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~ 4.2 J

THERMODYNAMICS Dr. J K SARMAH KU.

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  1. THERMODYNAMICSDr. J K SARMAH KU

  2. THERMODYNAMICS means flow of heat or any other form of energy into or out of a system as it undergoes a physical or chemical transformation and deals with the quantitative relationship between them.The subject matter of thermodynamics is based on three laws and all these laws are applicable to all phenomenon of nature.(The branch of science that deals with the study of different forms of energy and the quantitative relationships between them. )

  3. THERMODYNAMIC TERMS:SYSTEM, SURROUNDINGS AND BOUNDARY:A THERMODYNAMIC SYSTEM IS THAT PART OF THE UNIVERSE WHICH IS UNDER THERMODYNAMIC CONSIDERATION.REST OF THE UNIVERSE IS SURROUNDINGS.THE SYSTEM IS SEPERATED FROM THE SURROUNDINGS BY A REAL OR IMAGINARY SURFACE CALLED THE BOUNDARY.

  4. A system is said to be homogeneous if it is uniform throughout in all respects. E.g. A mixture of gases or a liquid or a solid.A system is said to be heterogeneous if it is not uniform throughout and consists of more than one phases separated from each other by sharp boundaries. E.g. A liquid and its vapour or a mixture of two immiscible liquids or two different solids.

  5. Types of Thermodynamic Systems:Depending on the interactions of the system and surroundings,three types of systems are there in thermodynamics:OPEN SYSTEM: A system which can exchange both matter and energy to and from the surroundings is known as an open system. (mass and energy do not remain constant).E.g. Hot water in a beaker placed on a laboratory tableThe water vapour(matter) and also the heat(energy) is transferred to the surrounding through the imaginary boundary.

  6. CLOSED SYSTEM: A system which can not transfer matter but can transfer energy in the form of heat, work and raditaion to and from the surroundings is known as a closed system. (mass remains constant and energy changes).E.g. Hot water in a sealed container/tube. No water vapour can escape from the systemit can transfer heat through the walls of the container/tube to the surroundings.A gas contained in a cylinder fitted with a pistonAs the piston is raised, the gas expands and transfers heat(energy) in the form of work to the surroundings.

  7. Isolated system: A system that prevents any interaction between the system and the surroundings (mass and energy remains constant). An isolated system is one that can transfer neither matter nor energy to and from the surroundings. E.g. Boiling water contained in a thermos flask. As the flask is sealed, matter(water vapour) can not escape from it and also it is insulated, no heat(energy) can be exchanged with the surroundings.

  8. Properties of a system: Measurable propertiesExtensive Property: A property that depends on the quantity of matter present in the system is called extensive property. Some extensive properties are mass, volume, internal energy, enthalpy, number of moles, free energy, entropy, heat capacity etc.e.g. Consider a glass of water. If we double the mass of water, the volume is doubled and so is the no of moles and the internal energy of the system.

  9. Intensive Property:A property which does not depend on the quantity of matter present in the system is called an intensive property. Some intensive properties are pressure, temperature, density, molar entropy, molar volume, molar energy, molar heat capacity, surface tension, viscosity, specific heat, thermal conductivity, refractive index, b.pts, etc.e.g. Concentration of NaCl in a salt solution is 0.1mole/litre, then any drop of solution also has the same NaCl concentration of 0.1mole/litre.

  10. State of a System:The state of a system means the conditions of the system where all its properties are fixed. State variables: State of a system is described in terms of certain observable properties which are called the state variables, for example, temperature (T), pressure (P), and volume (V). (a state with definite values of observable properties)PV=nRT , ideal gas equation.

  11. State function:The fundamental properties which determine the state of a system are P,T,V, mass and composition. A change in magnitude/value of such properties alters the state of the system and hence they are called the State Variables or the State Functions.Change of value during a process depends only upon the initial state and final state of the system and does not depend on the path by which the change has been brought about.

  12. # It is not necessary to state all the state variables to define a system completely? For a one mole of gas PV=RT, if P and T are specified, V is automatically fixed. Necessary variables to define the system: P & Tare called Independent Sate VariablesThe value of V depends on the value of P and T.V is a Dependent State Variable.

  13. Equillibrium and Non Equillibrium StatesA system in which the state variables have constant values throughout the system is said to be in a state of thermodynamic equillibrium. A system in which state variables have different values in different parts of the system is said to be in a non equillibrium state.

  14. Criterion of Equillibrium:1. the temperature of the system must be uniform and must be the same as the temperature of the surrounding (thermal eqbm)2. The mechanical properties must be uniform throughout the system (mechanical eqbm) i.e. no mechanical work is done in one part of the system 3. The chemical composition of the system must be uniform with no net chemical change (chemical eqbm)

  15. NATURE OF WORK AND HEATWhenever a system changes from one place to the other , it is accompanied by change of energy. The change of energy may appear as heat, work, light etc.The unit of energy is Joule. J =NmW ∞ JH, where, J is known as mechanical equivalent of heat.1 calorie = 4.184 J ~ 4.2 J

  16. FIRST LAW OF THERMODYNAMICSEnergy can neither be created nor be destroyed, although it can be transformed from one from to another. Also known as Conservation of Energy

  17. Basis:1. It is impossible to construct a perpetual motion machine i.e. a machine which can produce energy without expenditure of energy.Equivalence of Heat and Work(established by Joule experimentally):Suppose there are no such equivalence. Then it may be possible:i. Convert x joule to certain mechanical work and then ii. In reverse process transform the same mechanical work into heat producing y joules of heat, y>x. Then the original state has been restored but

  18. Heat energy eqvt to y – x joules, will have been created.By repeating the above cycle energy can be created continueously and in this way a perpetual machine may be fabricated. But this denied by human experience, hence there must be equivalence of heat and mechanical work.

  19. INTERNAL ENERGY(U) AND FIRST LAW OF TD:Suppose a system is subjected to change of P and V Δ U = UB – UAΔ U/ = UA – UBIf Δ U > Δ U/, continueously energy may be created. This is contrary to first law: hence, Δ U = Δ U/Thus the energy change accompanying a process is a function only of the initial and final state of the system and is independent of the path or the manner by which the change is brought about.

  20. ENERGY CHANGE IN RELATION TO WORK AND HEAT CHANGEs:Suppose a system is subjected to change of P and V Δ U = UB – UA = q - wWhere, q is heatabsorbed from surrounding and w is work done by system (mechanical or electrical). Two things happen:The absorption of heat increases the energy of the system and the performance of work done by the system lowers the energy of the system, hence negative sign.In general Δ U = q + w

  21. ENTHALPY(H): H = U + PV and Δ H = Δ U + PΔVSuppose the change of state of a system is brought about at constant P. In that case there will be change of volume. Let volm increase from VA to VB at const P. Then work done by the system will be: w = - P (VB – VA ) ---- 1We have from first law,Δ U = q + w ---- 2 Δ U = q - P (VB – VA )UB – UA = q - P (VB – VA ) (UB + P VB ) – (UA+ P VA ) = q ---- 3The quantity U + PV is known as enthalpy of the system and is denoted as H. It represents the total energy stored in the system at const P. H = U + PV

  22. Since U is a definite quantity, P and V are also definite quantities, hence H is also a definite quantity depending upon the state of the system. Therefore equation 3 implies HB – HA = qΔ H = q ----- 4further from eq 3(UB + P VB ) – (UA+ P VA ) = q ---- 3UB – UA + P (VB – VA ) = qtherefore eq 4 implies thatΔ H = Δ U + PΔV

  23. ENTHAPY OF VAPORISATION:The change in enthalpy, Δ H, when a liquid changes into vapour state or when vapours change into liquid state, is known as enthalpy of vaporisation. Evaporation of a liquid is accompanied by increase in enthalpy since it absorbs some heat from the surroundings. The increase for evaporation of 1 mole of water at 25 deg C is H2 O (l)  H2O (g), Δ Hvap = 40.7 KJmol-1When vapour condense to liquid state, it give out heat. Thus condensation of vapour is compensated by decrease in enthalpy.H2 O (g)  H2O (l), Δ Hvap = - 40.7 KJmol-1

  24. ENTHAPY OF FUSION:The change in enthalpy, Δ H, when a solid changes into liquid state or when liquid changes into solid state, is known as enthalpy of fusion (Δ Hfus ). When a solid melts and changes to liquid, it absorbs heat. Thus melting/fusion of solid to its liquid state is accompanied by increase in enthalpy since it absorbs some heat from the surroundings. The enthalpy of fusion for 1 mole of water at 0 deg C is H2 O (s)  H2O (l), Δ Hfus = +6.02 KJmol-1When liquid freezes and changes to solid state, it gives out heat. Thus freezing/solidification of a liquid is compensated by decrease in enthalpy.H2 O (l)  H2O (s), Δ H = - 6.02 KJmol-1

  25. HEAT CAPACITY: Molar heat capacity(C): C = q / (T2 - T1 ) Heat capacity of a system, between any two temperatures, is defined as the quantity of heat (q) required to raise the temperature of the system from the lower to the higher temperature devided by the temperature difference. If the mass of the system is 1gm, the heat capacity is called specific heat of the system. If the mass of the system is one mole, then the heat capacity is termed as molar heat capacity. It is ususally denoted by C. Thus the molar heat capacity of a system between temperatures T1and T2 is expressed as:C = q / (T2 - T1 )

  26. Molar Heat Capacity at constant Volume, CV: (CV = (∂U/∂T)V For a gaseous system at constant volume, work done is zero ( w = 0) since there is no change in volume.From the first law of TD we have,Δ U = q + w, i.e. q = Δ U ---- 1 CV = (q / T2 - T1 )V = Δ U / (T2 - T1)V or, CV = (∂U/∂T)V ------ 2Molar heat capacity of a gaseous system of mass one mole at constant volume is defined as the change in internal energy (U) of the system per degree rise in temperature. V

  27. Molar Heat Capacity at constant Pressure, CP: (CP = (∂H/∂T)P For a gaseous system at constant pressure, there is change in volume and some work is done. Suppose the change in volume is ΔV, and work done is w, then from the first law of TD we have,ΔU = q + w, or q = ΔU – w ---- 1 and CP = (q / T2 - T1 )PIncrease in volume means that work is done by the system on the surroundings, so w is negative. Hence, w = - P ΔVtherefore q = ΔU – (- P ΔV) = ΔU + P ΔV Therefore, CP = (ΔU + PΔV /T2 - T1)Por CP = (ΔH/T2 - T1)Por, CP = (∂H/∂T)P ------ 2Molar heat capacity of a gaseous system of mass one mole at constant pressure is defined as the change in enthalpy (H) of the system per degree rise in temperature.

  28. RELATION BETWEEN CP AND CvWhen a gas is heated at constant volume no external work is done since there is no change in volume. i.e. the heat supplied to the gas is utilised to increase its internal energy. Thus if temperature of one mole of gas is raised through 1 deg C say from T to T+1, the increase in its internal energy itself give Cv.However at constant pressure, the gas will change its volume when heated, it will expand and some work is done by it. Therefore some extra heat in addditon to the heat required to raise the internal energy of the gas molecules must be supplied to do some external work. Hence CP > CVThus, CP – CV = w, work done by one mole of gas in expansion when heated through 1 deg C at constant pressure.

  29. As we know work of expansion of a gas is w = P ΔVFor one mole of ideal gas PV = RT --- 1Now, when the temperature is raised through 1 deg C from T to T+1, so that the volume is V+ ΔV, thenP(V+ ΔV) = R(T+1) ---- 2Eqn 1 – 2 = P ΔV = R i.e. w = Ror CP – CV = R, the work done by one mole of an ideal gas in expansion at constant pressure when heated through 1 deg C is equal to R.Thus the difference in molar heat capacity of a gas at constant pressure and at constant volume is equal to gas constant R.

  30. THERMOCHEMISTRYDr. J K SARMAH KU

  31. ThermochemistryThe branch of chemistry which deals with energy changes involved in chemical reactions is called thermochemistry. It applies the first law of thermodynamics to chemical reactionsChemical reactions are accompanied by energy changes appear in the form of either evolution or absorption of heat. The chemical changes occur in a chemical reaction largely from the breaking of bonds in reactants and formation of new bonds in products.

  32. Changes in Internal Energy in a Chemical Reaction:As we know every substance has a definite amount of energy known as internal energy, E. Its exact value can not be determined but the change in internal energy, Δ U, can be accurately measured experimentally.Let us consider a chemical reaction taking place at a constant temperature and const volume. In such a case, work done, w = 0, hence from equation of 1st law: Δ U = q + wor Δ U = qvwhere q is heat exchanged at const vol.If UR and UP are the internal energies of the reactants and the products, thenΔ U = UP - URi.e. Δ U = UP - UR = qv = Heat of reaction at constant vol.

  33. Changes in Enthalpy in a Chemical Reaction:If q be the heat exchanged in a chemical reaction taking place at constant pressure. The heat exchanged at const press is termed as enthalpy change. ThusΔ H = qpIf HR and HP are the enthalpy of the reactants and the products, thenΔ H = HP - HRi.e. Δ H = HP - HR = qp = Enthalpy of reaction.

  34. Exothermic and Endothermic Reactions:Reactions that give out heat i.e such reactions are accompanied by evolution of heat, are called exothermic reactions. In case of exothermic reactions, HP < HR ,Δ H = HP - HR Δ H = negative,On the other hand reactions that absorbs heat, i.e. such reactions are accompanied by absorption of heat, are called endothermic reactions.In case of endothermicreactions, HP > HR , Δ H = HP - HR Δ H = positive

  35. Energy level diagram for exothermic and endothermic reactions:

  36. Relation between Enthalpy of Reaction at Const Pressure and Const Volume: qp = qv + Δ ng RTWe knowΔ H = Δ U + P Δ V, where Δ V is the change in volume that takes place in a reaction,Since, Δ U = qv, where q is heat exchanged at const vol.And Δ H = qp, heat exchanged at const pressqp = qv + P Δ V,

  37. Now for n moles of an ideal gas, PV = nRTLet n1 and n2 represent the no. of moles of gaseous reactants and gaseous products respectively and is n2>n1, thenn2 – n1 = Δ ng = increase in the no of gaseous moles ,The corresponding increase in volume, Δ V, will be given by, (V/n) Δ ng. Hence,P Δ V = P (V/n) Δ ng = RT Δ ngTherefore, qp = qv + P Δ V = qv + Δ ng RTi.e. qp = qv + Δ ng RT

  38. Example1: Dissociation of ammonia into nitrogen and hydrogen2NH3(g)  N2 (g) + 3H2(g) 2 moles 1 mole 3 moles Δ ng = 4 - 2 = 2 qp = qv + Δ ng RTi.e qp = qv + 2RT

  39. Example 2: Reaction involving hydrogen and oxygen to form water2H2(g) + O2(g) --------------- 2H2O(l) 2 moles 1 mole 2 moles Δ n= 2-3 = -1Therefore, qp= qv – RT

  40. Example 3: Establish the relationship between qpand qvin the Haber synthesis of ammonia, assuming that the gaseous reactants and products are ideal.Soln: N2(g) + 3H2(g) --------------- 2NH3(g) Here, Δ ng= 2 – (1+3) = - 2Hence, qp = qv - 2RT

  41. Example 4: Establish the relationship between qpand qvin the formation of water from the elements, assuming that the gaseous reactants and products are ideal.

  42. HESS’S LAW OF CONSTANT HEAT SUMMATIONThis law states that 'the heat change in a particular reaction is the same whether it takes place in one step or several steps'.

  43. For example, a reactant 'A' changes to a product 'B' in one step and the heat change during this process is  ΔH. If the reaction is carried out in two steps where 'A' first changes to 'C' an intermediate stage and then 'C' changes to 'B' in the following step then let the heat change during the formation of 'A' to 'C' be ΔH1 and that from 'C' to 'B' be Δ H2. From Hess's law the heat change for the reaction is given as:ΔH =  ΔH1 +  ΔH2

  44. This means that the amount of heat evolved or absorbed in a chemical reaction depends only upon the energy of initial reactants and the final products. The heat change is independent of the path or the manner in which the change has taken place.

  45. First Law of Thermodynamics • energy cannot be created nor destroyed. • Therefore, the total energy of the universe is a constant. • Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.

  46. Spontaneous Processes • Spontaneous processes are those that can proceed without any outside intervention. • The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously

  47. Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

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