Thermodynamica 1. college 5 boek hoofdstuk 3. Bendiks Jan Boersma Wiebren de Jong Thijs J.H. Vlugt Theo Woudstra. Process and Energy Department

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1 hermodynamia Bendiks Jan Boersma Wiebren de Jong hijs J.H. Vlugt heo Woudstra Proess and Energy Deartment February 9, 20 ollege 5 boek hoofdstuk 3

2 summary leture 4 roerties of matter -V- surfae isobar, isotherm, isohor hase transition oexisting hases working with tables, interolation S S + L S+G G February 9, 20 2 L L+G C G G aor L liquid S solid solid S S L ritial oint liquid L G aor/gas trile oint

3 equation of state of Ideal Gases he intensie roerties {,,} of matter (gas, liquid, solid) are not indeendent of one another. he relation between them is alled equation of state he simlest equation of state is the ideal gas equation of state It was found in stages, by realizing that V = C( n, ) if the Kelin-tem. sale is used, it was seen that V = C n ( ) the final equation of state is V = nr ressure [Pa] V olume [m 3 ] temerature [K ] n amount of substane [mol] uniersal gas onstant J R = mol K mol ontains N A = moleules February 9, 20 3

4 equation of state of Ideal Gases V = nr = R = V n R J = mol K ressure [Pa] V olume [m 3 ] temerature [K ] n amount of substane [mol] mol ontains N A = moleules February 9, 20 4

5 the iso -diagrams for ideal gases he relation between 2 of the 3 roerties {,,} an be isualized in state diagrams isotherm, =onst. isobar, =onst. isohor, =onst. February 9, 20 5

6 releane of the ideal gas model he ideal gas state is a limit for gases of low density Eery real omonent aroahes ideal gas behaior for ressures 0 bar and ρ 0 kg/m 3 Moleular interretation: a gas shows ideal gas behaior, when the moleular interations do not lay a role ( for low densities and/or high temerature) February 9, 20 6

7 dealing with real gases omressibility fator: Z = R ideal gas limit: lim Z = 0 redued ressure and redued temerature: R =, R = February 9, 20 7

8 February 9, 20 8

9 examles of equations of state Ideal gas equation of state Z = = R Virial equation of state (for non-ideal gasses) Z = = + B + C + D + R 2 3 ( ) ( ) ( )... Van der Waals equation of state (for non-ideal gasses) 2 na + 2 ( V nb) = nr V February 9, 20 9

10 the internal energy of an ideal gas exeriment of Joule and homson measurement: = 2 =!, V auum 2, V 2 ideal gas behaior first law: du = δq δw Δ 2U = Δ2Q Δ2W 0 0 Δ 2U = 0 U U = U(, V ) U(, V ) = u (, 2) u (, ) = 0 with 2 we onlude that u is only a funtion of temerature u() for an ideal gas February 9, 20 0

11 isothermal exansion of an ideal gas first law 2 isotherm, =onst. adiabati, ΔQ=0 du = δq δw 0 dv beause u() and =onst. V 2 2 V Δ 2Q =Δ 2W = dv = nr dv = nr ln V V V V ΔQ ΔW nr = ideal gas law V V note: an isothermal roess requires roiding energy in form of heat during exansion, otherwise the temerature would derease 2 February 9, 20

12 olytroi exansion n V = onstant (with n ) 2 V Δ W = dv = V 2 V V n 2 2 February 9, 20 2

13 Math: Differential relations of multiariable funtions onsider a differentiable funtion z(x,y) hink of z as the height oordinate in the Swiss Als, and x and y are the northerly diretion and the westerly diretion, resetiely 6 he total differential is z(x,y) dz = z x y dx + z y ( z/ x) y is the sloe with whih the height hanges with a hange dx in x-diretion. Subsrit y is for: in diretion of onstant y Note, that ( z/ x) y is a funtion of y, as seen in the illustration x dy z x 7 0 y x 3 z x 6 9 y S z y S8 y S5 x S22 February 9, 20 3

14 Seifi heat at onstant olume For an ideal gas: How does u ( ) hange with? Reall the first law: u (, ) = u ( ) du = δ q δw = δq d dq V = onstant +d At onstant olume: du = δ q Definition of heat aaity at onstant olume (here: in units of [J mol - K - ]) u : the amount of heat that needs to be added to inrease the temerature of mol material by Kelin, while keeing the olume onstant. Definition is alid also for nonideal gasses. February 9, 20 4

15 Seifi heat at onstant olume u dq V = onstant +d For an ideal gas, = ( ) Usually: V = a+ b Energy hange at onstant olume uon heating: u Δ = 2 = = 2 2 u u u d d February 9, 20 5

16 Energy hange uon heating at onstant V suose: V = a+ b 2 u 2 2 Δ u = u2 u = d = d = ( a+ b) d = 2 2 b 2 2 a + b = a ( ) ( ) 2 2 February 9, 20 6

17 he Enthaly H U + V h u+ dh = du + d ( ) = du + d + d February 9, 20 7

18 adding heat while keeing the ressure onstant Reall the first law: Enthaly: First law in terms of enthaly: du = δ q δw = δq d H U + V h u+ dq d = onstant +d dh = δ q + d Adding heat at onstant ressure dh = δ q Definition of heat aaity at onstant ressure: (here: in units of [J mol - K - ]) h February 9, 20 8

19 seifi heat at onstant ressure h dq d = onstant +d For an ideal gas, = ( ) Usually: = a+ b Enthaly hange at onstant ressure uon heating: h Δ = 2 = = 2 2 h h h d d February 9, 20 9

20 heating at onstant olume or ressure At onstant olume: du = δ q At onstant ressure: dh = δ q Definition of heat aaity at onstant olume (here: in units of [J mol - K - ]) Definition of heat aaity at onstant ressure (here: in units of [J mol - K - ]) u h : the amount of heat that needs to be added to inrease the temerature of mol material by Kelin, while keeing the olume onstant : the amount of heat that needs to be added to inrease the temerature of mol material by Kelin, while keeing the ressure onstant dq V = onstant +d dq = onstant d +d February 9, 20 20

21 Instrutions Chaters,2,3 hae been treated we disussed and illustrated the ideal gas law read thoroughly in book!!! memorize the definition of enthaly memorize the first law of losed systems in terms of the enthaly understand that u() and that h() for ideal gas understand the heat aaity and what does it tell you? make exerises onerning ideal gasses and non-ideal gasses; (omressibility fator Z, use ±5% deiation as a good aroximation to the ideal gas model). February 9, 20 2

22 relation between and for an ideal gas h= u+ = u+ R ( u R) h + = = = + R P P February 9, 20 22

23 adiabati exansion of an ideal gas ( = onstant) du = δ q d = d Rd d = d d = R d 2 2 ln = R 2 2 after some math... κ d = Rln = onstant (with κ = ) February 9, 20 23

24 Adiabati Exansion V κ = onstant κ = onstant κ = February 9, 20 24

25 Clement & Desormes Exeriment () O druk gebrahte fles is toestand ; na isohore exansie is toestand 2. Na de adiabatish eronderstelde snelle exansie juist oor het sluiten an de kraan is de druk 0 (omgeingsdruk) = 0 + ρgh () 0 = - ρgh (a) a 2 = 0 + ρgh 2 (2) 2 = ( - ρgh ) + ρgh 2 = -(h -h 2 )ρg (2a) Na o druk brengen an de fles olgt adiabatishe exansie: V = V γ γ 2 2 (3) Gas na o druk brengen en na de isohore exansie bereikt uiteindelijk dezelfde V = 2 V 2 (4) V 2 (5) = V2 Met (3) en V 2 =V 0 olgt dan: γ V (6) 0 = V2 γ February 9, γ

26 Clement & Desormes Exeriment (2) 0 ln γ 2 0 Uit (5) en (6) olgt: = (7) γ = (7a) 2 ln Verdere ereenoudiging mogelijk als niet teeel an 0 afwijkt. ( ( ) ) - h-h 2 ρg ρgh = (7) en (a) (8) ( ) h-h 2 ρg ρgh = Als (h -h 2 )ρg/ << dan mag (9), aangezien geldt oor een funtie (+x) α +αx met x<< (olgt uit f(x) = f(0+x) f(0)+x f (0) ), benaderd worden met γ γ (9) ( ) h-h γ ρg ρgh 2 = γ h = h h2 (0) February 9, 20 26