TRT 408 PHYSICAL CHEMISTRY

TRT 401 PHYSICAL CHEMISTRYPART 1: FIRST LAW THERMODYNAMIC1Basic concept:Work, heat and energyThe internal energyExpansion workHeat transactionAdiabatic changesThermochemistry:Standard enthalpy changesStandard enthalpies of formationThe temperature dependence of reaction enthalpies.State of function and exact differentials:Exact and inexact differentialsChanges in internal energyThe Joule Thomson effectFirst Law Thermodynamic2Work Work (w) is defined as the force (F) that produces the movement of an object through a distance (d):

Work = force distancew = F x d

Work also has units of J, kJ, cal, kcal, Cal, etc. The two most important types of chemical work are: the electrical work done by moving charged particles. the expansion work done as a result of a volume change in a system, particularly from an expanding or contracting gas. This is also known as pressure-volume work, or PV work.

PV work occurs when the force is the result of a volume change against an external pressure.3

The example of PV work in the cylinder of an automobile engine

The combustion of the gasoline causes gases within the cylinder to expand, pushing the piston outward and ultimately moving the wheel of the car.

The relationship between a volume change (V) and work (w):

W= -P V

Where P is external pressure

The units of PV work are Latm; 1 Latm = 101.3 J.

If the gas expands, V is positive, and the work termwill have a negative sign (work energy is leaving thesystem). If the gas contracts, V is negative, and the work termwill have a positive sign (work energy is entering thesystem). If there is no change in volume, V = 0, and there isno work done. (This occurs in reactions in whichthere is no change in the number of moles of gas.)

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If the gas expands, V is positive, and the work term will have a negative sign (work energy is leaving the system).

If the gas contracts, V is negative, and the work term will have a positive sign (work energy is entering the system).

If there is no change in volume, V = 0, and there is no work done. (This occurs in reactions in which there is no change in the number of moles of gas.)Expansion Work5b) Free expansion is expansion against zero opposing force. W=0No work is done when a system expand freely. Expansion of this kind occurs when a gasexpands into a vacuum.c) Expansion against constant pressure now suppose that the external pressure is constant throughout expansion.

Therefore, if we write the change in volume as V=Vf -Vi

6d) Reversible ExpansionA reverse change in thermodynamic is a change that can be reversed by an infinitesimal modification of a variable.The key infinitesimal sharpen the everyday meaning of the word reversible as something that can change direction.

7d) Isothermal reversible expansion

When the volume is greater than the initial volume, as in expansion, the logarithm in above equation is positive and hence w 0 (positive) If the reaction is reversed, energy is absorbed by the system from the surroundings.

The reactants have a lower E than the products, so U for the system is positive. This energy is lost by the surroundings, where U is negative.

U > 0 (positive)

27The difference , U is positive and energy glows into the system and out of surroundingsSummarizing:

if the reactants have a higher internal energy than a products, Usys is negative and energy flows out of the system into the surroundings.

if the reactants have a lower internal energy than a products, Usys is positive and energy flows into the system from the surroundings.SystemSurroundingsEnergy flowUsys > 0 (positive) Usurr < 0 (negative) 28A system can exchange energy with its surroundings through heat and work:SystemHeat (q)Work (w)SurroundingsAccording to the first law thermodynamic, the change in the internal energy of the system(U) must be the sum of the heat transferred (q) and the work done (w):U = q +wSign of conventions for q, w, and Uq (heat)+ system gain thermal energy- System loses thermal energyw (work)+ work done on the system- Work done by the systemU(change in internal energy)+ energy flows into the system- Energy flows out of the system29 For an isolated system, with no energy flowing in or out of the system, the internal energy is a constant.

First Law of Thermodynamics (restated): The total internal energy of an isolated system is constant.

It is impossible to completely isolate a reaction from its surroundings, but it is possible to measure the change in the internal energy of the system, U, asenergy flows into the system from the surroundings or flows from the system into the surroundings.

U = Uf - Ui30Most reactions are not done in sealed containers: they are carried out in open vessels at constant pressure, with the volume capable of changing freely, especially if the reactants or products of the reaction involve gases.

In these cases, V 0, and the energy change may be due to both heat transfer and PV work.

In order to eliminate the contribution from PV work, a quantity called enthalpy, H, is defined as internal energy (U) plus the product of pressure and volume:

p: pressure, V:volumeBecause p,V,U are all state function, the enthalpy is a state function too.

The change in enthalpy (H) is:

EnthalpyH = U + pVH = qP31The Measurement of Enthalpy Change1)CalorimeterMonitoring the temperature change that accompanies a physical or chemical change occurring at constant pressure. Example: thermally insulated vessel open to the atmosphere: the heat released in the reaction is monitor by measuring the change in temperature of the content.2) Bomb Calorimeter Measuring the internal energy change.

3)Differential Scanning CalorimeterThe most sophisticated way to measure enthalpy changes.

Hm = Um +pVm Um32Example 3:

When 2.0 mol CO2 is heated at the constant pressure of 1.25 atm, its temperature increase from 250 K to 277K. Given the molar heat capacity of CO2 at constant pressure is 37.11JK-1, calculate q, H and U.33

34Adiabatic Changes

The change in internal change of a perfect gas when the temperature is change from Vi to Vf.

Step 1:The volume changes and the temperature is constant at its initial value. The system expands at the constant temperature; there is no change in internal energy if the system consist of a perfect gas.

Step 2:The temperature of the system is reduced at constant volume.

Because the expansion is adiabatic, we know that q = 0; because U=q + w , it then follows that U = wad. Therefore, by equating the two expansion we have obtained for U we obtainedWad = Cv T35Renungan:

Termodinamik merupakan satu subjek yang aneh. Kali pertama anda mempelajarinya,anda tidak faham langsung. Kali kedua anda mempelajarinya, anda fikir anda memahaminyakecuali satu dua perkara. Kali ketiga anda mempelajarinya, anda tahu bahawa anda tidak memahaminya, akan tetapi anda sudah terbiasa dengan subjek itu maka subjek itu tidak akan menjadi masalah lagi kepada anda

Arnorld Sommerfeld (1868-1951) 36PART 2: FIRST LAW THERMODYNAMIC37THERMOCHEMISRTY38Thermochemistry: Basic TermsThermochemistry is the study of energy changes that occur during chemical reactions.System: the part of the universe being studied.Surroundings: the rest of the universe.39Types of systems: open (exchange of mass and energy)closed (exchange of energy)isolated (no exchange)

Isolated systemClose SystemOpen system40 Open: energy and matter can be exchanged with the surroundings. Closed: energy can be exchanged with the surroundings, matter cannot. Isolated: neither energy nor matter can be exchanged with the surroundings.Types of Systems

A closed system; energy (not matter) can be exchanged.After the lid of the jar is unscrewed, which kind of system is it?41ENTHALPY (H)The energy possessed by a system is called the enthalpy or heat content of the system and is given the symbol H. The change of the heat content is given by H,H = Hfinal - Hinitial

H = Hproduct - Hreactant 42Heat (q)Heat is energy transfer resulting from thermal differences between the system and surroundingsflows spontaneously from higher T lower TEOSflow ceases at thermal equilibrium43Enthalpy(H) vs heat (q)When reactions take place at constant pressure; heat change = enthalpy change q = H The quantity of heat energy liberated or absorbed during a reaction taking place at constant pressure is the enthalpy change of the reaction.44Exothermic reactionA chemical reaction that releases/gives off heat is called an exothermic reactionDuring an exothermic process, heat flowsout of the system and into the surrounding.Thus the energy of the system is negative. H = -veEg : C(s) + O2(g) CO(g) H = -393.5 kJ/mol45Exothermic Reaction

Surroundings are at 25 C Hypothetical situation: all heat is instantly released to the surroundings. Heat = qrxnTypical situation: some heat is released to the surroundings, some heat is absorbed by the solution.In an isolated system, all heat is absorbed by the solution. Maximum temperature rise.25 C 32.2 C 35.4 C 46Endothermic reactionA chemical reaction that absorbs/takes in heat is called an endothermic reaction. During an endothermic process, heat flows into the system from the surrounding. Thus the energy of the system is positive.H = +ve47For a system where the reaction is carried out at constant volume, DV = 0 and DU = qV.Internal Energy Change at Constant Volume

All the thermal energy produced by conversion from chemical energy is released as heat; no P-V work is done.

48lnternal Energy Change at Constant PressureFor a system where the reaction is carried out at constant pressure, DU = qP PDV or DU + PDV = qPMost of the thermal energy is released as heat.Some work is done to expand the system against the surroundings (push back the atmosphere).

49Enthalpy is the sum of the internal energy and the pressure-volume product of a system:H = U + PVEnthalpy and Enthalpy Change

Most reactions occur at constant pressure, so for most reactions, the heat evolved equals the enthalpy change.The evolved H2 pushes back the atmosphere; work is done at constant pressure.For a process carried out at constant pressure,qP = DU + PDV soqP = DH Mg + 2 HCl MgCl2 + H250Values of DH are measured experimentally.Negative values indicate exothermic reactions.Positive values indicate endothermic reactions.Enthalpy Diagrams

A decrease in enthalpy during the reaction; DH is negative.An increase in enthalpy during the reaction; DH is positive.51A themochemical equation is an equation that represents a reaction and shows the overall enthalpy changes during the reaction. The following are three important rules for writing thermochemical equations:1. The physical states (s, l, g) of all reactants and products must be specified. 2H2O(l) 2H2O(g) H= 88.0 kJ/mol

Thermochemical Equations 5232. When both sides of a thermochemical equation is multiplied by a factor, then H must also be multiplied by the same factor. H2O(s) H2O(l) H= 6.01 kJ/mol 2H2O(s) 2H2O(l) H= 12.02 kJ/mol

3. When a chemical equation is reversed, the value of H is reversed in sign53 DH changes sign when a process is reversed. Reversing a Reaction

Same magnitude; different signs.54ENTHALPY OF REACTIONStandard enthalpy of a reaction is the enthalpy change for a reaction in which the reactant in their standard states yield products in their standard states.

Enthalpy measured at this condition is called STANDARD ENTHALPY OF REACTION and is written as H55STANDARD STATES/ CONDITIONSTemperature fixed at 25 oC or 298 K Pressure fixed at 1 atm or 101 kPaConcentration of solution is 1 M Physical states of reactants and products are specified (s, l, g, aq) H2(g) + O2(g) H2O(g) H = -241 kJ/mol H2(g) + O2(g) H2O(l) H = -286 kJ/mol

56TYPES OF ENTHALPY OF REACTIONSTANDARD ENTHALPY OF FORMATION Hf

STANDARD ENTHALPY OF COMBUSTION Hc

STANDARD ENTHALPY OF NEUTRALIZATION Hn

STANDARD ENTHALPY OF SOLUTION Hsoln57Standard Enthalpies of FormationIt would be convenient to be able to use the simple relationshipH = Hproducts Hreactantsto determine enthalpy changes.Although we dont know absolute values of enthalpy, we dont need them; we can use a relative scale. We define the standard state of a substance as the state of the pure substance at 1 atm pressure and the temperature of interest (usually 25 C).The standard enthalpy change (H) for a reaction is the enthalpy change in which reactants and products are in their standard states.The standard enthalpy of formation (Hf) for a reaction is the enthalpy change that occurs when 1 mol of a substance is formed from its component elements in their standard states.

58DHrxn = Snp x DHf(products) Snr x DHf(reactants)

The symbol S signifies the summation of several terms.The symbol n signifies the stoichiometric coefficient used in front of a chemical symbol or formula.In other words Add all of the values for DHf of the products.Add all of the values for DHf of the reactants.Subtract #2 from #1(This is usually much easier than using Hesss Law!)Calculations Based onStandard Enthalpies of Formation59STANDARD ENTHALPY OF FORMATION Hf

Is the enthalpy change when one mole of compound forms from its elements in their standard states (298 K, 1 atm)C(s) + O2(g) CO2(g) Hf =-394 kJmol-1H2(g) + O2(g) H2O(l) Hf =-296kJmol-1Na(s) + C(s) + 3/2O2(g) NaCO3(s)

601 mole of compound formed from the most stable form of its element:C(s) + O2(g) CO2(g) Hf =-394 kJmol-1 element C exists naturally as solid atom element O2 exists as molecule in gaseous form. They do not need to be formed. Thus the enthalpy of formation of this element is given zero :Hf for Al(s), N2(g), Mg(s) H2(g) are all 0 kJ/mol61STANDARD ENTHALPY OF COMBUSTION Hc Is the enthalpy change when 1 mole of a substance is completely burnt in excess oxygen under standard conditions:i) CH4 (g) + 2O2(g) CO2(g) + 2H2O(l) Hc = -45 kJ/molii) Mg(s) + O2(g) MgO(s) Hc = -165 kJ/moliii) CH3OH(g) + 3/2O2(g) CO2(g) + 2H2O(l) Hc = -715 kJ/mol62Standard enthalpy of neutralizationIs the enthalpy change when 1 mole of water formed from the reaction between 1 mole of H+ and 1 mole of OH-

HCl + NaOH H2O + NaCl Hn =-57 kJ/molThe neutralization of any strong acid by any strong base gives about the same value ie ~ 57 kJ/mol63Standard enthalpy of solutionIs the enthalpy change when 1 mole of substance is completely dissolved in water.

NaOH (s) Na+(aq) + OH-(aq)NaNO3(s) Na+(aq) + NO3- (aq)64How to determine Hrxn 1. Direct method (calorimetry exp)2. From data of enthalpy of formation3. Using Hesss law4. Thermochemical cycle65How to measure enthalpy change 1. DIRECT METHOD ( through experiment) Calorimetry Heat absorb or evolve is measured using a calorimeter Two types of calorimeter : Coffee-cup calorimeter Bomb calorimeter 66CalorimetryCalorimetry is a technique used to measure heat exchange in chemical reactionsEOSCalorimetry measuring the heat flow associated with a chemical reaction by measuring the temperature.

67We measure heat flow using calorimetry.A calorimeter is a device used to make this measurement.A coffee cup calorimeter may be used for measuring heat involving solutions.Calorimetry

A bomb calorimeter is used to find heat of combustion; the bomb contains oxygen and a sample of the material to be burned.

68Heat evolved in a reaction is absorbed by the calorimeter and its contents.In a calorimeter we measure the temperature change of water or a solution to determine the heat absorbed or evolved by a reaction.The heat capacity (C) of a system is the quantity of heat required to change the temperature of the system by 1 C.C = q/DT (units are J/C)Molar heat capacity is the heat capacity of one mole of a substance.The specific heat (s) is the heat capacity of one gram of a pure substance (or homogeneous mixture).s = C/m = q/(mDT) q = s m DTCalorimetry, Heat Capacity, Specific Heat 69Heat Capacity: A Thought Experiment Place an empty iron pot weighing 5 lb on the burner of a stove.Place an iron pot weighing 1 lb and containing 4 lb water on a second identical burner (same total mass).Turn on both burners. Wait five minutes.Which pot handle can you grab with your bare hand?Iron has a lower specific heat than does water. It takes less heat to warm up iron than it does water.70Heat evolved in a reaction is absorbed by the calorimeter and its contents (water or solution).In a calorimeter the temperature change of water or a solution is measured to determine the heat absorbed or evolved by a reaction.

How does a calorimeter measure heat (q)71specific heat capacityThe specific heat capacity (or specific heat) is the heat required to raise the temperature of 1 gram of a substance by 1 oC. Specific heat capacity of water : 4.184 J /(g oC). 4.184 J energy is needed to raise the temperature of 1 g of water b 1 oC 72TO CALCULATE qq is calculated using q = mass x specific heat capacity x DT q = m c DTIf DT is positive (temperature increases), q is positive and heat is gained by the system.If DT is negative (temperature decreases), q is negative and heat is lost by the system

73Calculation of q - exampleCalculate the heat absorbed when the temperature of 15.0 grams of water is raised from 20.0 oC to 50.0 oC. (The specific heat of water is 4.184 J/g.oC.) using the equation q = mc DT T = (50oC 20oC ) = 30oC, therefore q = 4.184 x 15 x 30 = 188.28 J = 1.88 kJ743Calculating enthalpy using direct methodWhen 23.6 grams of calcium chloride, CaCl2, was dissolved in 500 mL water in a calorimeter, the temperature rose from 25.0 oC to 38.7 oC. Given the heat capacity of water is 4.18 J/g oC, and density of water is 1 g/mL, what is the enthalpy change per mole of calcium chloride?753Heats of Reaction: CalorimetryFirst, calculate the heat absorbed by the calorimeter. q = mc DT q= 500 g x 4.18 J/g oC x (38.7- 25) oC q= 28633 J = 28.633kJ Now we must calculate the heat per mole of calcium chloride. 763Heats of Reaction: CalorimetryCalcium chloride has a molecular mass of 111.1 g, so

Now we can calculate the heat per mole of calcium chloride. 773Calculating Hn by calorimetry25.0 cm3 of 1.00 M HCl at 21.5oC were placed in a polystyrene cup and 25.0 cm3 of 1.00 M NaOH at 21.5oC was added. The mixture was stirred, and the temperature rise to 28.2oC. The density of each solution = 4.18J/(K g). Calculate the standard molar enthalpy of neutralization.1. calculate heat change using

78 q= mct= 50.0 g x 4.18 J/(K g) x 6.7 OC= 1400 J = 1.4 kJ

2. Calculate heat change per mole (Hn ) mole of H2O formed in the reaction= 25.0 x10-3 dm3 x 1.0 M = 2.5 x10-2 mol 2.5 x10-2 mol H2O formed = 1.4 kJ 1 mol H2O formed = 1 x 1.4 2.5 x10-2 mol = 56 kJ/mol

792. Calculating enthalpy using data from Standard Enthalpies of FormationThe law of summation of heats of formation states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants.S is the mathematical symbol meaning the sum of, and m and n are the coefficients of the substances in the chemical equation.

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Using Hf to calculate enthalpy of reaction

Calculate the enthalpy of reaction for the following reaction:

2Al(s) + Cr2O3(s) Al2O3(s) + 2 Cr(s)

given Hf (Cr2O3(s)) = -1669 kJ/mol Hf (Al2O3(s)) = -1128 kJ/mol824Using Hf to calculate Hrxn 2Al(s) + Cr2O3(s) Al2O3(s) + 2 Cr(s) Hf = 0 Hf = -1669 Hf = -1128 Hf =0

Hrxn = of Hf of products - of Hf of reactants

= [2(0) + (-1128)] [(-1669) + 2(0)] = 541 kJ/mol 83Hesss LawHesss law states the enthalpy change of a reaction is constant whether the reaction is carried out directly in one single step or indirectly through a number of steps.System ASystem BSystem ASystem AH1H4H3H284Example 1 : Hesss lawHydrogen iodide can be prepared from hydrogen and iodine using two separate routes. Route I: Eqn 1: H2(g) + I2(s) 2HI(g) H = +52.1 kJRoute II: eqn 2: I2(s) I2(g) H= +61.3 kJ eqn 3: H2(g) + I2(g) 2HI(g)H= -9.2 kJ85 Eqn 1 can be obtained when eqn 2 is added to eqn 3; eqn 1 = eqn 2 + eqn 3 eqn 2: I2(s) I2(g) H= +61.3 kJ eqn 3: H2(g) + I2(g) 2HI(g) H= -9.2 kJ H2(g) + I2(s) 2HI(g

Hence H1 = H2 + H3 = 61.3 9.2 = +52.1 kJ The total enthalpy change for the route I is the same as that for route II.

86Hesss LawExample 2; H for formation of SO3 cannot be obtained directly but the enthalpy of these reactions are known:

The above data can be used to obtain the enthalpy change for the formation of SO3 according to the following reaction?

873The third equation can be obtained by multiplying the first equation by 2 and added to the reverse of the second equation, they will sum together to become the third.

883exerciseDetermine the heat of reaction;Fe2O3(s) + FeO(s) Fe3O4(s)Using the information below: i) 2Fe(s) + O2(g) 2FeO(s) Ho = -554.0 kJ ii) 4Fe(s) + 3O2(g) 2Fe2O3(s) Ho = -1648.8 kJ iii) 3Fe(s) + 2O2(g) Fe3O4(s) Ho = -1118.4 kJ(Answer : -22.0 kJ)89Thermochemical cycle / energy diagram Reactant ( A + B) route 2 Ho2

route 1 Ho route 3Ho3

Product (C + D ) 90 Enthalpy diagram illustrating Hesss law.

91Hesss Law: An Enthalpy Diagram

We can find DH(a) by subtracting DH(b) from DH(c)92Born Harber cycleBorn Haber cycle is an energy cycle used to calculate the lattice energy which cannot be obtained by direct experimental method. They only can be obtained by applying Hesss law in this cycle which involving breaking and forming bonds.93Lattice energyThe enthalpy change when one mole ofionic compound (crystalline substance) is formed from its gaseous ions.Na+(g) + Cl-(g) NaCl(s) Hlat = -788 kJ/mol 2Al3+(g) + 3O2-(g) Al2O3(s) Hlat = -1596 kJ/mol 94ii) Enthalpy of atomization The enthalpy change when one mole of gaseous atoms is formed from its elements under standard conditions Na(s) Na(g) H = +108 kJ/mol O2(g) O(g) H = +247 kJ/moliii) Electron AffinityThe enthalpy change when one mole of gaseous atom gains one mole of electron to form anion. O(g) + e- O-(g) H = -141.0 kJ/mol (1st electron affinity) O-(g) + e- O2-(g) H = -744.0 kJ/mol (2nd electron affinity)95iv) Ionization EnergyThe standard ionization energy is the enthalpy required to remove one mole of electrons from one mole of gaseous metallic atoms to form one mole of positively charged ions. The process is endothermic because energy is absorbed to release electrons from an atom. Mg(g) - e- Mg+(g) (1st ionization energy) Mg+(g) - e- Mg2+(g) (2nd ionization energy)96

97Born haber cycle for LiF

98Calculate the lattice energy for NaCl(s) using the following data. a. standard enthalpy of formation of NaCl(s) Hf = -411kJ/mol b. standard enthalpy of atomization of Na(s) Ha = +108 kJ/mol c. first ionization energy of Na(s) HIE = -+494 kJ/mol d. standard enthalpy of atomization of Cl2(s)Hf = +121 kJ/mole. first electron affinity of Cl2(g)HEA = -364 kJ/molf. Standard lattice enthalpy of NaCl(s)Hlat = x kJ/mol 99

100Measuring Enthalpy Kirchhoffs LawThe standard enthalpy of many important reactions have been measure atDifferent temperatures.

However, in the absence of this information standard reaction enthalpies at different temperature may be calculated from heat capacities and reaction enthalpy at some at some other temperature.

When temperature is increased, the enthalpy of the products and reactants both increase, but may do so to different extent.In each case, the change in enthalpy depends on the heat capacities of the substances.The change in reaction enthalpy reflects the difference in the changes of the enthalpy.101When a substance is heated from T1 to T2, its enthalpy changes from H(T1) to

Assumed that no phase transition take place in the temperature range of interest) because this equation applies to each substance in the reaction the standard reaction, the standard reaction enthalpy changes from Hr(T1) to

Where the is the difference of the molar heat capacities of product and Reactants under standard conditions weighted by the stoichiometri coefficient that Appear in chemical equation:

KirchhoffsLawEqn 1Eqn 2Eqn 3102Example:

The standard enthalpy of formation of H2O (g) at 298K is 241.82kJ mol-1.Estimate its value at 100oC given the following values of the molar heat capacitiesAt constant pressure: H2O (g) = 33.58JK-1mol-1H2(g) = 28.82 JK-1mol-1O2 = 29.36 JK-1 mol-1Assume that the heat capacities are independent of temperature.

Method: When is independent of temperature in the range T1 to T2, the Integral in eqn 2 evaluate to (T2-T1) . Therefore,

To proceed, write the chemical equation, identify the stoichiometri coefficients, and Calculate from the data.

103Answer: The reaction is H2 (g) + O2 (g) H2(g), so

It then follows that

104State Function and Exact Differentials105The internal energy and enthalpy are two example of state functions.

Physical quantities that do depend on the path between two state are called path function.

Example of path functions are the work and the heating that are done when preparing a state.

We do not speak of a system in a particular state as possessing work or heat.

In each case, the energy transferred as work or heat relates to the path being taken between states, not the current state itself.

A part of the richness to thermodynamic is that it uses the mathematical properties of state functions to draw far-reaching conclusions about the relations betweenphysical properties and thereby establish connections that may be completelyunexpected.

The practical importance of this ability is that we can combine measurements of different properties to obtain the value of a property we required.106Exact and Inexact Differentials

U: property of the state, w=property of the pathWork is done by the system as it expands adiabatically to a state f.

In these state the system has an internal energy Uf and the work done on the system along Path 1 from I to f of is w.Initial and final state are the same as those in Path 1 but in which expansion is not adiabatic.

Initial and final states are the same as before (because U is the state funtion).However, in the second path an energy q enter the system as heat and the work w is not the same as w. The work and the heat are path functions.Initial state of the system107If a system is taken along a path (for example, by heating it) U changes from Ui to Uf, and overall change is the sum (integral) of all the infinestimal changes alongthe path:The value of U depends on initial state and final state of the system but is independent of the path between them.

An exact differential is an infinestimal quatity that when integrated gives a resultthat is independent of the path between the initial and final state.

When the a system is heated the total energy transferred as heat is the sum of allIndividual contributions at each point of the path:

Eqn 4Eqn 5Inexact differential is an infinitesimal quantity that, when integrated gives a result that depends on the path between the initial and final state.

108The differences between Eqn 4 and Eqn 5:

We do not write q because q is not a state function and energy supplied as heat cannot be expressed as qf-qi.2. We must specified the path of integration because q depends on the path selected (example: and adiabatic path has q=0, whereas on the non-adiabatic path between the same two state would have q0).

In general, an inexact differential is an infinestimal quantity that, when integrated, gives a result that depends on the path between the initial and final state.

The work done on a system to change it from one state to another depends on thepath taken between the two specified states.example: in general the work is different if the change takes place adiabatically and non-adiabatically. It folows that dw is an exact differential.

109Changes in Internal EnergyIn internal energy U can be regarded as a function of V,T, and p, but becausethere is an equation of state, stating the values of two of the variables fixes thevalue of the third.

Therefore, it is possible to write U in terms of just two independent variables:V and T, p and T, or p and V.

Expressing U as a function of volume and temperature fits the purpose of ourdiscussion. 110General expression for a change in U with T and V

Definition of internal pressure

Eqn 6Eqn 7In terms of the notification Cv and ,eqn 5 can now write as

Eqn 8111The Joule ExperimentJames Joule thought that he could measure by observing the chenge inTemperature of a gas when it is allowed to expand into vacuum.

A schematic diagram of the apparatus used by Joule in an attempt to measure the change in internal energy when a gas expands isothermally. The heat absorbed by the gas is proportional to the change in temperature of the bath. Filled with air at about 22 atmWhen the stopcock opened and the air expanded into a vacuum.No change in temperature112The implication as follow:

No work was done in the expansion into a vacuum, so w=0.No energy entered of left the system (the gas) as heat because the temperature of the bath did not change, so q = 0. Consequently, within the accuracy of the experiment, U=0.

Joule concluded that:U does not change when a gas expand isothermally and therefore that =0.His experiment, how ever is crude.

The heat capacity of the apparatus was so large that the temperature change that gases do in fact cause was too small to measure.

Nevertheless, from his experiment Joule had extracted an essential limiting property of a gas, a property of a perfect gas, without a small deviations characteristic of a real gas.

113Changes in Internal Energy at Constant PressureAs an example, suppose we want to find out how the internal energy varies with temperature when the pressure rather than volume of the system is kept contant.

If we divided both sides of eqn 8 by dT and impose the condition of constant pressure on the resulting differentials, so that du/dT on the left becomes (U/ T)p we obtain:It is usually in thermodynamic to inspect the output of a manipulation like this to see if it contains any recognizable physical quantity. The partial derivative on the right in this expression is the slope of the plot of volume against temperature (at constant pressure). This property is normally tabulated as the expansion coefficient, , of a substance, which is defined as:Definition of the expansion coefficient

Eqn 9Eqn 10114A large value of means that the volume of the sample respond strongly to changes in temperature.

List some experimental values of 115Isothermal compressibility is a measure of the fractional change in a volumewhen the pressure is increase by a small amount; the negative sign in definitionensure that the compressibility is positive quantity; because an increase ofpressure, implying a positive dp, bring about reduction of volume, a negative dV.

Isothermal compressibility, KT( kappa) which define as

Eqn 11116Where the Joule-Thompson coefficient, (mu) is define as

This relation will prove useful for relating the heat capacities at constant volumeand for a discussion of liquefaction of gases.The Joule-Thompson EffectThe closed system of constant composition

Eqn 12Eqn 13117Observation of the Joule-Thompson EffectThe analysis of the Joule-Thompson coefficient is central to the technologicalproblems associate with the liquefaction of gases. We need to be able to interpret its physical and measured it.

The apparatus used to for measuringThe Joule-Thompson effect. The gas Expands through the porous barrier, Which acts as a throttle, and the whole apparatus is thermally insulated.

This arrangement corresponds to an Isenthalpic expansion (expansion at constant enthalpy).Whether the expansion results in a heating or a cooling of the gas depends on the conditions118

The isothermal Joule Thompson coefficients is the slope of the enthalpywith respect to changing pressure the temperature being held constant.Definiton of the isothermal Joule-Thompson coefficientBy compressing Eqn 12 and above equation, we can see that the two coefficients are related by

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A schematic diagram of the apparatus used for measuring the isothermal Joule-Thompson coefficient. The electrical heating required to offset the cooling arising from expansion is interpreted as H and used to calculate , which is then converted to .

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Real gases have nonzero Joule-Thompsoncoefficient. Depending on the identity of the gas, pressure, the relative magnitudes of the attractive and repulsive intermolecular forces, and the temperature, the sign of the coefficient may be either positive or negative.

+ve sign dT is ve when dp is ve (in which case the gas cool on expansion)

Heating effect (0) when the temperature is below their upper inversion temperature,T1.

A gas typically has two inversion temperatures, one at high temperature and the other at low.121

The principle of Linde refrigeratorThe gas at high temperature is allowed to expand through a throttle, it cools and is circulated past the incoming gas.

That gas is cooled, and its subsequent expansion cools it still further. 122THANK YOU123