remarks : It gas expands adiabatically then its temp, decreases & vice vers a, , P, , Practical Example, , isotherm, , •, , Propagation of sound waves in the form of compression &, rarefaction, , •, , Sudden bursting of a cycle tube., , v, , Slope of adiabatic curve , , 2.5.3 Isochoric Process, ,  rp, V, , Description : Volume remains constant, , As shown in graph adiabatic curve is steeper than, isothermal curve., U , , Condition : A gas being heated  or cooled inside a rigid, container., , nfRDT nR  T2  T1  P2 V2  P1 V1, , , 2, r 1, r 1, , Work Done by Gas : If a gas adiabatically expands from V1, to V2, V2, , W, , =, , =, , P2 V2  P1 V1 nR  T1  T2 , , 1 r, r 1, , First Law of Thermodynamics, , = constat, , Indicator Diagram, , •, , Equation of State : V = constant or, P, , P, , P, = constant, T, P, , dV, , V, , r, , V1, , cons tan t, , V2, , , , V1, , v, , dV, Vr, , U, ,  PV r  cons tan t , , ,   p  cons tan t , r, , , V, , =, , =, , V2, , =, , 1, , P1 V1r  P2 V2r  cons tan t, , Pv = constant, TV, ,  1, 1 ,  r 1  r 1 , V ,  V2, , Also we know, , Equation of State :, or, , cons tan t, 1 r, , =, ,  V  r 1 , cons tan t  , ,  1 r V, , 1, , U, , =, , T, , n f R T, 2, n R  T2  T1 , r 1, , P2 V2  P1 V1 nR T, , r 1, r 1, , T

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HEAT & THERMODYNAMICS, Work, , First Law of Thermodynamics, , W = O as gas does not expands, , Q, , = U + W, , , , Q, , =, , , , Q, ,  fR, ,  R  T, = , 2, , , , First Law of thermodynamics, Q = U + W, , , Q=, , n f R T, 2, , ...(7), , Remarks : Since we have studied earlier, that when heat is, supplied to any process. its temp increases according to, relation, Q = nCT, , , Q , Here it is refered as nT , v, , ...(8), , ...(11), , We getCv =, , Replacing, , fR, 2, , Description : When pressure remains constant, , Similar to molar specific heat at constant pressure and molar, specific heat at constant volume, we can define molar, specific heat for any process., , Mainly we have CP & CV, Specific Heat at Constant Volume : It is defined as the, amount of heat required to raise the temperature of 1g of a, gas through 1C°, when its volume is kept constant. It is, denoted as CV., , P, , Specific Heat at Constant Pressure : It is defined as the, amount of heat required to raise the temperature of 1g of a, gas through 1C° keeping its pressure constant. It is denoted, as CP., , T, , same as always, ,  PdV  PV, (as pressure is constant), PV2 – PV = nRT, , ...(13), , Basically gas does not possess a unique specific heat., , Indicator Diagram :, , =, , CV = R, , C isothermal = , , V, = constant, T, , =, , =, , C adiabatic = 0, , Equation of State : P = constant, , nf RT, 2, , ...(12), , For example :, , Condition : When in one container, the piston is free to, move and is not connected by any agent., , v, , fR, R, 2, , which is also called as Mayer’s Relation., •, , 2.5.4 Isobaric Process, , =, , fR, by CV we get, 2, , CP, , ...(9), , P, , CP, , From equation 9 & 12, , Comparing equation 7 and 8, , W, , Q , = nT , P, , We get, , i.e. Molar heat capacity at constant volume, , =, , CP, , From equation 10 & 11, , Now this C depends upon external conditions for gases., , U, , ...(10), , Remarks : Similar to CV, we can define molar heat capacity, at constant pressure, , , Q, C=, nT, , nfRT,  nRT, 2, , •, , Please Note CP, CV means Molar heat Capacity & CP, CV, means specific heat capacity, , •, , CV = MCV & CP = MCP where M stands for molar mass of any, sample., , •, , CP  CV , , R, M

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HEAT & THERMODYNAMICS, 2.7 Heat Engines, , 2.5.5 Melting Process, , It is a device that converts heat energy into mechanical, energy., , In any case first law is always applicable, Q = mLf, , as learned earlier., , Key Elements :, , W=0, (In the change of state from solid to liquid we ignore any, expansion or contraction as it is very small), According to first law of thermodynamics, U = Q – W, U = mLf, Remark : The heat given during melting is used in, increasing the internal energy of any substance, 2.5.6 Boiling Process, , •, , A source of heat at higher temperature, , •, , A working substance, , •, , A sink of heat at lower temperature., , Working :, •, , The working substance goes through a cycle consisting of, several processes., , •, , In some processes it absorbs a total amount of heat Q1 from, the source at temperature T 1., , •, , In some processes it rejects a total amount of heat Q2 to the, sink at some lower temperature T2., , •, , The work done by the system is a cycle is transferred to the, environment via some arrangement., , Here, Q = mLV, W = P[V2 – V1], , Schematic Diagram, , (Pressure is constant during boiling and it is equal to, atmospheric pressure), , , W, , U = Q – W, U = mLf – P(V2 – V1), Source or, Hot Reservoir, , 2.5.7 Cyclic Process, A cyclic process is one is which the system returns to its, initial stage after undergoing a serves of change, , Q1, , Working, Substamnce, , Q2, , Sink or, Cold Reservoir, , T1, , T2, , First Law of Thermodynamics, , Example Indicatgor Diagram, P, A, , , , Energy is always conserved, , , , Q1 = W + Q 2, , Thermal Efficiency of a heat engine is defined of the ratio, of net work done per cycle by the engine to the total amount, of heat absorbed per cycle by the working substance from, the source. It is denoted by ., , B, V, , U = mLV O, , W, = Q, 1, , W = Area under the loop., Q = W as per First Law of thermodynamics, , Q2,  = 1 Q, , 2.6 Limitations of the First Law of Thermodynamics, The first law does not indicate the direction in which the, change can occur., , •, , The first law gives no idea about the extent of change, , •, , The first law of thermodynamics gives no information about, the source gives no information about the source of heat., i.e. whether it is a hot or a cold body., , ...(15), , Using equation 14 and 15 we get, , Here W is positive if the cycle is clockwise & it is negative, if the cyclic is anti clockwise., , •, , ...(14), , 1, , ...(16), , Ideally engines shuld have efficiency = 1, Remarks : The mechanism of conversion of heat into work, varies for different heat engines., •, , The system heated by an external furnace, as in a steam, engine. Such engines are called as external combustion, engine.

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HEAT & THERMODYNAMICS, •, , Though all the statements are the same in their contents,, the following two are significant., , The system is which heat is produced by burning the fuel, inside the main body of the engine. Such an engine is called, as Internal Combustion Engine., , Kelvin Pnek Statement : No process is possible whose, sole result is the absorption of heat from a reservoir and the, complete conversion of the heat into work., , 2.8 Refrigerator and Heat Pumps, A refrigertor or heat pump is a device used for cooling, things., , Calcius Statement : No process is possible whose sole, result is the transfer of heat from a colder object to a hotter, object., , Key Elements :, •, , A cold reservoir at temperature T 2., , •, , A working substance., , •, , A hot reservoir at temperature T1, , Significance : 100% officiency in heat engines or infinite, CoP in refrigerators is not possible., 2.10 Reversible and Irreversible Process, , Working, •, , The working substance goes through a cycle consisting of, several process., , •, , A sudden expansion of the gas from high to low pressure, which cool it and converts it into a vapour-liquid mixture., , •, , Absorption by the cold fluid of heat from the region to be, cooled converting it into vapour., , •, , Heating up of the vapour due to external work done on the, working substance., , •, , Release of heat by the vapour to the sorroundings bringing, it to the initial state and completing the cycle., , Reversible Process : A thermodynamic process taking a, system from initial state i to final state f is reversible, if the, process can be turned back such that both, the system and, the surroundings return to their original states, with no, other change anywhere else in the universe., Conditions for reversibility :, 1., , Process should proceed at an extremely slow rate, i.e.,, process is quasistatic so that system is in equilibrium with, surroundings at every stage., , 2., , The system should be free from dissipative forces like, friction, inelasticity; viscosity etc., Examples : No process exactly reversible, though a slow, expansion of an ideal gas is approximately reversible., , Sychematic Diagram., W, , Hot Reservoir, T1, , Q1, , Q2, , Cold Reservoir, T2, , First Law of Thermodynamics, Q2 + W = Q 1, , ...(17), , Coefficient of Performance of refrigerator () is defined as, the ratio of quantity of heat removed per cycle from contents, of the refrigerator (Q2) to the energy spent per cycle (W) to, remove this heat, , , Q2, W, , ...(18), , Using equation 17 and 18 we get, , , , Irreversible Process : A process which does not satisfy, any of the conditions for reversible is called an irreversible, process., , Q2, Q1  Q 2, , Ideally heat pumps should have  = , 2.9 Second Law of Thermodynamics, There are number of ways in which this law can be stated., , Causes :, •, , Spontaneous process, , •, , Presence of friction, viscosity and such dissi-ptive forces, Significance of Reversibility :, , •, , Main concern of thermodynamics is the efficiency with, which the heat is converted into Mechanical Energy., , •, , Second Law of Thermodynamics rules out the possibility, of a perfect heat engine with 100% efficiency., , •, , It turns out that heat engine based on idealised reversible, processes achieves the highest possible efficiency., 2.11 Carnot Engine, Sadi Carnot devised on ideal cycle of operation for a heat, engine called as carnot cycle., Engine used for realising this ideal cycle is called as carnot, heat engine., Constructions : The essential parts of an ideal heat engine, or Carnot heat engine are shown in figure.

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HEAT & THERMODYNAMICS, , Cylinder, , Ideal, Gas, Source at, Temp. T1, , (i), , (ii), , Consider one gram mole of an ideal gas enclosed in the, cylinder. Let V1, P1, T1 be the initial volume, pressure and, temperature of the gas. The initial state of the gas is, represented by the point Aon P–V. diagram, We shall assume, that all the four processes are quasi-static and non, dissipative, the two conditions for their reversibility., , Insulating, Pad, , Sink at, Temp. T2, , Steps, 1., , The cylinder is placed on the source and gas is allowed to, expand by slow outward motion of piston. Since base is, perfectly conducting therefore the process is isothermal., , Source of heat : The source is maintained at a fixed higher, temperature T1, from which the working substance draws, heat. The source is supposed to possess infinite thermal, capacity and as such any another of heat can be drawn, from it without changing its temperature., , Now, U1 = O, , Sink of heat : The sink is maintained at a fixed lower, temperature T2, to which any amount of heat can be rejected, by the working substance. It has also infinite thermal, capacity and as such its temperature remains constant at, T2, even when any amount of heat is rejected to it., , (iii) Working substance : A perfect gas acts as the working, substance. It is contained in a cylinder with non-conducting, sides but having a perfectly conducting-base. This cylinder, is fitted with perfectly non-conducting and frictionless, piston., , Isothermal Expansion :, , V2, q1 = W1 = RT1l n . V = Area ABmKA, 1, , 2., , q1, , , , Heat absorbed by gas, , w1, , , , work done by gas, , Adiabatic Expansion :, The cylinder is now removed from source and is placed on, the perfectly insulating pad. The gas is allowed to expand, further from B(P2, V2) to C (P3, V3). Since the gas is thermally, insulated from all sides, therefore the processes is adiabatic, , Apart from these essential parts, there is a perfectly, insulating stand or pad on which the cylinder can be placed., It would isolate the working substance completely from the, surroundings. Hence, the gas can undergo adiabatic, changes., , q2 = 0, U 2 , , The Carnot cycle consists of the following four stages :, 1., , Isothermal exdpansion, , 2., , Adiabatic expansion, , 3., , Isothermal compression, , 4., , Adiabatic compression., , W2 , , 3., , The cycle is carried out with the help of the Carnot engine, as detailed below :, , Pressure (P), , r 1, , R  T3  T1 , r 1, , = Area BCNMB, , Isothermal Compression :, The cylinder is now removed from the insulating pad and is, placed on the sink at a temperature T 2. The piston is moved, slowly so that the gas is compressed until is pressure is P4, and volume is V4., U3 = 0, , P, , W2  RT2 l n, , A(V1, P1), Q1, D(V4, P4), , B(V 2, P 2), T1, , Q2, O, , R  T2  T1 , , K, , L, , M N, Volume (V), , C(V 3, P 3), T2, X, , q 3  RT2 l n, , V4, V3 = – Area CDLNC, , V4, V3, , q3 = Heat absorbed in this process, w3 = Work done by Gas

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HEAT & THERMODYNAMICS, 4., , Adiabatic Compression :, The cylinder is again placed on the insulating pad, such, that the process remains adiabatic. Here the gas is further, compressed to its initial P1 and V1., U 4 , , W4 , , Q1 T1, , Q 2 T2, , , R  T1  T2 , , r 1, , = – area DAKLD, , q4 = 0, w4 = work done by the gas, , •, , Carnot engine - depends only upon source temperature nd sink, temperature., , •, , Carnot engine =1 only when T 2 = 0 K or T 1 =  which is, impossible to attain., , •, , If T2 = T1   =  Heat cannot be converted to mechanical, energy unless there is same difference between the, temperature of source and sink., , Analysis :, Total work done by the engine per cycle., =, , W1 + W2 + W3 + W4, , =, , W1 + W3, , W, , =, , RT1l n, , Q1, , =, , Total heat absorbed = q1, , =, , RT1l n, , =, , Total heat released = –q3, , Q2, , 2.12 Carnot Theorem, Statement :, , V2, V1, , ...(19), , Working between two given temperatures, T 1 of hot, reservoir (the source) and T2 of cold reservoir (the sink, no, engine can have efficiency more than that of the Carnot, engine., , (b), , The efficiency of the Carnot engine is independent of the, nature of the working substance., , RT2 l n, , V3, V4, , Engine used for realising this ideal cycle is called as carnot, heat engine., Proof :, , ...(20), , Step - 1 : Imagine a reversible engine R and an irreversible, engine-I working between the same source (hot reservoir T1), and sink (cold reservoir T2)., Step - 2 : Couple two engines such that I acts like heat, engine and R acts like refrigerator., , We can see that for heat engine, , Step - 3 : Let engine I absorb Q1 heat from the source deliver, 1, 1, work W and release the balance Q1 – W to the sink in one, cycle., , W = Q 1 – Q2, =, , (a), , V2, V,  RT2 l n 4, V1, V3, , [q3 = Heat absorbed & not heat released], =, , T2, T1, , Division, , r 1, , R  T2  T1 , , Carnot  1 , , Area under ABCDA, , Efficiency of Carnot Engine, , W1, , Q, W, ,  1 2, Q1, Q1, , Q1, , Now steps 2 is adiabatic 2 step 4 is also adiabatic, , , T1 V2r 1  T2 V3r 1, , and, , T1 V1r 1  T2 V4r 1, , , , V2 V3, , V1 V4, From equation 19, 20 and 21 we get, , T1, , ...(21), , Q1, , 1, R, , Q1–W1, Q 1–W, , T2, , Step - 4 : Arrange R, such that it returns same heat Q, to, the source, taking Q 2 from the sink and requiring work, W = Q1 – Q2 to be done on it., Step - 5 : Supppose R < I (i.e.) If R were to act as an, engine it would give less work output than that of I (i.e.), 1, 1, W < W for a given Q1 and Q1 – W > Q1 – W

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HEAT & THERMODYNAMICS, Step - 6 : In totality, the I-R system extracts heat (r1 – W) –, 1, 1, (Q1 – W ) = W – W & delivers same amount of work in one, cycle, without any change in source or anywhere else. This, is against second Law of Thermodynamics. (Kelvin - Planck, statement of second law of thermodynamics), , •, , 3.2.2 Real Gas, All gases are referred to as real Gases., All real gas near the ideal gas behaviour at low pressures, and temperatures high enough, where they cannot be, liquified., , Hence the assertion q1 > qR is wrong., •, , Similar argument can be put up for the second statement of, carnot theorem, (ie) Carnot efficiency is independent of, working substance., , , , 3.2.3 Ideal Gas Laws, , We use ideal gas to else are calculating but the relation., , Avogadro Hypothesis : Equal volumes of all gas under, identical conditions of pressure and temperature would, contain equal number of molecule., , Q1 T1, , Q 2 T2 will always hold true for any working substance, , Perfect Gas Equation :, , used in a carnot engine., , 3. KINETIC THEORY OF GASES, , PV = nRT, and, , In this topic, we discuss the behaviour of gases and how, are the various state variable like P, V, T, moles, U are interrelated with each other, , SRT, M, n = Number of moles., r = Universal Gas constant = NxkB, , where, , ND = Avagadro No., kB = Boltzman constant, , Same as Atomic Theory given by Delton. According to, him, atoms are the smallest constituents of elements. All, atoms of one element are identical, but atoms of different, element are different., , R = 8031 J/mol K., R = 1.98 Cal/mol K., Boyle’s Law : When temperature of a given mass of gas is, kept constant, its pressure varies inversely as the volume, of gas., , In solids : Atoms tightly packed, interatomic spacing about, Å . Interatomic force of attraction are strong., , (i.e.), , In liquids : Atoms are not as rigidly fixed as in solids., Interatomic spacing is about the same 2Å. Interatomic for a, attraction are relative weaker., , PV = constant, , Charles Law : When pressure of a given mass of kept, constant, volume of the gas varies directly as the, temperature of the gas., , In Gases : Atoms very free. Inter atomic spacing is about, tens of Angstroms. Interatomic forces are much weaker in, gases than both in solids and liquids., , (i.e.), , V T, , Dalton’s Law of Partial Pressures : The total pressure of a, mixture of ideal gases is the sum of partial pressures exerted, by the individual gases in the mixture., , In this chapter, we mainly focus on gases., 3.2 Molecular Nature of Matter, , That gas which strictly obeys the gas laws, (such as Boyle’s, Law, Charles, Gay Lussac’s Law etc.), , P, , where, , 3.1 Molecular Nature of Matter, , 3.2.1 Ideal Gas, , There is no force of attraction or repulsion amongst the, molecules of an ideal gas., , P–V = (n1 + n2 + n3 + ...)RT, , , P = (n1 + n2 + n3 + ...), P = P1 + P2 + ......, , Characteristics, •, , The size of the molecule of an ideal gas is zero., , RT, V, , where P1 =, , n1 RT, Pressure of gas, V

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HEAT & THERMODYNAMICS, Neviation of Real Gas from Ideal Gas :, , volume occupied by the molecules is negligible in, comparison to the volume of the gas., , Ideal Gas, , 1, , T1, T2, , 4., , The molecules do not exert any fore of attraction or, repulsion on each other, except during collision., , 5., , The collisions of the molecules with themselves and with, the walls of the vessel are perfectly elastic. As such, that, momentum and the kinetic energy of the molecules are, conserved during collisions, though their individual, velocities change., , 6., , There is not concentration of the molecules at any point, inside the container i.e. molecular density is uniform, throughout the gas., , 7., , A molecule moves along a straight line between two, successive collisions and the average straight distance, covered between two successive collisions is called the, , T1 > T2 > T3 >, , T3, 200, , 0, , 400, 600, P (atm), , 800, , 1.6, 1.4, P 1.2, , mean free path of the molecules., , T1, , 1.0, , T1 > T2 > T3, , 0.8, , 8., , T1, , 0.6, 0.4, , T2, , T3, , T2, , 3.4 Pressure of an Ideal Gas and its Expression, , 0.2, 0, 20, , T3, 60, , 100, , The collisions are almost instantaneous, i.e., the time of, collision of two molecules is negligible as compared to time, interval between two successive collisions., , 140, , 160, V, , 220, , Pressure exerted by the gas is due to continuous, bombardment of gas molecules against the walls of the, container., Expression :, , 1.2, , P1 > P2 > P3, , 1.0, T 0.8, , P1, , 0.6, P2, , 0.4, , P3, , 0.2, 0, , 0, , 100, , 200, , 300, , 400, V, , 500, , 3.3 Kinetic Theory Postulates, 1., , A gas consists of a very large number of molecules (of the, 23, other of Avogadro’s number. 10 ), which are perfect elastic, spheres. For a given gas they are identical in all respects,, but for different gases, they are different., , 2., , The molecules of a gas are in a state of incessant random, motion. They move in all directions with different speeds.,, ( of the order of 500 m/s) and obey Newton’s laws of motion., , 3., , The size of the gas molecules is very small as compred to, the distance between them. If typical size of molecule is 2, Å, average distance between the molecules is  20 Å. Hence, , Consider a gas enclosed in a cube of side 1. Take the axes, to be parallel to the sides of the cube, as shown in figure. A, molecule with velocity (x, y, z) hits the planar wall parallel, 2, to yz-plane of area A (= l ). Since the collision is elastic, the, molecule rebounds with the same velocity; its y and z, components of velocity do not change in the collision but, the x-component reverses sig. That is, the velocity after, collision is (–x,vz, vy). The change in momentum of the, molecule is : –mx – (mx) – 2mx. By the principle of, conservation of momentum, the momentum imparted to the, wall in the collision = 2mx., To calculate the force (and pressure) on the wall, we need, to calculate momentum imparted to the wall per unit time, if, it is within the distance x t from the wall. that is, all, molecules within the volume, Ax t only can hit the wall in, time T is 1/2A x t n where n is the number of molecules, per unit volume. The total momentum transferred to the, wall by these molecules in time t is :, Q = (2mx) (1/2 n Avxt), The force on the wall is the rate of momentum transfer Q/t, and pressure is force per unit area :, 2, , P = Q/(A t) = nm x

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HEAT & THERMODYNAMICS, Actually, all molecules in a gas do not have the same, velocity; there is a distribution in velocities. The above, equation therefore, stands for pressure due to the group of, molecules with speed x in the x-direction and n stands for, the number density of that group of molecules. The total, pressure is obtained by summing over the contribution due, to all groups:, , P  nm, , 2, x, , where  2x is the average of  2x . Now the gas is isotropic,, , , , 1, nRT  M  2, 3, nRT , , , , n 3RT 1,  m 2, N 2, 2, , Also N = nNA, , , 3 R, 1,  m 2, 2 NA 2, Average KE of translation per molecule of the gas, , where  is the speed and  2 denotes the mean of the squared, speed. Thus, , 1, P    nm 2, 3, 2, 1, 1M 1 2, mn  ,  S, 2, 2V 2, , 1, Nm  2, 3, , , , 1, 1,     2x   2y   2z      2, , , 3,  ,  3, , P, , ...(22), , M = Total mass of gas moleculus, V = Total volume fo gas molecules, 3.4.1 Relation between Pressure and KE of Gas Molecules, , From above equations, we can easily see that KE of one, molecule is only dependent upon its Temperature., , , KE of molecule will cease if, the temperature of the gas, molecules become absolute zero., , , , Absolute zero of a temperature may be defined as that, temperature at which the root mean square velocity of the, gas molecule reduces to zero., All the ideal gas laws can be derived from Kinetic Theory, of gases., , 3.6 Derivation of Gas Laws from Kinetic Theory, 3.6.1 Boyle’s Law, , 1, P  S 2, 3, , We know that PV , , 2, E, 3, , , , P, , , , Pressure exerted by an ideal gas is numerically equal to two, third of mean kinetic, , From equation 22, We know, P, , 1M 2, , 3 , , 3, k BT, 2, , 3.5 Kinetic Interpretation of Temperature, , From equation 22, , 3.4.2 Average KE per molecule of the Gas, , ...(23), , with equation 23 and Ideal gas equation, , i.e. there is no preferred direction of velocity of the molecules, in the vessel. Therefore by symmetry,,  2x   2y   z2, , 1, PV  M  2, 3, , 2, NK, 3, , where K is the average kinetic energy of translation per, gas molecule. At constant temperature. K is constant and, for a given mass of the gas. N is constant., Thus, PV = constant for given mass of gas at constant, temperature, which is the Boyle’s Law., 3.6.2 Charle’s Law, We know that PV , , 2, NK, 3, , For a given mass of gas, N is constant.

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HEAT & THERMODYNAMICS, Since K , , r rms, , 3, k B T, K  T and as such PV  T., 2, , But as  rms , , If P is constant, V  T, which is the Charles’ Law., 3.6.3 Constant Volume Law, We know that PV , , 2, NK, 3, , r1, 2, , r2, 1, , 3, k B T, K  T, 2, , The rates of diffusion of two gases are thus inversely, proprotional to the square roots of their densities which is, Graham’s Law of diffusion., , Thus, PV  T, If V is constant, P  T, which the constant volume law., , 3.6.7 Dalton’s Law of Partial Pressures, , 3.6.4 Ideal Gas Equation, , The kinetic theory of gases attributes the gas pressure to, the bombardment of the walls of the containing vessel by, molecules. In a mixture of ideal gases, we might therefore, expect the total pressure (P) to be the sum of the partial, pressures (p1, p2, ...) due to each gas, i.e.,, , 3, 2, As PV  NK and K  k B T, 2, 3, , PV , , 2 3, , N  k B T  or PV = Nk T, B, 3 2, , , which is the ideal gas equation., , P  p1  p 2  ... , , 3.6.5 Avogadro’s Law, Consider two gases 1 and 2. We can write, P1 V1 , , 2, 2, N1 K1 , P2 V2  N 2 K 2, 3, 3, , If their pressures, volumes and temperatures are the same,, then, P1 = P2, V1 = V2, K1  K 2 ., , or, , P, , 2 N1, 2 N2, K1 , K 2  ..., 3 V1, 3 V2, , , N, 2  N1, K1  2 K 2  ... , , 3  V1, V2, , , In equilibrium, the average kinetic energy of the molecules, of different gases will be equal, i.e.,, , K1  K 2  ...K , , 3, k BT, 2, , Thus,, , Clearly, N1 = N2 Thus :, Equal volumes of all ideal gases existing under the same, conditions of temperature and pressure contain equal, number of molecules which is Avogadro’s Law or, hypothesis., This law is named after the Italian physicist and chemist,, Amedeo Avogadro (1776–1856)., Alliter : As PV  Nk B T, N , , 3P, 1, or r , , , , Therefore, if r1 and r2 are the rates of diffusion of two gases, of densities 1 and 2 respectively,, , For a given mass of gas, N is constant. Since, , K, , 3P, ,r, , , PV, k BT, , P, , 2, 3,  n1  n 2  ...  k B T    n1  n 2  ... k B T, 3, 2, , , ...(24), , N1, N2, where n1  V , n 2  V ..., 1, 2, Equation (24) represents Dalton’s Law of partial pressures, which states that :, , 3.6.6 Graham’s Law of Diffusion, , The resultant pressure exerted by a mixture of gases or, vapours which do not interact in any way is equal to the, sum of their individual (i.e., partial) pressures., , The rate (r) of diffusion of a gas through a porous pot or, into another gas is determined by the rms speed of its, molecules, i.e.,, , Figures shows a model explaning kinetic theory of gases. It, has been constructed in accordance with theory on one, hand and real experimental observations on the other hand., , If P, V and T are constants, N is also constant.

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HEAT & THERMODYNAMICS, , I, , 1, mv2  K, 2, 3,  kBT, 2, , PV = nRT, , Based on, experimental, observations, of gases, , II, , 1  Nm  2, P , v, 3 V , , Based on, Newtonian, mechanocal, treatment of, gas as, collection of, particles, , N = Number of degrees of freedom of the system, III, , Based on combining, two mathematical, models of a gas in, boxes I and II, , 2, PV  NK, 3, , N = 3A– R, 3.8.1 Monoatomic Gases, The molecules of a monoatomic gas (like neon, argon,, helium etc) consists only of one atom., , , R=0, , , N=3, Here 3 degrees of freedom are for translational motion, , 3.8.2 Diatomic Gases, A=2, , = NkBT, , Assuming the distance between the two molecules is fixed, then R = 1, , Based on substituting, and rearranging the expression, for K back into equation in box II, , The above piece of logic is tempting but false. This is due, to the reason that though the equation in box IV is useful,, it does not tell us anything new, since it results from, combining equations in boxes II and III., , A=1, , , , N=3×2–1=5, Here 5 degrees of freedom implies combination of 3, translational energies and 2 rotational energies., , 3.7 Internal Energy, As studied in thermodynamics, Internal Energy of any, substance is the combination of Potential Energies &, Kinetic Energies of all molecules inside a given gas., •, , In real gas, Internal Energy = P.E of molecules + K.E of Molecules, , •, , Y, , In real gas, , Y, , Internal Energy = K.E of Molecules, Here PE of molecules is zero as assumed in Kinetic theory, postulates; There is no interaction between the molecules, hence its interactional energy is zero., 3.8 Degree of Freedom, , X, Z, , •, , •, , A particle moving in straight line, say along X-axis need, only x coordinate to define itself. It has only degree of, freedom., A particle in a plane, needs 2 co-ordinates, hence has 2, degree of freedom., In general if, A = number of particles in the system, R = number of independent relations among the particles, , A2, , A1, , X, , (i), , Z, (a), , A2, (ii), , (b), , Y, A1, , The number of degrees of freedom of a dynamical system is, defined as the total number of coo-ordinates or independent, quantities required to describe completely the position &, configuration of the system., Example :, , A1, , A2, Z, , A3, , X, , (c), , If vibrational motion is also considered then [only at very, high temperatures, N=7, where 3 for translational, 2 for rotational, 2 for vibrational, 3.8.3 Triatomic Gas

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HEAT & THERMODYNAMICS, Linear, R=2, , , dy,  vibrational velocity, dt, , where, , A=3, N=3×3–2=7, , ky 2, = Energy due to configuration, 2, , and, , Non-Linear, , According to Law of Equipartition, Energy per degree of freedom , , , , , , 1, 1, k B T  K B T  K B T is energy for, 2, 2, complete one vibrational mode., , A=3, , 3.10 Specific Heat Capacity, , R=2, , , N=3×3–3=6, , •, , Here again vibrational energy is ignored., 3.8.4 Polyatomic Gas, A polyatomic gas has 3 translational, 3 rotational degrees, of freedom. Apart from them if there v vibrational modes, then there will be additional 2v vibrational degrees of, freedom., , , , 1, k BT, 2, , Total degree of freedom, , With the knowledge of law of equipartition, we can predict, the heat capacity of various gases., 3.10.1 Monoatomic Gas, Degree of freedom = 3., , , Average Energy of a molecule at temperature T, , , , 1, , E  3  k BT , 2, , , n = 3 + 3 + 2V = 6 + 2V, 3.9 Law of Equipartition of Energy, Statement : According to this law, for any dynamical system, in thrmal equilibrium, the total energy is distributed equally, amongst all the degrees of freedom, and the energy, associated with each molecule per degree of freedom is, , Energy for one mole  E × NA, , , U, , 3,  k B NA  T, 2, , , , U, , 3, RT, 2, , In thermodynamics, we studied, , 1, k B T , where kB is Boltzman constant and T is temperature, 2, of the system., k BT, where = Total degree of freedom., 2, This law is very helpful in determining the total internal, energy of any system be it monoatomic, diatomic or any, polyatomic. Once the internal energy is know we can very, easy predict Cv & CP for such systems., Application : U  f, , Remark : In case vibrational motion is also there in any, system, say for diatomic molcule, then there should be, energy due to vibrational as well given by, , CV , , Q , U, , , T  V T, , , , CV , , 3R, 2, , , , CP , , 5R, &, 2, , r, , [ W = O for constant v], , CP 5, , CV 3 ., , 3.10.2 Diatomic Gases, When no vibration, Degree of freedom = 5, , 2, , EV , , 1  dy  1 2, m    ky, 2  dt  2, , Average energy for one mole =, , 5, RT, 2

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HEAT & THERMODYNAMICS, , , CV , , U 5,  R, T 2, , CP , , 7R, 2, , Therefore the atoms do not possess any translational or, rotational degree of freedom., On the other hand, the molecules do possess vibrational, motion along 3 mutually perpendicular directions., Hence for 1 mole of a solid, threre are NA number of atoms., , C, 7, r P , CV 5, , The energy associated with every molecule, ,  1, ,  3  2  k B T   3K B T,  2, , , When vibration is present., There is only one mode of vibration between 2 molecules., , , , U = 3 Rt for one mole, , , , 7, U  RT, 2, , , , C, , •, , The above equation is called as Dulong & Petit’s Law., , , , 7, CV  R, 2, , •, , and, , CP , , At low temperatures the vibrational made may not be that, active hence, heat capacity is low at low temperatures for, solids., , , , Degree of freedom = 7, , and, , r, , 9, R, 2, , Q U, ,  3R, T T, , C, 3R, , 9, 7, , 3.10.3 Polyatomic Gases, Degree of freedom, , 300 K, , = 3 for translational, , 3.10.5 Specific heat capacity of Water, , + 3 for rotational, , Water is treated like solid., , + 2v for vibrational, = 6 + 2v, if v = Number of vibrational modes, , , U   6  2v  K, , , , CV = (3 + V)R, , RT, 2, , CP = (4 + V) R, and, , r, , •, , , , Water has three atoms, 2 of hydrogen and one of oxygen, , , In solids, there is very less difference between heat capacity, at constant pressure or at constant volume. Therefore we, do not differentiate between CP & CV for solids., , Q U, C, , T T, {As solids hardly expand or expansion is negligible}, Now in solid the atoms are arranged in an array structure, and they are not free to move independently like in gases., , Total degree of freedom for every atom, =3×2=6, , , , Total degree of freedom for every molecule of water, = 3 × 2 = 18, , , , 4V, 3 V, , 3.10.4 Specific heat capacity of solids, , T, , 1, , , 18  RT , Q U , 2, , C, , , T T, T, C = 9R, , 3.11 Maxwell Law of Distribution of Molecular, Assumptions of Maxwell Distribution, •, , Molecules of all velocities between 0 to  are present., , •, , Velocity of one molecule, continuously changes, though, fraction of molecules in one range of velocities is constant., Result,  M , N V  4N , ,  2k B T , , 3/ 2, , V2e, , , , mv 2, 2k BT

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HEAT & THERMODYNAMICS, Expression :, , dN V, NV , dV, , Where, , Where, , d, , V, , N, , =, , Mean Free Path, , Total number of molecules, with speeds between V & V + dV, V, , N, , =, , , , t, , Total number of molecules., , NV, , d, d, , V, , Vmp Vav Vrms, , Based on this we define three types of speed f molecules of, gas, ,  , , Vrms  V 2, , Vrms , , 1/ 2, , 1, ,    V 2 dN  V  , N, , , 3RT, M, , Where, , M = Molecular Mass of Gas, , Similarly Vav  V , , , , 1, VdN V, N, , 8RT, M, , But VMP is velocity at which, , , , VMP , , 1/ 2, ,  = 1/(n <v> d2), The average distance between two successive collisions,, called the mean free path 1, is :, l = <v> = 1/(nd2), , dN V, 0, dv, , 3RT, M, , Physically VMP is velocity possessed by Maximum number, of molecules., Remarks :, Vrms > Vav > VMP >, 3.12 Mean Free Path, The path tranversed by a molecule between two successive, collisions with other molecule is called the free path, , l, , Suppose the molecules of a gas are spheres of diameter d., Focus on a single molecule with the average speed <v>. It, will suffer collision with any molecule that comes within a, distance d between the centres. In time t, it sweeps a, volume d2 <v> t wherein any other molecule will collide, with it (as shown in figure). If n is the number of molecules, per unit volume, the molecule suffers nd2 <v> t collisions, in time t. thus the rate of collisions is nd2 <v> or the time, between two successive collisions is on the average., , Total distance travelled by a molecule, No. of collisions it makes with other molecules., , In this derivation, we imagined the other molecules to be at, rest. But actually all molecules are moving and the collision, rate is determined by the average relative velocity of the, molecules. Thus we need to replace <v> by <vr> in equation., A more exact treatment., , l=, , , , 1, 2nd 2, , , , Result, , l=, , , , 1, 2nd 2, , , , Remark : Mean free path depends inversely on the number, density and size of the molecule., 3.13 Brownian Motion, The irregular movement of suspended particles like tiny, dust particles or pollen grains in a liquid is called Brownian, Motion.