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# First Law of Thermodynamics State Functions A State Function is a thermodynamic quantity whose value depends on ly on the state at the moment i

e the temperature pressu re volume etc The value of a state function is independent of the history of the system Temperature is an example of a state function The fact that temperature is a state function is extremely useful because it we can mea

## First Law of Thermodynamics State Functions A State Function is a thermodynamic quantity whose value depends on ly on the state at the moment i

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## Presentation on theme: "First Law of Thermodynamics State Functions A State Function is a thermodynamic quantity whose value depends on ly on the state at the moment i"â€” Presentation transcript:

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First Law of Thermodynamics State Functions - A State Function is a thermodynamic quantity whose value depends on ly on the state at the moment, i. e., the temperature, pressu re, volume, etc - The value of a state function is independent of the history of the system. - Temperature is an example of a state function. - The fact that temperature is a state function is extremely useful because it we can measure the temperature change in the system by kno wing the initial temperature and the final temperature. - In other words, we don’t need all of the nitty-g ritty detail of a

process to measure the change in the value of a state function. - In contrast, we do need all of the nitty-gritty details to measure the heat or the work of a system. Reversibility A reversible process is a process where the effects of following a ther modynamic path can be undone be exactly reversing the path. An easier definition is a process that is always at equilibrium even when undergoing a change. Phase changes and chemical equilibria are examples of reversible processes. Ideally the composition throughout the system must be homogeneous. - This requirement implies that the no gradients,

currents or eddys can exist . - To eliminate all inhomogeneities, a reversible process must occur infinitely slow ! - Thus no truly reversible processes exist. Howev er, many systems are approximately reversible. And assuming reversible processes will greatly aid our calculations of various thermodynamic state fun ctions. Reversibility during pressure changes ensures that p = p ex That is, the pressure on the inside of the containe r is always equal to the pressure exerted on the outside of the container.
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Theorem of Maximum Work The maximum amount of work that can be extracted

fr om an expansion process occurs under reversible condition s. Thus the theorem implies that during an irreversibl e expansion, some of the energy is lost as heat rather than work. The inhomogeneities (currents, gradients and eddys) in pressure that occur during an irreversible process are responsible for the heat. Initial Definitions of Energy and Energy Transfer Internal Energy - Sum of the kinetic and potential energy in a sam ple of matter. Microscopic modes of Internal Energy (20 th century view) - degrees of freedom for energy storage translational rotational librational – vibrations

caused by intermolecula r forces vibrational electronic nuclear, etc - more specifics in second semester - unnecessary for understanding of thermodynamics but sometimes helpful. Macroscopic view of internal energy (19 th century view) - A reservoir of energy within the sample. - The details of the energy are unknown. Internal energy is a state function. Work Recall the definition of work from classical mechan ics. w F dl = × Microscopically, work involves the concerted motion of molecules (that is, molecules moving in one direction.) Work is a transfer of energy, not a quantity of ene rgy.

Work is not a state function. -calculation of work depends on thermodynamic path (as seen from definition)
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A more thermodynamically useful rearrangement (let F = p A and dV = dl A) is ex w p dV = - × - the negative sign is a sign convention (discusse d below) - p ex is the external pressure (the pressure outside the container) - p ex is used in definition because we really do not kno w what kind of pressures develop during a thermodynamic change. - i.e., pressure currents and eddies are likely f or an arbitrary change. Other forms of work are possible w E dq = × Electrical

work w E d VP = × Polarization work (e.g., piezoelectricity) w dA = s× Surface tension w d = t× q Twisting work Sign conventions for work -w – work done by system (expansion for pV work) +w – work done on the system (compression for pV wo rk) Heat - macroscopically heat is a thermal energy transfe r - energy transfer usually characterized by temper ature and the zeroth law of thermodynamics. - microscopically heat transfer comes from - inelastic collisions - energy transfer from one molecule to another in multiple (every) direction Sign conventions for heat -q – heat transferred away from body

(heat lost) +q – heat transferred into body (heat gained)
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First Law of Thermodynamics The physicists that studied energy changes recogniz ed that the energy of an object could be changed via heat or work. James Joule demonstrated experimentally the law of conservation of energy , that is, that energy is not gained or lost, but merely trans ferred from one object to another. The law of conservation of energy is also known as the first law of thermodynamics . Since energy changes can be expressed only as he at or work, the first law of thermodynamics has the mathematical ex

pression U q w D = + Subtle points to make: 1) q and w are energy changes, writing q and w is improper. - We can’t refer to object having an amount of he at or work. The object has internal energy, enthalpy, free energy, etc 2) The form of the first law of thermodynamics de pends on the sign convention chosen for heat and work. Older texts state that U q w D = - ; however, the sign convention for work is different, ex w p dV = + × Differential Form of the First Law The differential form of the first law is written a s dU q + The differential forms of heat and work are written with the slash to

emphasize that they are inexact differentials. Constant volume heat During a constant volume process, w = 0. Therefore U q D = That is, constant volume heat is equal to a change in the free energy.
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Measureable thermodynamic quantities Thermal expansivity The thermal expansivity is the measure of how a mat erial changes its volume as the temperature changes. 1 V V T a = Isothermal compressibility 1 V V p k = - The isothermal compressibility is the measure of ho w a material changes its volume as the pressure changes. Almost always, vol ume will decrease with increasing

pressure, therefore, a negative sign is included in the definition to allow for the tabulation a positive values. Both definitions modify an extensive change, or , into an intensive change by dividing the change by the volu me. Enthalpy Background In order to fully explain his ideas of free energy, Josiah Willard Gibbs needed to construct an energy state function that had the def inition H = U + PV Gibbs named the quantity the heat content because a change in the quantity corresponded to heat gained or lost by a system at constant pressure provided no non- pV work is being done, that is. H q D

= Kamerlingh Onnes eventually named the function, H t he enthalpy and the name stuck.
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Heat capacity - Heat is proportional to mass. - Heat is proportional to temperature difference. - Proportionality constant for a specific substanc e is the heat capacity or specific heat . q mC T = D Definition of Constant Pressure Heat Capacity – C Since constant pressure heat is a change in enthalp y, the constant pressure heat capacity can be rewritten as This quantity is one of the easily measurable quant ities that is very important in thermodynamics. Definition of Constant Volume Heat

Capacity – C Since constant volume heat is a change in internal energy, the constant volume heat capacity can be rewritten as This quantity is another one of the easily measurab le quantities that we will use a great deal in thermodynamics studies. Relationship between C and C One way of examining the difference between interna l energy and enthalpy is by examining the difference between the constant press ure heat capacity and the constant volume heat capacity. p V p V V p V p p V U PV PV H U U U U C C T T T T T T T U V U T T T ¶ + ¶ ¶ ¶ ¶ ¶ ¶ - = - = - = + - ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ = + - ¶ ¶ ¶

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At this point, we want to find another expression f or since it is not an easily measurable quantity. We can find an alternative ex pression for based the total differential of the internal energy. The internal energy is a function of temperature an d volume, thus the total differential is V T U U dU dT dV T V ¶ ¶ = + ¶ ¶ Dividing the total differential by dT under constan t pressure conditions yields p V p T p V T p U U T U V U U V T T T V T T V T ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ = + = + ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ Two of the partial derivatives in this expression s hould be familiar: and = a . Remember

the thermal expansivity is 1 V V T a = Let us substitute our result into the original equa tion for C – C , p V p p V V T p p V T p p T p T U V U U U V V U C C P T T T T V T T T U V V U V U P P V P V T T V T V ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ - = + - = + + - ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ = + = + = a + ¶ ¶ ¶ ¶ ¶ ¶ What is the relationship between C – C for an ideal gas? p p 1 V 1 nRT 1 nR nR V T V T p V p pV ¶ ¶ a = = = = ¶ ¶ p V T T T U nR U nR U C C V p V p p V pV V p V nR U nR 1 U p nR 1 p V p p V ¶ ¶ ¶ - = a + = + = + ¶ ¶ ¶ ¶ ¶ = + = + ¶ ¶
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Remember that for an ideal gas , therefore p V 1 U C

C nR 1 nR p V - = + = The relationship can be rewritten in terms of molar heat capacities. p V C C R - = This relationship is very important and worth memor izing. Why is C greater than C ? - Heat capacity is substance’s capacity to store e nergy. - A substance that can store energy via work has a greater heat capacity than a substance that has its volume kept constant. Heat Reexamined Isothermal Heat dq CdT q 0 Isochoric Heat V V dq C dT q C dT If C is independent of temperature, (fair assumption fo r small temperature changes), then V V V 2 1 V q C dT C dT C T T C T = = = - = D Isobaric Heat

p p dq C dT q C dT Adiabatic Heat By definition, q = 0 All of the above results are free from assumptions (unless noted).
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Thermochemistry Extent of Reaction To monitor the progress, a variable known as the extant of reaction is defined as i i,0 n n x = where n is the number of moles of chemical species i, n i,0 is the initial number of moles of chemical species i and is the stoichiometric coefficient of the chemical species i for the specific reaction occurring. When the extent of reaction is defined this way, th e value of the extant of reaction does not depend on the

chemical species used to exa mine the state. Example: Consider the reaction P (s) + 5 O (g) 2 P (s). Suppose we start with 3 moles of phosphorus and 15 moles of oxygen. a) What is the extant of reaction when 4 moles of has been created? i i,0 n n 4 mol 0 mol x = = = Note: How much P has been consumed to create 4 moles of P ? A: 2 moles. b) What is the extant of reaction when 2 moles of phosphorus has reacted? i i,0 n n 1mol 3mol x = = = n - The definition includes the stoichiometric coeffici ent to put the changes of the all the chemical species on an equal footing. Soon we will be dealing

reaction energies and other quantities specific to reactions. The reaction energies need to be intensive quantiti es and yet each chemical species may have a different amount. Thus reaction energie s are defined with respect to the extent of reaction so that the specific chemical sp ecies is not important, but the reaction as a whole is important.
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10 Standard States Standard Pressure - p Thermodynamic quantities are measured with respect to a pressure of 1 bar (100.000 kPa) Standard Temperature - T Thermodynamic quantities are measured with respect to a temperature of 25 C

(298.15 K) Standard Concentration - c Thermodynamic quantities are measured with respect to a concentration of 1m (that is 1 molal). - Molal is used as a concentration unit rather th an molar because molal is independent of temperature. Biological Standard State The biological standard state is a pressure of 1 ba r, a temperature of 37 C and pH of 7, (that is [H ] = 1.0 10 -7 ).
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11 Hess’ Law Since enthalpy is a state function, we can choose m ore than one thermodynamic path to calculate a state function. For chemical changes, Hess’ Law states that the ent halpy of a reaction

can be calculated from the enthalpies of all of the chemic al step processes needed for the chemical reaction. In other words, if reactions can be added together to form a resultant reaction, then enthalpies of the step reactions can be added to fi nd a resultant enthalpy of reaction. Example: Calculate the enthalpy of reaction for t he following reaction: 2 Al(s) + 3 Cl (g) 2 AlCl (s), given the reactions below. 2 Al(s) + 6 HCl(aq) 2 AlCl (aq) + 3 H (g) = -1049 kJ/mol HCl(g) HCl(aq) = -74.8 kJ/mol (g) + Cl (g) 2 HCl(g) = -185 kJ/mol AlCl (s) AlCl (aq) = -323 kJ/mol Rearrange the equations

such that their sum is the reaction of interest. 2 Al(s) + 6 HCl(aq) 2 AlCl (aq) + 3 H (g) = -1049 kJ/mol 6[HCl(g) HCl(aq)] = 6[-74.8 kJ/mol] 3[H (g) + Cl (g) 2 HCl(g)] = 3[-185 kJ/mol] 2[AlCl (aq) AlCl (s)] = 2[+323 kJ/mol] 2 Al(s) + 6 HCl(aq) + 6 HCl(g) + 3 H (g) + 3 Cl (g) + 2 AlCl (aq) 2 AlCl (aq) + 3 H (g) + 6 HCl(aq) + 6 HCl(g) + 2 AlCl (s) = -1049 kJ/mol + 6[-74.8 kJ/mol] + 3[-185 kJ/mol] + 2[+323 kJ/mol] = -1406 kJ/mol 2 Al(s) + 3 Cl (g) 2 AlCl (s) = -1406 kJ/mol
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12 Enthalpy of Formation The enthalpy of formation is the enthalpy of a form ation reaction for a

particular substance. A formation reaction is that where a compound is fo rmed from elements in the naturally occurring state. 2 C(s) + 3 H (g) + ˝ O (g) C OH (l) ˝ P (s) + 5/2 O (g) P (s) 7 Fe(s) + 18 C(s) + 9 N (g) Fe [Fe(CN) (s) Enthalpy of formation of the elements By definition, the enthalpy of formation of an elem ent in its natural state is zero. (B(s)) = 0 kJ/mol (C(graphite)) = 0 kJ/mol (Br (l)) = 0 kJ/mol (S (s)) = 0 kJ/mol (Ag(s)) = 0 kJ/mol (Xe(g)) = 0 kJ/mol Reaction enthalpies can be calculated as a stoichio metric sum of enthalpies of formation. This technique is an application

of Hes s’ law. Example: Calculate the enthalpy of reaction for t he following reaction: CH COOH(l) + C OH(l) C OOCCH (l) + H O(l), given the reactions below. 2 C(s) + 2 H (g) + O (g) CH COOH(l) = - 483.52 kJ/mol 4 C(s) + 5 H (g) + ˝ O (g) C OH(l) = - 328 kJ/mol 6 C(s) + 6 H (g) + O (g) C OOCCH (l) = - 609.6 kJ/mol (g) + ˝ O (g) H O(l) = - 285.83 kJ/mol Note all of the above reactions are formation react ions. Rearrange the equations such that their sum is the reaction of interest. CH COOH(l) 2 C(s) + 2 H (g) + O (g) = + 483.52 kJ/mol OH(l) 4 C(s) + 5 H (g) + ˝ O (g) = + 328 kJ/mol 6 C(s) + 6 H

(g) + O (g) C OOCCH (l) = - 609.6 kJ/mol (g) + ˝ O (g) H O(l) = - 285.83 kJ/mol CH COOH(l) + C OH(l) + 6 C(s) + 6 H (g) + O (g) + H (g) + ˝ O (g) 2 C(s) + 2 H (g) + O (g) + 4 C(s) + 5 H (g) + ˝ O (g) + C OOCCH (l) + H O(l) = + 483.52 kJ/mol + 328 kJ/mol - 609.6 kJ/mol - 28 5.83 kJ/mol = - 84 kJ/mol CH COOH(l) + C OH(l) C OOCCH (l) + H O(l) = - 84 kJ/mol
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13 Calorimetry Constant Pressure Calorimeter - coffee cup calorimeter - open to atmosphere - appropriate for solution chemistry Constant Volume Calorimeter - bomb calorimeter - sealed and isolated - appropriate for gas phase

chemistry Adiabatic Calorimeter - heat measured by temperature needed to keep ther mal energy constant. Process of measuring heat of combustion with bomb c alorimeter 1. Mass wick used to start combustion. 2. Mass water used to absorb heat. 3. Mass standard (benzoic acid) to be combusted 4. Add water to combustion chamber to ensure that w ater from combustion will be in liquid phase. 5. Load standard and seal 6. Fill combustion chamber with oxygen ( 25 atm) 7. Begin combustion and record temperature change o f water. 8. Calculate heat capacity of calorimeter from stan dard - need to account

for heat of wick, and heat capa city of water. 9. Repeat process with sample. - with heat from wick, heat from the water and he at from the calorimeter, the heat of combustion of sample can be calculated. Hess’ law is often used to the enthalpy of formatio n of a substance after its enthalpy of combustion has been calculated. Example: 0.523 g of the military explosive, cyclo tetramethylenetetranitramine (HMX), C is combusted in a bomb (!) calorimeter and an inte rnal energy change of – 4.620 kJ is measured. Calculate the enthalpy of formation for HMX. First write a balanced equation for its

complete co mbustion. (s) + 2 O (g) 4 CO (g) + 4 H O(l) + 4 N (g) Next calculate the molar internal energy of reactio n from the experimental data. 4.62 kJ 296.155g 2616 kJ mol 0.523g mol × = -
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14 To calculate the enthalpy of the reaction, we need to return the definition of enthalpy. H U PV U nRT U n RT D = D + D = D + D = D + D Reexamining the balanced chemical equation, we see that 8 moles – 2 moles = 6 moles of gas has been produce d. (s) + 2 O (g) 4 CO (g) + 4 H O(l) + 4 N (g) ( ) 0.008314kJ H U n RT 2616kJ mol 6 298.15K mol K 2.478kJ 2616 kJ mol 6 2616 kJ mol 14.87 kJ

mol 2601kJ mol mol D = D + D = - + = - + = - + = - The molar enthalpy of reaction can also be written in terms of the molar enthalpies of formation. 4 (CO (g)) + 4 (H O(l)) - (C (s)) = Rearranged, the equation becomes, (C (s)) = 4 (CO (g)) + 4 (H O(l)) - From table, we find the enthalpies of formation for CO (g) and H O(l) (C (s)) = 4 (-393.5 kJ/mol) + 4 (-285.8 kJ/mol) - (-2601 kJ/mol) = -116 kJ/mol Temperature Dependence of Internal Energy and Entha lpy Recall that and These relationships imply that we can find the inte rnal energy or enthalpy at a nonstandard temperature as long as we know

the heat capacity. dU C dT and dH C dT Integrating both sides of these equations yields U C dT D = and H C dT D =
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15 For large temperature changes, we can’t assume that the heat capacity is independent of temperature. Thus to calculate the internal ene rgy or enthalpy at a nonzero temperature, we need the temperature dependence of the heat capacity. Example: Calculate the change in enthalpy of H (g) from 373 K to 1000 K, given that the constant pressure heat capacity has the fo rm C d eT fT = + + where d = 27.28 J/K mol, e = 0.00326 J/K mol and f = 0.00050 J K/mol. ( ) ( ) (

) ( )( ) ( ) 1000 1000 1000 1000 K 373 K p 373 373 373 1000 1000 1000 2 2 373 373 373 H H H C dT d eT fT dT d dT eT dT fT dT eT f e 1 1 d T | | | d 1000 373 1000 373 2f 2 T 2 1000 373 0.00326J K mol 27.28J K mol 627 K 860871K 2 0.00050 D = - = = + + = + + = + - = - + - - - = × + ( ) ( ) J K mol 0.0016810K 17105J mol 1403J mol 1.68 10 J mol 18508J mol 18.51kJ mol × - = + + ´ = = Kirchoff’s law ( ) ( ) r r i p,i 298 H T H 298K C dT D = D + n That is by taking a stoichiometric sum of the heat capacities and integrating over the temperature range, we can find the correction to th e reaction

enthalpy at a non standard state temperature.
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16 Example: 471 kJ/mol is the reaction enthalpy unde r standard conditions for the following reaction: 2 Fe (s) + 3 C(s) 4 Fe(s) + 3 CO (g). What is the reaction enthalpy at 1000 K? According to the NIST Webbook Internet site, the co nstant pressure heat capacities of the chemical species in the above smelting process can be fitted according to the Shumate equation, 2 3 2 C A BT CT DT E T = + + + + The values for A, B, C, D and E for each species ar e given below. A B C D E Fe 93.43834 0.108358 -5.08645 10 2.56 10 -1610000 C 10.68

0 0 0 0 CO 24.99735 0.055187 -3.36914 10 7.95 10 -137000 Fe 18.42868 0.024643 -8.91 10 9.66 10 -12600 Sum -70.2099 0.047418 -3.50 10 1.13 10 2762174 ( ) ( ) ( ) ( ) 1000K 2 3 2 r r i i i i i i 298K 0 1000 2 1000 3 1000 3 1000 1 1000 i i i r i i 298 298 298 298 i 298 H 1000K H 298K A B T C T D T E T dT B C D H 298K A T | T | T | T | E T | 2 3 4 D = D + n + + + + = D + n + + + - ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 3 4 4 0.047418 471kJ H 1000K 70.2099 1000 298 1000 298 mol 2 3.50 10 1.13 10 1 1 1000 298 1000 298 2762174 3 4 1000 298 D = + - - + - - ´ ´ + - + - - - ( ) 471kJ 49.3kJ 21.6kJ 34.1kJ

11.2kJ 6.5kJ 413.9k H 1000 K mol mol mol mol mol mol mol D = - + - + - = Kirchoff’s Law can be stated with in a differential form as well. Include ¶D = D and ¶D = D Bond Enthalpies A bond enthalpy is the energy needed to separate two atoms. Tabulated bond enthalpies are average values calcul ated from the dissociation of many different compounds. Thus tabulated bond enthalpies are approximate valu es. *However, they can be useful for approximating enth alpies of reaction because most of the chemical energy of a compound is held i n its bonds.*
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17 Work Reexamined Isochoric

Work pV work involves a volume change. ex w p dV = - × Since a constant volume process has no volume chang e, w = 0 if there is no other work (no non-pV work). No other assumptions have been made about the syste m. Isobaric Work Since the pressure is constant, it has no dependenc e on the volume. Thus the pressure can pulled out of the integral. Subsequently, the integral of dV is V – V = V ex ex ex 2 1 ex w p dV p dV p V V p V = - × = - = - - = - D The only other assumption made is that the system d oes not have any non-pV work. Isothermal Work In general we need to know the relationship

between pressure and volume to perform the calculation, i.e, we need an equation of state. Let us do the calculation for an ideal gas under reversible conditions. ex nRT dV V w p dV p dV dV nRT nRT ln V V V = - × = - × = - × = - = - Note restrictions on the applicability of the calcu lation. 1. no other non-pV work 2. ideal gas 3. reversible conditions Adiabatic Work dU q + w dU w w U = D No assumptions made!
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18 Refrigeration and the Joule-Thomson coefficient Joule coefficient - m = - internal pressure q = 0, w = 0 U = 0 isoenergetic process Joule-Thomson coefficient - JT m =

- isoenthalpic, adiabatic - = 0 for an ideal gas Classic Experimental Apparatus - measure JT directly. f i i f i i f f V 0 f i i i f f i U U U p dV p dV p V p V U p V U p V H H D = - = - - = - - = -
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19 Modern Experimental Apparatus - measure JT via - use isothermal conditions rather adiabatic condi tions. isothermal conditions: T = T T T T p T p p H H T p H T 1 T T p H H p H T p C H T T ¶ ¶ - - ¶ ¶ ¶ ¶ - ¶ ¶ = - = = = ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ JT m =
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20 Isoenthalps on T- p plot. - inversion temperature - > 0 for substance to be used as a refrigerant - temperature

must decrease as pressure decreases .
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21 Real Gases Reexamined Calculations on van der Waals gas - pV work ex w p dV = - × nRT an V b V = - ex 1 2 1 nRT an nRT an w p dV p dV dV dV dV V b V V b V V b 1 1 nRT ln an V b V V = - × = - × = - - × = - × + × - - = - - - - Difference in heat capacities p v p v H U C C T T ¶ ¶ - = - ¶ ¶ T V p T V p T p V p T p V V p T U U 1 U U dU dV dT dU dV dT V T T V T U U V U T U V U T V T T T V T T U U U T T V ¶ ¶ ¶ ¶ = + = + ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ = + = + ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ = - ¶ ¶ ¶ p p p p pV H U U V H U pV T T T T T ¶ ¶ ¶ ¶ = + = + = +

¶ ¶ ¶ ¶ ¶ [ ] p v p p p T p p T p p T p T T U V U U V C C p T T T V T V U V V U p p T V T T V V U U p V p V p T V V ¶ ¶ ¶ ¶ ¶ - = + - - ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ = + = + ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ = + = a + = a + p ¶ ¶ ¶
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22 Internal Energy Reexamined Why is U = q under isochoric conditions? ex dU q + w dU q - p dV dU q U q D = Assuming no non-pV work. Isobaric Internal Energy v ex v ex v dU q + w dU C dT p dV U C dT p dV C dT p V = - D = - = - D Note: no assumptions about the equation of state. Isochoric Internal Energy ex v v dU q + w dU q p dV dU q U C dT q = - D = = Note: assumption of no

non-pV work Isothermal Internal Energy dU q + w = C dT dT 0 dU 0 ˛ =
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23 Enthalpy Reexamined Why is H = q under isobaric conditions? H U pV dH dU d pV = q + w + p dV + v dp dH = q + -pdV + p dV + v dp dH q H q = + = + D = Assuming no non-pV work and reversible conditions Isobaric Enthalpy p p H U pV dH dU d pV = q + w + p dV + v dp dH = q p dV + p dV dH q H C dT q = + = + D = = Assuming no non-pV work and reversible conditions Isochoric Enthalpy H U pV dH dU d pV = q + w + p dV + v dp dH = q p dV + pdV + v dp dH q + v dp H C dT v dp = + = + D = - For ideal gas under

reversible conditions p p nRT H C dT vdp C dT dp D = - = - Why not H C dT nRT ln D = - ? - temperature has a dependence on the pressure. Isothermal Enthalpy H U pV dH dU d pV = C dT H 0 = + = + D = Important: All of the above enthalpies assume that the system has no non-pV work and reversible conditions (so that P = P ex )
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24 Summary Isothermal T = 0 Isobaric p = 0 Isochoric V = 0 Adiabatic q = 0 Definition of work : ex w p dV = - × Definitions for heat : q = H q = U Sign conventions for heat and work . -w – work done by system (expansion for pV work) +w – work done on the

system (compression for pV wo rk) -q – heat transferred away from body (heat lost) +q – heat transferred into body (heat gained) Reversibility p = p ex Extent of Reaction i i,0 n n x = Definition of Heat Capacities p p q H T T ¶ ¶ = = ¶ ¶ v v q U T T ¶ ¶ = = ¶ ¶ First Law : ex dU w U q w U CdT p dV = + D = + D = - Partial Derivatives to be acquainted with : , , C , C , , J, JT 1 V V T a = 1 V V p k = - m = JT m = and internal pressure p = (equals zero for ideal gas)
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25 Natural variables of U and H and total differential Internal energy can be expressed as a function of V and

T U(V,T) Enthalpy can be expressed as a function of p and T H(p,T) - both of these functional dependences are not uni que - will consider other natural variables in the nex t chapter. Total differential can be expressed as V T U U dU dT dV T V ¶ ¶ = + ¶ ¶ H H dH dT dp T p ¶ ¶ = + ¶ ¶ Hess’ law Temperature Dependence of Enthalpy and Internal Ene rgy U C dT D = and H C dT D = Kirchoff’s law ( ) ( ) r r i p,i 298 H T H 298K C dT D = D + n ¶D = D and ¶D = D Miscellaneous Applications of Heat Capacities p V C C R - = g =