3. THE SECOND AND THIRD LAWS OF THERMODYNAMICS

Transcription

1 3. HE ECOND AND HIRD LAW OF HERMODYNAMIC n he Carnot Cycle 3.1. a. Efficiency = h c = h 1000 b. Heat rejected = c. Entropy increase = d. Entropy decrease = e. = f. = 150 J K 1 ; H = 0 = 30 kj = 150 J K 1 = 150 J K 1 = 0.8 = 80% G = H = = J = 150 kj 3.. Efficiency = w q h = = = 7010 J must be withdrawn.

2 HE ECOND AND HIRD LAW OF HERMODYNAMIC n a. C B D A b. q h + q c h = 0 = q h c + h q c 300 K = 0 Work performed by system, w = q h + q c = 10 kj h = q h h = 100 J K 1 q 100 J K 1 c K = 0 q c = J = 30 kj q h = 40 kj h = K = 400 K 3.4. a. Efficiency = ( h c )/ h = 0.5 = 5% b. Heat absorbed at 400 K = 800/0.5 = 300 J = 3. kj c. Heat rejected at 300 K = = 400 J =.4 kj d. Entropy change for A B = 300 J/400 K = 8.0 J K 1 e. Zero f. H = 0 and thus G = = = 300 J = 3. kj g. he efficiency is 5%, so to produce 000 J of work, 000 J 0.5 = 8000 J = 8.0 kj must be absorbed he heat transferred from the water to the ice in melting it is

3 50 n CHAER 3 (10 1 g/18 g mol 1 ) 6.05 kj mol 1 = kj he fraction of this that can be converted into work is (Eq. 3.1) h c 0 = h = he available work is therefore = kj One day is = s. he power, or the rate of doing work, is thus / = kj s 1 = J s 1 = W = 65 MW 3.6. A thermodynamic equation of state (Eq. 3.18) is U = +. From the ideal gas law, we have = R/ m ; therefore, (/) = R/ m. ubstituting above, and recognizing that R/ m =, we obtain U R = + m = 0. n Entropy Changes 3.7. C 6 H 6 : CHCl 3 : H O: C H 5 OH: /353 = 87.3 J K 1 mol /334 = 88.0 J K 1 mol /373 = J K 1 mol 1 } Hydrogen-bonded structure in liquids /351 = J K 1 mol 1 H O and C H 5 OH a. he reaction is C(graphite) + H (g) + 1 O (g) CH 3 OH(l) f H = 16.8 [ ( ) + 1 (05.14) ] = 4.9 J K 1 mol 1 b. he reaction is C(graphite) + H (g) + 1 O (g) + N (g) H NCONH (s) f H = [ ( ) + 1 (05.14) ] = J K 1 mol 1

4 HE ECOND AND HIRD LAW OF HERMODYNAMIC n a. r = [(19.77) (130.68)] J K 1 mol 1 = J K 1 mol 1. b. r = (40.1) 304. J K 1 mol 1 = J K 1 mol First, calculate the value of r at K. hen, using this value, calculate the increase in entropy for the increase in temperature. Assume C is constant throughout the range. For the dissociation H (g) H(g), r (98.15 K) = [( ) ] J K 1 mol 1 = J K 1 mol 1. he value of r at different temperatures may be calculated from r () = r ( K) rc d, where r C is νic i, i for the reaction. If r C is independent of temperature, the quantity r C can be taken out of the integral. Using the C values at K, we get r ( K) = [(0.784) 8.84] = ln(173.15/98.15) = J K 1 mol 1. A more accurate value could be obtained if values of or C were available at K a. m = b. m = C,m d = R ln 98 = 3.5 J K 1 mol 1 C,m d = R ln 98 =.11 J K 1 mol For N and O, the volume increases by a factor of.5, increase of entropy for each = R ln.5 = 7.6 J K 1 For H, volume increases by a factor of 5: = ln 5 = 6.69 J K By Eq. 3.63, otal = = 1.9 J K 1 d

5 5 n CHAER 3 = ln ln 3 = = 5.99 J K he mole fractions are x N = 0.79 x O = 0.0 x Ar = 0.01 he entropy change per mole of mixture is thus, by Eq. 3.65, = (0.79 ln ln ln 0.01) = ( ) = 4.61 J K 1 mol he equation for the formation of ethanol from its elements is C(graphite) + 3H (g) + 1 O (g) C H 5 OH(l) he standard molar entropy of formation is thus f m = product Σ reactants = [( 5.74) + ( ) + 1 (05.14) ] = J K 1 mol a. gas = R ln 1 = ln10.0 = 19.1 J K 1 mol 1 surr = 19.1 J K 1 mol 1 b. For an adiabatic process, Eq..90 applies: h γ 1 = c γ 1 where γ = 5 3 γ = c 10 c = 64. K Because gas must be calculated from a reversible process, gas = 19.1 J K 1 mol 1 ; surr = 0 since q = 0 Net = 19.1 J K 1 mol a. aking into account the entropy change for heating liquid water and the entropy change during vaporization, we obtain C m, (l) (373.0 K) /(J K mol ) = d = 75.48ln + = b. We recognize that changing the pressure on the surface of a liquid dies not affect the entropy. herefore, we take into account the entropy changes for vaporization, heating the vapor at

6 HE ECOND AND HIRD LAW OF HERMODYNAMIC n 53 constant pressure and, finally, compressing the vapor to the final pressure at constant temperature (see Eq. 3.51). his yields Cm, (g) (373.0 K) /(J K mol ) = + d ln ubstituting C,m = and replacing / 1 in the last term by 1 / = /1.00, we get (373.0 K) /(J K mol ) = ln d = ln (373 7) ln 1.00 = hus the entropy change for the system depends only on the initial and final states and not the path taken to go from one to the another a. ositive (increase in number of molecules) b. ositive (decrease in electrostriction) c. Negative (increase in electrostriction) d. ositive (decrease in electrostriction) Heat absorbed when temperature rises = dq = C d Corresponding entropy change: = C d Entropy increase when the temperature rises from 1 to : C 1 d = n C,m 1 d If the gas is ideal, C is constant and = C ln 1 = nc,m ln By Eq the entropy change is = ln ln 10 5 = = 4.4 J K Let the value of the final Celsius temperature be u. hen

7 54 n CHAER (100 u ) = ( u 0) u = u 0 u = = 5.0; = 5.0 C a. mercury = d = ln = 6.8 J K 1 b. water + vessel = d = ln = 7.10 J K 1 c. Net = = 0.8 J K Heat required to melt 0 g of ice is = 6689 J Heat required to heat 0 g of water from 0 C to C is ( ) J Heat required to cool 70 g of water from 30 C to C is (30 ) J Heat balance equation: = (30 ) = 5.57 C Reversible processes: a. Cool 70 g of water to 0 C: system = ln = 30.5 J K 1 surr = 30.5 J K 1 b. Melt 0 g of ice at 0 C: system = = 4.49 J K surr = 4.49 J K 1 c. Heat 90 g of water to 5.57 C:

8 HE ECOND AND HIRD LAW OF HERMODYNAMIC n 55 system = ln surr = 7.60 J K 1 = 7.60 J K 1 Net system = 1.57 J K 1 ; surr = 1.57 J K m = 1000 C,m 300 d m /J K 1 mol 1 = /K = 8.58 ln 1000 = = m = 36.8 J K 1 mol (/K) 3 d(/k) ( ) In an isothermal reversible expansion E = 0 = q + w and q rev = If the process were reversible, the work done by the system would be w rev = = = J = 15 kj ince the actual work was less than this, the process is irreversible. Degree of irreversibility = 6 15 = It is necessary to devise a process in which the freezing occurs reversibly: tep 1: Heat the supercooled water reversibly from 3 C to 0 C: 1 = 75.3 ln = 0.83 J K 1 mol 1 tep : Freeze the water at 0 C: = =.04 J K 1 mol 1 tep 3: Cool the ice reversibly from 0 C to 3 C: 3 = 37.7 ln = 0.4 J K 1 mol 1 he net entropy change in the system is therefore = = 1.63 J K 1 mol 1 o calculate the entropy change in the environment, we calculate the heat that has been gained by the environment in the three steps: tep 1: = 5.9 J mol 1 tep : 600 J mol 1 tep 3: = J mol 1

9 56 n CHAER 3 he net heat gained by the environment is thus = J mol 1 his heat was gained by the environment at 3 C, and the entropy change is therefore = 1.87 J K 1 mol 1 he net entropy change in the system and environment is thus = 0.4 J K 1 mol he solution contains 0.1 mol, and the volume ratio is 1000/00 = 5. hen, by Eq. 3.51, n = = 0.1 = ln 5 = J K 1 = 1.34 J K mol of substance A is present and the volume increases by a factor of 4: (A) = ln 4 = J K mol of B is present and the volume increases by a factor of 4/3: (B) = ln (4/3) = J K 1 Net = 1.51 J K he final temperature is 40 C. he water at 60 C can be cooled reversibly to 40 C, and the water at 0 C can be heated reversibly to 40 C. 1 = ln = 46.6 J K 1 = ln = J K 1 he net entropy change is thus = 3.08 J K For O : = ln = 8.8 J K 1 he entropy change for the expansion of the N is the same, and the net entropy change is thus 57.6 J K he process can be imagined as occurring by the following reversible processes: tep 1: he water freezes to ice at 0 C: = =.04 J K 1

10 HE ECOND AND HIRD LAW OF HERMODYNAMIC n 57 tep : he ice is cooled reversibly to 1 C: 3 = 37.7 ln = 1.69 J K 1 Net (system) = = 3.73 J K 1 he heat gained by the freezer = = J his was gained at 1 C, and therefore (surroundings) = = 4.78 J K 1 he net entropy change is (system) + (surroundings) = 1.05 J K For the freezing of 1 mol of water at 0 C: = For the reversible cooling of the ice from 0 C to 10 C: = 37.7 ln = 1.41 J K 1 Net (system) = = 3.45 J K 1 =.04 J K 1 he heat gained by the freezer = ( ) = 6397 J his was gained at 10 C, so that the entropy change in the freezer is = 4.31 J K 1 he net entropy change is 0.86 J K he final temperature is calculated in terms of the heat balance: 75.3 (60 ) = ( 0) = C = K = 75.3 ln = = 0.8 J K ln here are obviously several reversible paths that can be constructed between the initial and final states in this case. Let us consider four of them. 1. Isothermal expansion to the final volume followed by constant volume cooling to the final temperature. 3 1 m, f i 5 1 m, f i U = 0 + C ( ) = R( ) = J mol H = 0 + C ( ) = R( ) = 931. J mol f = Rln + C ln = Rln + Rln = J K mol i f m, i

11 58 n CHAER 3. Isothermal expansion to the final pressure followed by constant pressure cooling to the final temperature. 3 1 m, f i 5 1 m, f i U = 0 + C ( ) = R( ) = J mol H = 0 + C ( ) = R( ) = 931. J mol i = Rln + C ln = Rln + Rln = J K mol f 3. Isothermal expansion to ( 0, 0 ) followed by adiabatic expansion to the final state. f m, i Note that we need to find the intersection of the isotherm that passes through the initial state and the adiabat that passes through the final state. his intersection is ( 0, 0 ), at i. Using the relationships for adiabatic processs (Eq..90), 0 i (3/) = f f (3/) ; herefore, 3/ = dm = 8.44 dm U = 0 + C ( ) = R( ) = J mol 3 1 m, f i 5 1 m, f i H = 0 + C ( ) = R( ) = 931. J mol Rln Rln J K mol 0 = + = = i Constant ressure heating to the final volume followed by constant volume cooling to the final pressure. he gas will have to be heated to 0 = K in order for it to reach the volume of dm 3 at 10.0 bar pressure. herefore, 3 1 U = C ( ) + C ( ) = R( ) = J mol m, 0 i m, f m, ( 0 i) m, ( f 0) ( ) 931. J mol H = C + C = R + = 0 f = Cm, ln + C, mln = Rln i + R ln = J K mol Yet another path we can try is constant volume cooling to the final pressure followed by constant pressure heating to the final temperature. In each of these cases, we have verified that U, H and are the same, thus proving that they are independent of the path taken, as any state property should be. We now have to find the entropy change of the surroundings. Entropy change of the surroundings. he actual process is the expansion of the gas against a constant external pressure of bar. For this process, according to the first law, U = q act ext ( f i ); herefore, q act = U + ext ( f i ) = ( ) (8.3145/ )

12 HE ECOND AND HIRD LAW OF HERMODYNAMIC n 59 = J mol 1. surr = q act / surr = J mol 1 /98 K = 3.56 J K 1 mol 1 univ = + surr = J K 1 mol 1. his is, therefore, a spontaneous process (system) = ln = J K 1 he heat accepted by the refrigerator is (50 3) = J (refrigerator) = /76.15 = J K 1 Net = = 4.90 J K a. In this case all of the ice melts and the final temperature is 1 C. (ee the solution to roblem.3.) he entropy changes are 1. Reversible melting of the ice at 0 C: 1 = 605 J mol 1 (100/18) mol/73.15 K = 1.5 J K 1. Reversible heating of 100 g of water from 0 C to 1 C: = 75.3 J K 1 mol 1 (100/18) mol ln (85.15/73.15) = 18.0 J K 1 3. Reversible cooling of 1 kg of water from 0 C to 1 C: 3 = (1000/18) mol 75.3 J K 1 mol 1 ln (85.15/93.15) = J K 1 he net entropy change is therefore = = 4.8 J K 1 b. In this case only 50 g of the ice melts, and the final temperature of the water is 0 C. (ee the solution to roblem.3.) he entropy changes are now 1. For the reversible melting of 50 g of ice at 0 C: 1 = (50/18) mol 605 J mol 1 /73.15 K = J K 1. For the cooling of 1 kg of water from 0 C to 0 C: = (1000/18) mol 75.3 J K 1 mol 1 ln (73.15/93.15) = 95.6 J K 1 he net entropy change is = = 10.8 J K 1.

13 60 n CHAER Using the expression for C,m given in able.1, d / (800.0 K) /(J K mol ) = = 17.3 J K 1 mol 1 n Gibbs and Helmholtz Energies From Eq. 3.91, G = ( ) + ( ) ( ) = kj mol 1 H = ( 88.3) + ( ) ( ) = kj mol 1 = H G = = 459. J K 1 mol Work done by system = = dm 3 atm mol 1 = 3059 J mol 1 H = J mol 1 U G = 0 = = H () = = J mol = J K 1 mol From Example 3.6, H and are available. a. At 0 C, H = 600 J mol 1 =.04 J K 1 mol 1 G = = 0 b. At 10 C, H = 5644 J mol U = 0 = 0.64 J K 1 mol 1 G = = 13 J mol 1

14 HE ECOND AND HIRD LAW OF HERMODYNAMIC n 61 H = U + () = 0 = q rev = 1 1 d = R ln 1 = ln 10 = J K 1 mol 1 A = U = = kj mol 1 G = kj mol 1 he quantities are all state functions, and the preceding values therefore do not depend on how the process is carried out a. (a) G = ( ) J mol 1 = kj mol 1 (b) G = ( ) J mol 1 = 16.9 kj mol 1 (c) G = ( ) J mol 1 = kj mol 1 b. At = 85 00/170. = K 3.4. From Eq. (3.168), we write o o cg 1 cg cg 1 ch, = 1 1 Where is the mid-point of the temperature range ( 1, ). In the limit 0, this will yield Eq. (3.168). ubstituting, we get ch = = herefore, c H = (33.0) = kj mol Conversion of water to vapor at atm: H = kj mol 1 = = J K 1 mol 1 Reversible isothermal expansion from atm to 10 5 atm: H = 0 = ln = 66.9 J K 1 mol 1 Net H = kj mol 1

15 6 n CHAER 3 = = 14.5 J K 1 mol 1 G = = J mol 1 = kj mol a. U, H e. None b. f. G c. H g. U d. H h. None G = herefore G = d he molar volume of mercury is m = 00.6 g mol g cm 3 = m 3 mol 1 hen G m = m bar 10 5 a bar 1 = 1485 J mol 1 = kj mol 1 G m = m = (A + B ln ) where A = J K 1 mol 1 B = 0.79 J K 1 mol 1 G m = (A + B ln ) d = [A + B( ln )] = (A B) 50 B (33.15 ln 33.15) ln 98.15) = J mol 1 = 4.8 kj mol he entropy at K is calculated as ( K) /(J K mol ) = d d d = J K - mol -1

16 HE ECOND AND HIRD LAW OF HERMODYNAMIC n U and H are zero for the isothermal expansion of an ideal gas. q = w = 500 J mol 1 = R ln = 5.76 J K 1 mol 1 G = = 178 J mol 1 = 1.73 kj mol 1 w rev = kj mol 1 ; q rev = 1.73 kj mol ince ext = 0 (evacuated vessel), no work is done; w = 0. U = q + w = q = 30 kj mol 1. For H O(l) H O(g), n = 1 (see Example.5) H = U + () = U + nr = = J mol 1 = 33.1 kj mol 1 o obtain the entropy change, consider the reversible processes: (1) H O (l, 100 C) H O (g, 100 C, 1 atm) 1 = = J K 1 () H O (g, 100 C, 1 atm) H O (g, 100 C, 0.5 atm) = R ln 1 = R ln = 5.76 J K 1 mol 1 Net = J K 1 mol 1 Net G = H = = 9658 J mol 1 = 9.66 kj mol Allow the process to occur by the following reversible steps: (1) H O (g, 100 C, atm) H O (g, 100 C, 1 atm) H 1 = 0 1 = R ln / 1 = R ln = 5.76 J K 1 mol 1 G 1 = 1 = 150 J mol 1 () H O (g, 100 C, 1 atm) H O (l, 100 C, 1 atm) H = J mol 1 = = J K 1 mol 1 G = 0 (reversible process at constants and )

17 64 n CHAER 3 (3) H O (l, 100 C, 1 atm) H O (l, 100 C, atm) he H,, and G changes are negligible for this process. he overall changes are thus H = J mol 1 = 40.6 kj mol 1 = = 103 J K 1 mol 1 G =.15 kj mol U = 1 C,m d C,m = C,m R = ( /K) J K 1 mol 1 q = 0; U m = 300 = C,m d = dw = d 1 ( / K) d R 300R 1 1 d = = ( 300) ( 300 ) = = 0 = 8.58 ± ( )1/ = = 55.3 K U m = 0.7 ( ) ( ) = J mol 1 H m = 8.58 ( ) ( ) = J mol 1 d = dq du + d = For 1 mol of ideal gas, m = R d( m ) = Rd = d m + m d d m = Rd m d = Rd R d d m = du m + Rd Rd = C,m d R d

18 HE ECOND AND HIRD LAW OF HERMODYNAMIC n 65 m /J K 1 mol 1 = = 8.58 ln m =. J K 1 mol 1 ( ) d ln H = ( 85.85) = kj mol 1 G = ( 37.13) = kj mol ( ) =. = H G ( ) 1000 = = 43. J K 1 mol a. and b. rue only for an ideal gas. c. rue only if the process is reversible. d. rue only for the total entropy. e. rue only for an isothermal process occurring at constant pressure First vaporize the water at 1 atm pressure: 1 = = J K 1 hen expand from 1 bar to 0.1 bar: = R ln 10 = 19.1 J K 1 he net entropy change is = 1 + = = 17.9 J K 1 he Gibbs energy change is G = H = ( ) = J = 7.13 kj H = = kj mol 1 G = = 76.5 kj mol 1 = H G = = 80.0 J K 1 mol 1

19 66 n CHAER 3 n Energy Conversion atm engine: h = 585 K; c = 303 K Efficiency = atm engine: h = 45; = 48.% c = 303 K Efficiency = = 8.7% w q = c 69 = w = = 890 J min 1 = 14.8 J s 1 At 40% efficiency, power = 14.8/0.4 = 37.0 J s 1 = 37.0 W h erformance factor = h c a. 59.6%; b. 11.9%; c. 6.6% Efficiency = = g 1 liter = 8.0 kg = 114. g mol 1 = mol Energy produced = mol 5500 kj mol 1 = kj Work = = kj Let q h be the heat supplied to the building at 0 C and q c be the heat taken in by the heat pump at 10 C: q h q = c Work supplied to heat pump: w = q h q c = q h = q h Let q h be the heat produced by the fuel at 1000 C and q c be the heat rejected at 0 C. Work performed by the heat engine and supplied to the heat pump is

20 HE ECOND AND HIRD LAW OF HERMODYNAMIC n 67 hus w = q h q c = q h q h = q h and the performance factor is q h = q =.6 h = q h he heat that must be removed from 1 kg (= 55.5 mol) of water in order to freeze it is = 334 kj = q c his is the heat gained by the refrigerator. a. If the efficiency were 100%, h = c = q q h c = q h 334 kj hus, the heat discharged at 5 C is q h = 365 kj Work required to be supplied to the refrigerator is = 31 kj With an efficiency of 40%, the actual work will be = 78 kj b. he heat discharged at 5 C will be = 41 kj. n hermodynamic Relationships 3.6. a. Using the given relationship and the definitions of α and κ, we have = ( / ) ( / ) α =. κ ubstituting into Eq. (3.18), we obtain U α α κ = + =. κ κ b. Using the chain rule of partial differentiation, we obtain U U = α κ = ( κ ) = ( κ α ). κ

21 68 n CHAER 3 H = +. ince =, we get 1 ince α = H = H, we get ( α) = a. m = R/. herefore, ( m / ) = R/, and the cubic expansion coefficient is 1 m α = = R/( m ) = 1/. m b. ince m = R/, ( m / ) = R/. herefore, 1 m κ = = R /( m ) = 1/, m since (R/ m ) = From the first and second laws du = d d herefore U = = using the Maxwell equation, Eq From the van der Waals equation = m b = 1 + a m U = a m µ H = ( H / ) ( H / ) ince dh = d + d, 1 H = C

22 HE ECOND AND HIRD LAW OF HERMODYNAMIC n 69 H = + = + from the Maxwell equation, Eq hus µ = C = m m m C, for 1 mol he equation ( m b) = R applies to 1 mol of gas, and it follows that = R herefore µ = R m C,m a. G = (Eq ) herefore G = d = q rev = C d = C C = G b. C = H = H H = + (see roblem 3.66)

23 70 n CHAER 3 H = + C = A = U da = du d d At constant temperature da = du d = dq + dw d But dq = d, and therefore da = dw du = d d (Eq ) Dividing by d at constant : U = = 0 But = (Eq. 3.14) and therefore = Integrating, ln = ln + const, or. hus = const By Euler s reciprocity theorem (Appendix C), U = U U ) / ( ) / ( U = (Eq ) U = (Eq ) U = For an ideal gas, U depends only on so that U =

24 HE ECOND AND HIRD LAW OF HERMODYNAMIC n 71 For an isothermal process involving n mol of an ideal gas d = nrd ln = nrd = d hus, (/) = / and therefore (/) U = / a. Eq. (3.159) gives the following relationship for the fugacity of a gas: f R ln Dividing both sides by R and using the definition of the compression factor Z, we get f ln Which is the desired result. b. Dividing both sides of the given equation of state by R, we get Z = 1 + [b A/(R /3 )](/R) = 1 + [b/(r) A/( R 5/3 )], which means (Z 1)/ = b/(r) A/( R 5/3 ). Now, setting 1 = 0 and integrating, we obtain 3.7. Using the expression for fugacity derived in roblem (3.71), we get which yields 1 = = 1 f ln = 0 m m d = Z 1 d, R Z 1 d = herefore, the fugacity of the gas f = exp( ) = 68 bar. b R 1 A R R. d. f Z 1 1 a b ln = d = b d + d 0 R R 0 R 0 ln f 1 = b R a R 1 + b R 5/ 3 =