Chapter 2: The First Law: Concepts

1. Chapter 2: The First Law: Concepts Homework: Exercises(a only):4, 6, 9, 18, 24, 25, 26, 29, 30, 35, 40 Problems: 2.2, 2.6, 2.10, 2.24, 2.34

2. Basic Definitions System - volume of interest (reaction vessel, test tube, biological cell, atmosphere, etc.) Surroundings volume outside system Open system - matter can pass between system & surroundings Closed system - matter cannot pass between system & surroundings Isolated system - Neither matter nor energy can pass between system & surroundings No mechanical or thermal contact

3. Work, Energy and Heat Work - generalized force (intensive factor) over a generalized displacement (extensive factor) Examples Mechanical work - force x distance; -fdx Expansion work: pressure x volume; -pdV Electrical: emf x charge displacement; EdQ Magnetic: field strength x magnetization; HdM Sign Work done by a system is negative Work done on a system is positive Work is the result of organized motion of molecules Energy is the capacity to do work For an isolated system doing work reduces the energy, having work done on it increases the energy Heat is the change in energy in a system that is produced by a change in its temperature Recall adiabatic boundaries don�t allow heat flow; diathermic do In an adiabatic container endothermic and exothermic processes produce temperature change in the system In diathermic container they result in heat flow across the boundary Heat is the result of disordered (thermal) motion of molecules

4. First Law of ThermodynamicsYou can�t win! Total energy of a system (kinetic + potential) is called the internal energy (U) The change in energy from some initial state, i, to some final state, f, DU is: DU = Uf - Ui Internal energy is a state function Value depends only on current state of the system, not how you got there - path independent Any change in state variable (e.g. pressure) results in a change in U Unit of measure for energy (U, work, heat) is the joule (J) [kg m2 s-2] Calorie (cal) = 4.184 J eV = 1.6 x 10-17 J (chemical processes on atomic scale are a few eV) On a molecular scale most of the internal energy of a gas is due to motion of atoms (3/2 kT {~3.7 kJ/mole @ 25�C}from Boltzman distribution) so as temperature is raised U increases Internal energy is changed by either work or heat acting on system If work is done on the system (heat in), U increases If system does work (heat out), U decreases This implies that for an isolated system U is constant {1st Law of Thermodynamics} Mathematical summary of 1st Law: DU = q + w where q = heat transferred to a system and w = work done on the system No perpetual motion machines! (You can�t win!) For an isolated system doing work lowers U Isolated system DU = 0

5. First Law of ThermodynamicsExpansion Work In differential form, the first law is: dU = dq + dw Expansion work is work done by a change in volume Consider piston at right, change in volume is Apistondz (dV); force is pexternal If pex = 0 (e.g. vacuum),w = 0 If pex = constant (e.g. atmosphere), pex is removed from integral and w = - pex DV Indicator diagram is a plot of p vs V where area between volumes is work Note: other types of work (non-expansion work) have similar formulations (see earlier or Table 2.1). Total work is sum of individual components.

6. Reversible Change A reversible process is one that is carried out along a path in which all intermediate states are equilibrium states An equilibrium state is one in which an infinitesimal changes in opposite directions result in opposite changes in state Thermal equilibrium between two systems at the same temperature Reversible changes are typically slow Reversible expansions pex set equal p (pressure of gas) along the path of the expansion. Thus, If we know how p varies with V we can evaluate integral Variation of p with V is the equation of state of the gas [e.g., p =nRT/V (ideal gas)]

7. Isothermal Reversible Change In an isothermal reversible expansion, temperature does not change T is not a function of V and can be removed from the integral For ideal gas, w = -nRT ln (Vf/Vi) If final volume > initial (expansion), w < 0 System has done work on surroundings internal energy(U) decreased If final volume < initial (compression), w > 0 System has work done on it, U increased Note that as T increases |w| increases

8. Reversible vs. Irreversible Change Irreversible expansion: w= = -pexDV Reversible (isothermal) expansion: w=-nRTln{Vf/Vi} Which is greater? Consider indicator diagram Work(rev.) = Area under curve Work (irrev.) = Area under rectangle Work (rev.) > W (irrev.) Reversible work is the maximum which can be done True of PV work True of all work

9. Heat Changes - Calorimetry Recall: dU = dq + dw If expansion work is dwexp and dwother is other work done (electrical, magnetic, etc.), then dw = dwexp + dwother dU = dq + dwexp + dwother If V is constant, dwexp is zero since no PV work can be done Assume dwother is also zero, then dU = dq = qv qv is the change in q at constant volume Thus, measuring the heat change in a system at constant volume gives a measure of the change in internal energy This process of measuring heat change is known as calorimetry One constant volume container for measuring heat change is a bomb calorimeter Experiment 1 in lab Typically, substance is burned in calorimeter and temperature rise is measured (dV is constant, but P does change in bomb) In a bomb calorimeter, DT a qv This proportionality is quantified by calibration, typically by combusting a known substance

10. Heat Changes - Heat Capacity If V constant, U increases as temperature increases The rate of change of U at any temperature, (dU/dT)V is called the heat capacity, CV CV(A) is not equal to Cv(B) generally smaller if TA < TB Note volume is const. If that changes CV(T may change) CV is an extensive property (2x the amt gives twice the heat capacity) Molar heat capacity, CVm, is the Rate of change of CV with T typically small around room temperature, you can assume it�s constant This means that dU = qV= CV dT � CV DT You can estimate CV by determining the amount of heat supplied to a sample Because qV � CV DT, for a given amount of heat the larger CV, the smaller DT At a phase transition (boiling point) DT = 0 so CV = 8

11. If PV work can be done by a system, at constant pressure, V changes Some of the internal energy lost through work U=q-pV or q = U+ pV dU < dq, if work is done; dU=dq, if no work is done Define new variable, H, such that H= U + pV H is called enthalpy Like U, H is a state function At constant p, dH = dU +pdV This is an important function because most experiments are carried out at constant pressure, e.g., atmospheric pressure Enthalpy is the heat supplied at constant pressure If pdV=0, dH=dq or DH=qP As with Cv, Cp is defined as dqp/dt = (dH/dT)p Heat Changes - Enthalpy

12. Measuring Enthalpy Changes Isobaric calorimeter measures heat changes at constant pressure Unlike bomb calorimeter in which volume constrained You can use a bomb calorimeter to measure enthalpy by converting DU to DH See Experiment 1 Assume molar enthalpy and molar internal energy are nearly the same True for most solids and liquids If process involves only solids and liquids DU � DH Calculate pV term based on density differences (Example 2.2) Differential scanning calorimeter (see Atkins, p. 46) Themogram: plot of Cp vs. temperature Peaks correspond to changes in enthalpy Area in the peak = DH Typically small samples (mg) and high temperatures reached

13. Typical DSC

14. Enthalpy Calculations Ideal Gas Since PV term for ideal gas is nRT, DH = DU +pV = DU +nRT If a reaction changes the amount of gas, PV term becomes (nfinal -ninitial)RT or DH = DU +DngRT, where Dng= nfinal -ninitial DH - DU = DngRT Note: sign Dng of is important Enthalpy changes can be directly related to heat input Example 2.3: Calculate molar internal energy change and enthalpy changes at 373.15 when 0.798 g of water is vaporized by passing 0.5A @ 12V through a resistance in contact with the water for 300 seconds

15. Change of Enthalpy with Temperature (Cp) The instantaneous slope of the plot of enthalpy vs temperature is the heat capacity at constant pressure Extensive property Cp,m, the molar heat capacity is an intensive property As with U and Cv, dH = Cvdt, or for measurable intervals DH=CvDT Also, qp = CvDT, because at constant pressure DH= qp This assumes Cv is constant over the temperature range True if range is small, esp. for noble gases More generally Cv = a + bT +(c/T2) a, b,and c are constants depending on the substance If Cv is not constant with T, DH ? CvDT Now what happens????

16. Variation of Cp with Temperature If Cp varies with temperature, you must integrate over the temperature limits of interest. Example: Assume Cv = a + bT +(c/T2) Try Self Test 2.5

17. Relationship between Cv and Cp(More about this in Chapter 3) In most cases Cp > Cv For ideal gases, Cp = Cv +nR This amounts to ~8 JK-1mol-1 difference

18. Adiabatic expansion of an ideal gas Work is done, U decreases, hence T decreases To calculate the final state (Tf and Vf) assume a two step process Remember U is a state function, it doesn�t matter how you get there Step 1: Volume is changed and temperature held constant Internal energy change = 0 since it�s independent of volume the molecules occupy Step 2: Temperature changed from Ti to Tf Adiabatic so q = 0 If Cv is independent of temperature, adiabatic work = wad= CvDT DU = q + wad = 0 + CvDT= CvDT This says that for an adiabatic expansion the work done is only proportional to the initial and final temperatures The relationship between initial and final volumes can be derived using what we�ve learned about reversible adiabatic expansions and perfect gases

19. Adiabatic Expansion of an Ideal GasRelationship between T and V Given dq = 0 (adiabatic) and dw = -pdV (reversible expansion) dU = dq + dw =-pdV [1] For perfect gas dU = CvdT [2] Thus, combining [1] and [2], CvdT = -pdV [3] For an ideal gas, pV = nRT, so [3] becomes CvdT = -(nRT/V)dV [4] or, rearranging, (Cv/T)dT = -(nR/V)dV [5] To obtain the relationship between Ti, Vi and Ti, Vi [5] must be integrated Ti corresponds to Vi Cv independent of T The adiabatic work, wad= CvDT, can thus be calculated once this relationship is found

20. Integration of (Cv/T)dT = -(nR/V)dV

21. Adiabatic Work, wad, and Temperature

22. pV Changes During Adiabatic Expansions This relationship means that the product, pVg doesn�t change as you go through an adiabatic expansion For ideal gas, g, the ratio of the heat capacities is >1 since Cp,m = Cv,m +R g = Cp,m /Cv,m = (Cv,m+R) /Cv,m g = 1 +(R /Cv,m) For monotomic gas, Cv,m= 3/2 R, so g = 5/2 For monotomic gas, Cv,m= 3 R, so g = 4/3 Plot of P vs V for an adiabatic change is called an adiabat p a Vg, where g>1 Recall isotherm, p a V Adiabats fall more steeply than isotherms

23. The First Law and Chemistry The study of heat produced from chemical reactions is called thermochemistry We can apply what we learned from our 1st law considerations to the system of reactants and products in a chemical reaction In a reaction, heat either flows into the system (endothermic reactions) or out of the system (exothermic reactions) At constant pressure (reactions done in a beaker for example), the first law tells us for Endothermic reactions DH>0 {heat flows in} Exothermic reactions DH<0 {heat flows out} Standard enthalpy change, DH�, is the change in enthalpy for substances in their standard state Standard state is the form of the substance at 1bar (~1atm) and specified temperature DH�298 is the standard enthalypy change at 1bar and 298K (~RT)

24. Enthalpies of Physical Change Changes in state (liquid to gas, solid to liquid, solid to gas, etc.) have enthalpies associated with the transition, DtransH� Liquid-to-Gas, vaporization: DvapH� Solid-to-Liquid, fusion: DfusH� Solid-to-Gas, sublimation: DsubH� Table 2.4 gives symbols for a variety of transitions Because transition temperature is an important point, transition temperatures are often listed at this point May like to know them at convenient temperature as well Recall that enthalpy is a state function. Thus, value is path independent. Can construct information if you don�t have it by making a path and summing the steps Example: You need the enthalpy of sublimation, but can�t find it Path: Step 1 - solid to liquid (DfusH�) Step 2 - liquid to gas (DvapH�) Step 3 - sum of steps1 & 2 Reverse of process involves only change of sign

25. Enthalpies of Chemical Change Like state transitions reactions have enthalpies associated with them Can write a thermochemical equation Standard chemical equation plus enthalpy change Reactants and products in standard states � enthalpy is standard enthalpy of reaction, DrH� Typically, you must calculate DrH� Again because enthapy is a state function you create a path using enthalpies you know or can find and sum them Usually a table of standard free energies of formation of components are available Given per mole of substance, DrH�m Recall that decomposition is the same as formation except sign is reversed For aA + bB � cC + dD, DrH�m = cDrH�m(C) + dDrH�m(C) - aDrH�m(A) - bDrH�m(D) Generally, Remember state is important! More generally, the standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which it may be divided - Hess�s Law Consequence of the fact that enthalpy is a state function Component reactions could be hypothetical

26. Standard Enthalpies of Formation , DfH� As above, the std enthalpy of formation refers to compound in standard state (1bar, typically 298K although other temperature could be specified) Enthalpies of reaction can be determined from enthalpies of formation as before

27. Temperature Dependence of DrH� Kirkoff�s Law To be accurate DfH� needs to be measured at temperature of interest If not available, can be estimated as follows: