Chapter 2 Heat, Work, Internal Energy, Enthalpy and the First Law of Thermodynamics. Physical Chemistry 2 nd Edition Thomas Engel, Philip Reid

Transcription

1 Chapter 2 Heat, Work, Internal Energy, Enthalpy and the Frst Law o Thermodynamcs Thomas Engel, Phlp Red

2 Objectves Introduce nternal energy, U. Introduce the rst law o thermodynamcs. Dscuss the concepts o the heat capacty, the derence between state and path unctons, and reversble versus rreversble processes. Introduce enthalpy, H, as a orm o energy. Show how U, H, q, and w can be calculated or processes nvolvng deal gases.

3 Chapter Outlne 1. The Internal Energy and the Frst Law o Thermodynamcs 2. Work 3. Heat 4. Heat Capacty 5. State Functons and Path Functons 6. Equlbrum, Change, and Reversblty 7. Comparng Work or Reversble and Irreversble Processes

4 Chapter Outlne 8. Determnng U and Introducng Enthalpy, a New State Functon 9. Calculatng q, w, U, and H or Processes Involvng Ideal Gases 10.The Reversble Adabatc Expanson and Compresson o an Ideal Gas

5 2.1 The Internal Energy and the Frst Law o Thermodynamcs Internal energy s total energy or the system o nterest. Frst law o thermodynamcs states that energy can be nether created nor destroyed, both system and the surroundngs are taken nto account. The nternal energy, U, o an solated system s constant. U = system U surroundn gs

6 2.1 The Internal Energy and the Frst Law o Thermodynamcs For any decrease o U system, U surroundngs must ncrease by exactly the same amount. Ths derve a second and useul ormulaton o the rst law: U = q + w where q = heat w = work

7 2.2 Work Work s dened as any quantty o energy that lows across the boundary between the system and surroundngs When compresson work s beng done on a gas, the work done n movng a mass s gven by w = = = F d P P l external external dv dadl

8 2.2 Work From electrostatcs, the work or a constant current, I, that lows or a tme, t, s w electrcal = Iφt

9 2.2 Work Table 2.1 shows the expressons or work or our derent cases.

10 Example 2.1 a. Calculate the work nvolved n expandng 20.0 L o an deal gas to a nal volume o 85.0 L aganst a constant external pressure o 2.50 bar. b. A water bubble s expanded rom a radus o 1.00 cm to a radus o 3.25 cm. The surace tenson o water s N m -1. How much work s done n the process? c. A current o 3.20 A s passed through a heatng col or 30.0 s. The electrcal potental across the resstor s 14.5 V. Calculate the work done on the col.

11 Soluton a. w = P external dv = P external 5 10 Pa = 2.50bar bar ( V V ) ( 85.0L 20.0L) = 16.3kJ b. A actor o 2 s ncluded as a bubble has an nner and an outer surace: 10 3 L m 3 w = γdσ = 2γ 4π = 8π ( 2 2 r r ) 4 10 m 2 cm ( ) = 1.73J 2

12 Soluton c. = φ w φdq = φq = I t = 14.5V 3.20 A 30.0s = 1. 39kJ

13 2.3 Heat Heat s the quantty o energy that lows across the boundary between the system and surroundngs because o a temperature derence

14 Example 2.2 A heatng col s mmersed n a 100-g sample o H 2 O lqud at 100 C n an open nsulated beaker on a laboratory bench at 1 bar pressure. In ths process, 10% o the lqud s converted to the gaseous orm at a pressure o 1 bar. A current o 2.00 A lows through the heater rom a V battery or s to eect the transormaton. The denstes o lqud and gaseous water under these condtons are 997 and kg m -3, respectvely.

15 Example 2.2 a. It s oten useul to replace a real process by a model that exhbts the mportant eatures o the process. Desgn a model system and surroundngs that would allow you to measure the heat and work assocated wth ths transormaton. For the model system, dene the system and surroundngs as well as the boundary between them.

16 Example 2.2 b. Calculate q and w or the process. c. How can you dene the system or the open nsulated beaker on the laboratory bench such that the work s properly descrbed?

17 Soluton a. The model system s shown n the ollowng gure. The cylnder walls and the pston orm adabatc walls. The external pressure s held constant by a sutable weght.

18 Soluton b. The heat nput to the lqud water can be equated wth the work done on the heatng col. Thereore, As the lqud s vaporzed, the volume o the system ncreases at a constant external pressure. Thereore, the work done by the system on the surroundngs s w = P q = Iφt = = 24. 0kJ ( V V ) = 10 Pa + 1. kj external =

19 Soluton c. Dene the system as the lqud n the beaker and the volume contanng only molecules o H 2 O n the gas phase. Ths volume wll consst o dsconnected volume elements dspersed n the ar above the laboratory bench.

20 2.4 Heat Capacty The heat capacty s a materal-dependent property dened by the relaton C = lm T 0 T q T = dq dt where C s the SI unt o J K -1 Most common are constant volume or constant pressure, C V and C P, respectvely.

21 2.4 Heat Capacty At constant pressure, heat low between the system and surroundngs can be wrtten as q p = T sys, T sys, C system P T surr, surroundngs ( T ) dt = C ( T )dt T surr, P The molar heat capacty, C m, s used n calculatons wth the unts o J K -1 mol -1

22 Example 2.3 The volume o a system consstng o an deal gas decreases at constant pressure. As a result, the temperature o a 1.50-kg water bath n the surroundngs ncreases by 14.2 C. Calculate q P or the system. surr, surroundngs Soluton: q p = CP ( T ) T T surr, = 1.50kg 4.18Jg dt 1 K = C 1 surroundngs P 14.2K T = 89.1kJ

23 2.4 Heat Capacty For an deal gas: CP CV = nr or CP m CV, m =, R const. p ex Not all heat low nto the system can be used to ncrease U n a constant pressure process, because the system does work on the surroundngs as t expands.

24 2.5 State Functons and Path Functons U s ndependent o the path between the ntal and nal states and depends only on the ntal and nal states. For any state uncton, U must satsy the equaton, U = For cyclc path such that the ntal and nal states are dentcal, du = U U du = U U = 0

25 2.5 State Functons and Path Functons As the heght o the mass n the surroundngs s lower ater the compresson, w s postve and U s ncreased.

26 2.5 State Functons and Path Functons As both q and w are path unctons, there are no exact derentals or work and heat. q dq q q and w w w

27 2.6 Equlbrum, Change, and Reversblty A quas-statc process s where rate o change o the macroscopc varables s neglgbly small rom the ntal to nal state When an nntesmal opposng change n the varable that drves the process n reversal drecton, the process s reversble.

28 2.6 Equlbrum, Change, and Reversblty I an nntesmal change n the drvng varable does not change the drecton o the process, the process s rreversble.

29 2.7 Comparng Work or Reversble and Irreversble Processes Indcator dagram shows the relatonshp between P and V or the process graphcally. Total work n yellow area. Expanson work n red area. Compresson work n total red and yellow area

30 2.7 Comparng Work or Reversble and Irreversble Processes When the process s reversed and the compresson work s calculated, the ollowng result s obtaned: w compresson V = nrt ln 1 V 2

31 Example 2.4 In ths example, 2.00 mol o an deal gas undergoes sothermal expanson along three derent paths: (1) reversble expanson rom P = 25.0 bar and V =4.50 L to = 4.50 bar; (2) a sngle-step expanson aganst a constant external pressure o 4.50 bar, and (3) a two-step expanson consstng ntally o an expanson aganst a constant external pressure o 11.0 bar untl P=P external, ollowed by an expanson aganst a constant external pressure o 4.50 bar untl P=P external. Calculate the work or each o these processes. For whch o the rreversble processes s the magntude o the work greater?

32 Soluton We rst calculate the constant temperature at whch the process s carred out, the nal volume, and the ntermedate volume n the two-step expanson: T V V PV = = = 677K 2 nr nrt = = = P 4.50 nt = nrt P nt = L = 10.2L

33 Soluton The work o the reversble process s gven by w V 25.0 = nrt1 ln = ln = V J For rreversble processes: w w sn gle two step = P = P external V external V = = = ( ) ( ) ( ) J = J

34 2.8 Determnng U and Introducng Enthalpy, a New State Functon When process under constant volume condtons and that non-expanson work s not possble. U = q v U+PV s a state uncton and s called enthalpy, H. H U + PV At constant pressure, H = q p

35 2.9 Calculatng q, w, U,, and H or Processes Involvng Ideal Gases Change o nternal energy or a temperature range over whch C V s constant, U = q v = C V ( T T ) 2.1 Ideal Gas under Constant Pressure or Volume

36 2.9 Calculatng q, w, U,, and H or Processes Involvng Ideal Gases 2.2 Reversble Isothermal Compresson o an Ideal Gas 2.3 Reversble Isobarc Compresson and Expanson o an Ideal Gas 2.4 Isochorc Heatng and Coolng o an Ideal Gas 2.5 Reversble Cyclc Processes

37 Example 2.5 A system contanng 2.50 mol o an deal gas or whch C V,m =20.79 J mol 1 K 1 s taken through the cycle n the ollowng dagram n the drecton ndcated by the arrows. The curved path corresponds to PV=nRT, where T=T 1 =T 3.

38 Example 2.5 a. Calculate q, w, U and H or each segment and or the cycle. b. Calculate q, w, U and H or each segment and or the cycle n whch the drecton o each process s reversed.

39 Soluton Segment 1 2 U 1 2 = nc V, m nc ( ) V, m T T = ( PV PV ) = = 99.6kL 2 1 nr 2 2 ( )

40 Soluton Segment 1 2 The process takes place at constant pressure, so ( V V ) = ( ) = 39. kj w = Pexternal 8 Usng the rst law 2 1 q = U w = = kJ We next calculate T 2, and then U 1 2 : PV T2 = = = nr K

41 Soluton Segment 1 2 We next calculate T 3 = T 1, PV T1 = = = 79. 9K nr H 1 2 = U = ( PV ) = U1 2 + nr( T2 T1 ) ( ) = 139.4kJ +

42 Soluton Segment 2 3 As noted above, w=0, and U 2 3 = q 2 3 = C = 99.6kJ V = ( T3 T2 ) ( ) The numercal result s equal n magntude, but opposte n sgn to U 1 2 because T 3 =T 1. For the same reason, H = H

43 Soluton Segment 3 1 For ths segment, U3 1 = 0 and H3 1 = 0 as noted earler and w3 1 = q3 1. Because ths s a reversble sothermal compresson, 3 V w3 1 = nrt ln = ln = 5. 35kJ 3 V

44 2.10 The Reversble Adabatc Expanson and Compresson o an Ideal Gas I C V s constant over the temperature nterval T - T, then C U V As C p -C v =nr or an deal gas, ln T T = w, C v dt V = nr ln V = P ex dv C V T ln T = ( γ 1) V ln V or, equvalently, T T = V V 1 γ where γ = C p, m CV, m

45 2.10 The Reversble Adabatc Expanson and Compresson o an Ideal Gas Substtutng T T = P V / P V n the prevous equaton, we obtan γ P V = P V γ Reversble adabatc compresson o a gas leads to heatng, and reversble adabatc expanson leads to coolng.

46 2.10 The Reversble Adabatc Expanson and Compresson o an Ideal Gas 2.6 Reversble Adabatc Heatng and Coolng o an Ideal Gas

47 Example 2.7 A cloud mass movng across the ocean at an alttude o 2000 m encounters a coastal mountan range. As t rses to a heght o 3500 m to pass over the mountans, t undergoes an adabatc expanson. The pressure at 2000 and 3500 m s and atm, respectvely. I the ntal temperature o the cloud mass s 288 K, what s the cloud temperature as t passes over the mountans? Assume that C P,m or ar s J K 1 mol 1 and that ar obeys the deal gas law. I you are on the mountan, should you expect ran or snow?

48 Soluton Soluton As the process s adabatc, q =0, and ( ) ( ) ( ) ( ) ( ) K T T atm atm P P R C C R C C P P P P T T P P T T V V T T m P m P m P m P ln ln 1 ln 1 ln 1 ln 1 ln 1 ln 1 ln,,,, = = = = = = = = = γ γ γ γ γ γ U = w snow