U = 4.18 J if we heat 1.0 g of water through 1 C. U = 4.18 J if we cool 1.0 g of water through 1 C.

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Lionel Manning
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1 CHAPER LECURE NOES he First Law of hermodynamics: he simplest statement of the First Law is as follows: U = q + w. Here U is the internal energy of the system, q is the heat and w is the work. CONVENIONS (he System Centered Point of View): Heat (q): In numerical calculations, when the numerical value of q turns out to be positive, heat is absorbed by the system, if the value turns out to be negative, heat is given out by the system. If we supply q as input to a problem, the sign of q has to be determined based on whether the heat is absorbed by the system or is lost from the system. Example: It takes 4.18 J of heat to heat 1.0 g of water through 1 C. If no work is done, according to the first Law, Work (w): U = 4.18 J if we heat 1.0 g of water through 1 C. U = 4.18 J if we cool 1.0 g of water through 1 C. In numerical calculations, when the numerical value of w turns out to be positive, work is done on the system, if the value turns out to be negative, work is done by the system. Example: See Fig... It takes 5.0 J of energy to push a 5.0 kg piston through 1.0 m. If no heat is involved, according to the first Law, w = 5.0 J if the piston is pushed in (by us) through 1.0 m. w = 5.0 J if the piston is pushed out (by the gas) through 1.0 m. herefore, the process determines whether heat is absorbed or given out, and whether work is done on the system or by the system. Examples.1,., Problem.7.

2 State Functions and Path-Dependent Functions: A thermodynamic property that depends only on state of the system and not on how it got there is called a state function. he internal energy, U, is a state function. herefore, a change in the internal energy, U, can be calculated simply by knowing the initial and final values of U: U = U f U i. On the other hand, q and w are path-dependent functions. We generally denote state functions with upper case variables and path-dependent functions with lower-case variables. Reversible and Irreversible processes: A reversible process is one in which the system and surroundings are continuously in equilibrium. his implies that the process takes place infinitesimally slowly so that there is plenty of time at each stage for equilibration with the surroundings. Reversible processes are NO cyclic processes, i.e., the system is not required to return to the initial state at the end of the process. Fig..4(a) Fig..4(b) w rev = V1 V PdV w irr = P (V V 1 )

3 3 First Law: du = dq + dw. Constant Volume (Isochoric) Processes: At constant volume, dv = 0; therefore, du = dq V, where the subscript implies that volume is being held constant. he heat required to heat an object (system) through 1 C is called the heat capacity C of the object (system). herefore, du = dq V = C V d, or C V = dq V d = U. V Constant Pressure (Isobaric) Processes: From the first law, dq P = du dw = du + PdV. herefore, U q P = V U1 du + V1 PdV =(U U 1 )+P(V V 1 ) his yields q P = (U + PV ) (U 1 + PV 1 ) and suggests the definition of a new state function, enthalpy, defined as H = U + PV. It follows that dq P = dh and, therefore, dh = dq P = C P d or C P = dq P d = H P. he molar heat capacities are denoted as C V,m and C P,m, respectively. For an ideal gas, dh m = du m + d(pv m ) = du m + d(r). herefore, H m or C P,m = C V,m + R. = U m P + R, V Here we have exploited the fact that H and U are both state-functions. A more rigorous derivation is given on p. 83 (see Eq..119). Problems:.3,.4,.5,.6,.8,.9,.10 Assignment: Review basic thermochemistry and Hesse s Law from Gen. Chem. Example.3,.4, Problems.14,.16,.18,.1,.3.6,.30,.31.

4 4 Measurement of Enthalpy Changes: Calorimetry Basic principle: (a) heat lost = heat gained. Constant pressure calorimetry: coffee-cup calorimeter: his simple device directly gives the enthalpy change since the process studied takes place at constant pressure. Example: heat of neutralization (Gen. Chem.): the neutralization reaction generates heat which is lost to the surrounding water, thermometer, stirrer etc., and results in an increase in their temperature. If the heat capacity of the calorimeter (water + thermometer + stirrer, etc.), C (J C 1 ), is known, then heat lost = C (t f t i ). herefore, the heat generated by the neutralization can be calculated and the molar enthalpy of neutralization determined. (b) Constant Volume Calorimetry: the bomb calorimeter: his device yields the U for the process from which H has to be calculated, as H = U + (PV) = U + (nr) = U + nr, where n = no. of moles of gaseous products no. of moles of gaseous reactants for the reaction of one mole of the substance of interest. For instance, C H 6 O(l) + 3O (g) CO (g) + 3 H O(l); n = 3 = 1. C H 6 (g) + (7/)O (g) CO (g) + 3 H O(l); n = 4.5 =.5. Problem.55 Standard State: By general agreement, the standard state of a substance is defined as the most stable form of that substance at K and 1 bar pressure. herefore, oxygen is a gas in its standard state and water is a liquid. Standard state of a solution is defined as 1 mol kg 1 or a 1 m solution. Calorimetry: Example.5, Problems.5,.10,.15,.17,.3,.35

5 5 emperature dependence of Enthalpies Since dh = C P d, the temperature dependence of enthalpy is related to the temperature dependence of C P, which is commonly expressed as C P () = d + e + f, where the constants d, e and f are experimentally determined for a large number of gases (See able.1, p. 69). hen,,h = 1 d + e + f d = d( 1 ) + e 1 f Problems.19,.,.46. emperature Dependence of Enthalpies of Reaction Enthalpy changes during reaction are normally calculated under standard conditions using tables of standard enthalpies of formation. Consider the reaction n A A + n B B n C C + n D D he standard enthalpy of reaction is calculated as:, r H =(n C, f H C + n D, f H D ) (n A, f H A + n B, f H B ). If the temperature dependence of the C P,m s of each reactant and product are also given in terms of the parameters d, e and f, we can define,d =(n C d C + n D d D ) (n A d A + n B d B ), and similarly, e and f. hen, to evaluate the enthalpy of reaction at a certain temperature θ, we can simply evaluate, r H D =, r H D K (,d +,e +,f )d More generally, if we know the enthalpy of reaction at a temperature 1, we can calculate the enthalpy at another temperature. See Eq. (.51) and (.5). Example.6.

6 6 Reversible Processes Involving Ideal Gases (Section.6) (a) Reversible isobaric expansion/compression: w rev = V1 V PdV = P(V V 1 ). Fig..6 (a) q rev = 1 C P d = C P ( 1 ),,U = q rev + w rev, (first law!) and,h = q rev = q P. if C P is constant. (b) Reversible isochoric expansion/compression: w rev = 0! q rev = 1 C V d = C V ( 1 ) if C V is constant.,u = q rev = q V.,H = 1 C P d = C P ( 1 ), by definition! Fig..6 (b) (c) Reversible isothermal expansion/compression: w rev = V1 V PdV = nr V1 V dvv = nr ln V V = nr ln P 1 P 1 Fig..6 (c),u =,H = 0! since = 1. from first law. q rev = w rev,

7 7 (d) Reversible adiabatic expansion/compression: In adiabatic process, by definition, q = 0. herefore, in an adiabatic expansion, the internal energy decreases (i.e., the temperature decreases) and in an adiabatic compression, internal energy increases (i.e., the temperature increases).. Fig..6 (d),u = C V ( 1 ), q rev = 0, and therefore, w rev =,U = C V ( 1 ),H = C P ( 1 ). An isotherm is characterized by the equation PV = constant, or P 1 V 1 = P V. We now try to derive an equation that characterizes an adiabat. From First Law, we have du = PdV, or C V,m d = R for ideal gas. V dv, Rearranging, we get C d V,m which, when integrated, yields = R dv V, C V,m ln = R ln V 1 1 V his immediately gives the result (.86) V 1 or using the fact that C P,m C V,m = R, = R/CV,m 1 V, where γ = CP,m/CV,m. (.90) = (CP,m CV,m)/CV,m? 1 V 1 1 V = V 1 V, We now eliminate temperature by using ideal gas law to get P V P 1 V =? 1 V 1 P 1 V,or P =? V 1 1 V, (.9) which is equivalent to P 1 V 1 γ = P V γ or PV γ = constant. Reading Assignment: Bond Enthalpies (p. 73)

8 8 Reversible Isothermal Expansion of a Real Gas We will focus on the van der Waals equation. P = V nr nb + n a V. herefore, V w rev = V V1 PdV = nr V1 V nb + n a V dv, which, upon integration, yields w rev = nr ln From Chapter 3, we use the relationship du = n a V dv, and thus get V,U = V1 Now, using the first law, V nb V 1 nb n a 1 V 1 V 1 n a V dv = n a 1 V 1 V 1 q rev =,U w rev = nr ln V nb V 1 nb. (.13) (.14) (.16) Also, since initial and final pressures and volumes are known (or can be found),,h =,U +(P V P 1 V 1 ) In order to calculate U, H etc. for isochoric and isobaric processes, we need to be given the values of C V,m and C P,m for the gases. It is not easy to derive the equation for an adiabat for a real gas Example.10, Problems.37.45,.49.51,.54,.56,.60,.63,.64

Enthalpy and Adiabatic Changes

Enthalpy and Adiabatic Changes Chapter 2 of Atkins: The First Law: Concepts Sections 2.5-2.6 of Atkins (7th & 8th editions) Enthalpy Definition of Enthalpy Measurement of Enthalpy Variation of Enthalpy