Nhiet Dong Hopc

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TRNG I HC LT KHOA VT L

Bin son: LNG DUYN PHU

Bi ging tm tt

NHIT NG HCDng cho sinh vin ngnh vt l

(Lu hnh ni b) LT - 2008

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MC LC Chng I. CC KHI NIM C BN CA NHIT NG HC 1.1. i tng v phng php ca nhit ng hc 1. i tng ca nhit ng hc 2. Phng php nghin cu nhit ng hc 1.2. Chuyn ng nhit 1. Chuyn ng nhit 2. Trng thi nhit 1.3. Nhit 1.4. Cc tham s trng thi 1. Th tch v p sut 2. Cc tham s trng thi 3. Phng trnh trng thi 1.5. Cng v nhit 1. Cng 2. Nhit 3. Nhn xt chung v cng v nhit 1.6. Kilomol 1.7. Nguyn l th nht ca nhit ng hc 1. Ni nng 2. Nguyn l th nht nhit ng hc Chng II. KH L TNG 2.1. Kh l tng 2.2. p sut ca kh l tng 2.3. Phng trnh trng thi ca kh l tng 2.4. Phn b phn t theo vn tc 2.5. Phn b phn t theo cao trong trng trng lc 2.6. Ni nng ca kh l tng 2.7. Cng v nhit trong cc qu trnh ca kh l tng 1. Qu trnh ng tch 2. Qu trnh ng p 3. Qu trnh ng nhit 4. Qu trnh on nhit 2.8. Qung ng t do trung bnh Chng III. KH THC 3.1. Kh thc 3.2. Phng trnh trng thi ca kh thc 3.3. Kim tra thc nghim 3.4. Ni nng ca kh thc Chng IV. CHT LNG 4.1. Cht lng 4.2. Cc hin tng b mt ca cht lng 1. Ni p sut 2. Sc cng mt ngoi 3 5 5 5 5 6 6 6 6 7 7 8 8 8 8 9 10 10 11 11 11 12 12 12 14 15 17 17 19 19 19 20 20 21 22 22 22 25 26 28 28 28 28 29

3. Nng lng mt ngoi 4. Gii thch mt vi hin tng mt ngoi 4.3. Hin tng dnh t 4.4. Hin tng mao dn 1. p sut ph di mt khum 2. Hin tng mao dn Chng V. CHIU HNG CA QU TRNH NHIT 5.1. Qu trnh thun nghch v khng thun nghch 5.2. Nguyn l th hai ca nhit ng hc 5.3. Entropy 5.4. My nhit 1. My nhit 2. Chu trnh Carnot 5.5. Cc hm th nhit ng lc Chng VI. CN BNG PHA V CHUYN PHA 6.1. Cc pha ca h v m 6.2. Cn bng pha 1. Cn bng hai pha 2. Cn bng ba pha 3. Cn bng nhiu pha 6.3. Chuyn pha 1. Chuyn pha loi mt 2. Chuyn pha loi hai Chng VII. CC QU TRNH KHNG CN BNG 7.1. Qu trnh khng cn bng 7.2. Khuch tn 5.3. Ni ma st 7.4. Truyn nhit 7.5. Nhit ng hc xa cn bng TI LIU THAM KHO

30 30 31 32 32 33 34 34 35 36 38 38 39 41 43 43 43 43 44 44 45 45 46 47 47 48 49 49 50 53

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Chng I

CC KHI NIM C BN CA NHIT NG HC

1.1. I TNG V PHNG PHP CA NHIT NG HC 1. i tng ca nhit ng hc Vt cht quanh ta c cu to t cc phn t, bn thn phn t c cu to t mt hay nhiu nguyn t. Kch thc ca cc phn t nm trong khong t 10 nm xung n 0,1 nm. Cc ht vt cht c kch thc t khong 10 nm tr xung c gi chung l cc ht vi m. Cc h vt cht quanh ta m chng ta c th cm nhn c trc tip bng gic quan gi l cc h v m. Cc h ny bao gm mt s rt ln cc phn t. Th d, trong iu kin bnh thng, 1 cm3 khng kh cha khong 2,4.1019 phn t. Cc kiu h vt cht thng thy l cht kh, cht lng, cht rn. T nhng nm 40 ca th k 20, vt l cn nghin cu mt kiu h vt cht mi l plasma. Plasma l khi vt cht nhit rt cao, hng ngn 0C tr ln, l hn hp cc ion dng ca cc nguyn t v cc electron. Sau y l th d v cc h vt cht kiu khc, trong cc h ny cc ht thnh phn khng phi l cc phn t: - Cc electron trong mt khi kim loi hoc dng cc electron trong chn khng, - Cc photon trong mt bnh cha kn, thnh bnh khng hp th m ch phn x, - Cc neutron trong ngi sao neutron... tin pht biu, sau ny ta s thng gi cc ht thnh phn l phn t, song cc lp lun vn ng cho cc h vt cht kiu khc trong ht thnh phn khng phi l phn t. Khi xt ring cho cc h vt cht kiu khc th ht thnh phn s c ni r. Mn hc chng ta nghin cu y c tn l Nhit ng hc, hay Vt l nhit, cng cn gi l Nhit hc. i tng ca nhit ng hc l cc h v m, tc l cc h vt cht c cha mt s ln cc ht thnh phn. Cc h v m cng c gi l vt th hay vt. Cc h ny c kho st trong iu kin c chuyn ng nhit nn cn gi l cc h nhit. Sau ny khi ni v h vt l m khng ni c th, ta s hiu ngm nh l h nhit. Mc ch ca nhit ng hc nh vy l nghin cu cc tnh cht ca cc h nhit. 2. Phng php nghin cu nhit ng hc C hai phng php nghin cu h nhit: - Phng php nhit ng: m t cc tnh cht v m ca h v xc nh cc tnh cht y; nu tnh cht c biu th bng i lng vt l th nu cch o chng. - Phng php thng k: t chuyn ng v tnh cht ca cc ht vi m thnh phn, tng hp thng k rt ra cc tnh cht v m. Hai phng php ny b sung cho nhau.

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1.2. CHUYN NG NHIT 1. Chuyn ng nhit Tin c bn nghin cu cc h nhit l quan im sau y: Cc phn t trong h chuyn ng khng ngng. Trong qu trnh chuyn ng nh th chng truyn nng lng cho nhau thng qua tng tc. Hnh nh n gin nht ca tng tc l va chm. Do c mt mt s ln cc phn t v lun xy ra va chm nn chuyn ng phn t tr nn hn lan. Chuyn ng khng ngng v hn lon nh th c tn l chuyn ng nhit. Chuyn ng nhit nh hng hu nh n tt c cc tnh cht v m ca h. Do cc phn t c vai tr bnh ng nhau trong mt h v m v do chuyn ng nhit nn nu khng c tc ng t ngoi th h s cn bng nhit. c trng ca cn bng nhit l cc i lng vt l phn b ng u trong ton h: cc ht phn b ng u, nng lng phn b ng u, ... 2. Trng thi nhit Mt h c gi l mt trng thi xc nh khi cc tnh cht ca h l xc nh. Ni ring, nu tnh cht xt c biu th bng i lng vt l th i lng vt l y c gi tr xc nh trong trng thi xc nh nu. V chuyn ng nhit gi vai tr trung tm trong trng thi ca h nhit nn trng thi ca h nhit cn gi l trng thi nhit. i lng vt l c trng tnh cht ca h cn gi l tham s trng thi hay tham s nhit. Qu trnh nhit l tp hp cc trng thi nhit k tip nhau. Nu qu trnh l cn bng th thng phi chm ti mi thi im, trng thi kp thit lp s cn bng. Trong mn hc ny, cc chng I IV v VI s ch kho st cc qu trnh cn bng. Chng V s xt cc qu trnh cn bng ln khng cn bng. Chng VII dnh ring cho cc qu trnh khng cn bng.

1.3. NHIT i lng vt l c ngha trung tm trong vt l nhit l nhit . Nhit l i lng biu th mc nng lnh ca vt th. Khi nim nng lnh y phi hiu mt cch khch quan, khng chi phi bi cm gic ch quan ca con ngi, mc d n xut pht thc s t cm gic nng lnh. Nhit c k hiu bng ch t hoc T. Bn cht ca mc nng lnh chnh l mc chuyn ng nhit. Trong c hc ta bit i lng biu th mc chuyn ng l ng nng. Nh vy mc chuyn ng nhit chnh l ng nng chuyn ng nhit ca cc phn t, hiu theo ngha gi tr trung bnh. Ta k hiu l ng nng tnh tin trung bnh ca mt phn t trong h: = m0 v 2 / 2 (m0 l khi lng mt phn t, v 2 l trung bnh ca bnh phng vn tc phn t). Mt tnh cht c bn ca s nng lnh l lm dn n cc vt th. Phn tch chng t rng a s cc cht lng v cht kh dn n th tch theo nhit mt cch tuyn tnh. C th li dng tnh cht ny ch to nhit k, l dng c o nhit . Nhit k thng dng l nhit k Celcius: cht dn n l nc, ru hoc thy ngn, tt nht l thy ngn. Cht lng Hg c cho vo mt ng rt ht kh, gn kn, qui c nhit nc ang tan l 0 Celcius (00C), ang si l 1000C. Thang t 00C n 1000C c chia u lm 100 khong, mi khong ng vi 10C. Sau c th chia thang ngoi suy di 00C v trn 1000C. Thang o nh th c gi l 6

thang nhit Celcius. Ngy ny c nhiu loi nhit k o c nhng nhit rt thp (n - 2730C) v rt cao (n hng ngn 0C). Khi o nhit ta phi cho nhit k tip xc vi vt th (h ang xt). nng lnh s truyn t vt sang nhit k hoc ngc li cho n khi cn bng. Lc cn bng cng l lc ta c c s o nhit . Mt thang nhit khc thng s dng l thang Kelvin, khong chia thc hin nh thang Celcius nhng gc tnh khc. Nhit trong thang Kelvin c n v l kelvin, vit tt l K. K hiu t l nhit Celcius, T l nhit Kelvin th lin h gia hai thang nh sau: T (K) = t (0C) + 273,15.0

(3.1)

ngha quan trng ca thang Kelvin l ch khi T = 0 K th t = -273,15 C: y l nhit ng vi cc phn t ng yn, khng cn chuyn ng nhit, l iu khng th t ti. Vo nm 1992, vt l to c nhit thp k lc: Tmin = 2.10-9 K. Mt Tri l mt thin th c nhit cao: nhit trn b mt l 104 K, nhit trong lng l 107 K. Theo phn tch ca ngnh v tr hc, nhit ca v tr ti thi im st sau V N ln l 1039 K. mt s nc cn dng mt thang nhit c tn l Fahrenheit. Mt h cn bng th trc ht th hin ch T = const trn ton h. Ta cng c th dng trc tip ng nng tnh tin trung bnh ca phn t lm s o nhit . Thang o nh th gi l thang nng lng, n v l joule (J). Thang nng lng v thang Kelvin lin h vi nhau bng mt h s hng s

=B

3 kB T , 2

(3.2)

trong kB = 1,38.10-23 J/K gi l hng s Boltzmann. H s 3/2 trong cng thc (3.2) chn cho tin v sau.

1.4. CC THAM S TRNG THI

1. Th tch v p sut Th tch V ca h l mt tham s trng thi, biu th khong khng gian m h chim. p sut trong mt h l lc tng cng tc dng ln mt n v din tch b mt

p =

F (S l mnh din tch b mt ca h). S

(4.1)

H cn bng th p sut phi ng u trong ton h, tr trng hp tc dng ngoi nh hng ln s phn b, chng hn khi t khi kh trong trng hp dn. Trong h o SI n v ca p sut l newton/mt vung (N/m2), 1 N/m2 cn c tn l 1 pascal (1 Pa). Ngoi ra cn thng dng mt s n v khc sau: - atmosphere k thut, k hiu at: 1 at = trng lng ca 1 kg nn ln 1 cm2 = 98 066 Pa 98 100 Pa.

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- atmosphere vt l, k hiu atm: 1 atm = p sut khng kh trn mt t 00C = 101 325 Pa = 1,033 at. - milimet thy ngn, k hiu mm Hg: 1 mm Hg = p sut ng vi lm dng ct thy ngn ln cao 1mm = 133,32 Pa. Theo thang ny, p sut khng kh trn mt t l 760 mm Hg. 2. Cc tham s trng thi Cc i lng nhit T, th tch V v p sut p nu trn l nhng tham s trng thi, v chng l nhng i lng c trng tnh cht ca h nhit v c gi tr xc nh khi trng thi l xc nh. C th nu thm mt s tham s trng thi khc: s ht N (s phn t), th ha (l nng lng thm vo h khi h tng thm mt ht), entropy ( hn lon trong h). Nhng i lng ny s ni sau. Khi xt trng in t trong mi trng vt cht v c xt n chuyn ng nhit th cng trng in E v cm ng t B cng l nhng tham s trng thi. Cc tham s trng thi c phn lm hai loi: - Loi qung tnh, gm cc tham s c ph thuc khong khng gian m h chim, nh th tch V, s ht N,... Cc i lng ny c s dng nh nhau trong h cn bng cng nh khng cn bng. - Loi cng tnh, khng ph thuc vo khong khng gian h chim m c xc nh ti tng im trong h, nh nhit T, p sut p, ... Cc i lng cng tnh trong h cn bng th nh nhau ti mi im, cn trong h khng cn bng th c th khc nhau t im ny qua im khc. 3. Phng trnh trng thi Cc tham s trng thi c th ph thuc vo nhau. H thc lin h gia cc tham s trng thi khi chng c ph thuc vo nhau c gi l phng trnh trng thi. Th d, mt khi kh thng thng c c trng bi ba tham s trng thi l V, p v T. Trong chng ch c hai l c lp nn c mt phng trnh trng thi, vit tng qut nh sau:

f (p, V, T) = 0.

(4.2)

Tm phng trnh trng thi l mt trong nhng nhim v ch yu ca nhit ng hc.

1.5. CNG V NHIT

1. Cng Trong c hc, s truyn nng lng c thc hin bng cng. Trong nhit hc, s truyn nng lng phc tp hn do lin quan n chuyn ng ca nhiu ht thnh phn.

Cng l nng lng truyn to nn dch chuyn c hng ca cc phn t.Hy xt mt th d v dn nn mt khi kh trong mt bnh tr c pitng. Khi t ln pitng mt lc F, ta nn khi kh vo mt khong dx. Cng thc hin l

A = - Fdx = - pSdx = - pdV,8

trong S l din tch mt pitng, cng l tit din ca bnh, dV l bin i ca th tch khi kh (khi nn vo th dx < 0 tc dV < 0, lm cng thc xut hin du tr).

A = - pdV

(5.1)

Khi nn nh th tt c cc phn t u dch chuyn theo cng mt hng (Hnh 1.1a). Ta qui c du ca cng A nh sau: cng m h nhn vo l dng, cng do h sinh ra (ln vt khc) l m. Trong cng thc (5.1) khi nn khi kh th n nhn cng A > 0, khi dn th n sinh cng ra ngoi, dV > 0, nn A < 0. Cng trong mt qu trnh hu hn l(2)

A =

(1)

p dV

(5.2)

trong tch phn ly t trng thi (1) n trng thi (2). Cng thc (5.1) v (5.2) ng cho mi h.

Hnh 1.1a Cng y S mt an dx 2. Nhit

Hnh 1.1b Nhit truyn qua S

Trong cc h nhit cn mt hnh thc truyn nng lng na l nhit. Nhit (hay lng nhit) l nng lng truyn ca chuyn ng nhit v lm thay i mc chuyn ng hn lon ca cc phn t (Hnh 1.1b). Gi th Q l nhit h nhn trong mt qu trnh v cng b no . Nhit ny nu lm tng nhit ca h mt lng dT, th Q ~ dT. Ngoi ra Q phi t l vi khi lng m ca h: Q ~ m. K hiu h s t l l c, ta c:

Q = mcdT.

(5.3)

H s t l c l nhit dung ca vt: n bng nhit lng cn thit cho 1 kg ca vt lm tng nhit ln l K. Du ca nhit lng cng qui c nh du ca cng: h nhn nhit th Q > 0, khi h truyn nhit cho vt khc th i vi n Q > 0. Nhit trong mt qu trnh hu hn l(2)

Q = m

(1)

c dT .

(5.4)

n v ca nhit l n v ca nng lng: J. Ngoi ra cn dng mt n v khc khng thuc h SI: calo (vit tt: cal), 1 cal = 4,186 J. Nh vy n v ca nhit dung s l J/kgK hoc cal/kgK. 9

C nhng trng hp nhit cung cp cho h nhng khng lm thay i nhit ca h. l khi cht rn tan chy hoc khi cht lng ha hi. lm tan chy 1 kg cht rn cn mt nhit lng xc nh F, gi l nhit tan chy. Trong sut qu trnh tan chy nhit lng c cung cp nhng nhit khng thay i, cht rn chuyn dn thnh cht lng. Tng t lm ha hi 1 kg cht lng cn mt nhit lng V, gi l nhit ha hi. Trong qu trnh ha hi nhit cng khng thay i. n v ca F v V l J/kg hoc cal/kg. Nh vy nhit lng cn thit lm tan chy hoc lm ha hi m kg ca mt cht l

QF = mF, QV = mV .3. Nhn xt chung v cng v nhit

(5.5)

Cc i lng cng A v nhit Q khng phi l cc tham s trng thi: chng khng c gi tr xc nh trong trng thi xc nh. Nhng i lng nh th thng l cc s gia trong mt qu trnh. Chng khng ch ph thuc vo cc trng thi u v cui ca qu trnh m cn ph thuc cch thc din bin ca qu trnh trung gian, nh s thy trong cc chng sau.

1.6. KILOMOL

Xt khi vt cht c cu to t mt loi phn t, phn t c khi lng m0 v c phn t lng l . 1 mol cht c nh ngha l lng cht c khi lng g, v 1 kmol cht l kg ca cht . Ta c th tnh s phn t trong 1 kmol, k hiu l NA. Ta bit rng 1 n v khi lng nguyn t l 1,66.10-27 kg, nn khi lng mt phn t l m0 = 1,66.10-27 kg,

NA = khi lng 1 kmol / khi lng 1 phn t =

=

1, 66.1027

= 6,023.1026 phn t/kmol.

(6.1)

S NA c tn l s Avogadro. S ny nh nhau vi mi cht. Nu h c khi lng m th s kmol ca cht l = m/. K hiu C l nhit dung kmol, tc l nhit lng lm tng 1 kmol cht ln 1 K th

C = c hay c = C/.Cc cng thc v nhit lng trong 1.5 nu vit theo nhit dung kmol s c dng sau:

Q =

m

C dT ,

Q =

m

(2)

(1)

C dT .

(6.2)

Tng t, nu k hiu F l nhit tan chy tnh theo kmol, V l nhit ha hi tnh theo kmol th F = F v QV = V nn cc cng thc (5.4) tr thnh

QF =

m

F , QV =

m

V .10

(6.3)

1.7. NGUYN L TH NHT CA NHIT NG HC

1. Ni nng Ni nng ca mt h nhit l tng nng lng ni ti ca cc ht trong h. ng nng chuyn ng ca ton h cng nh th nng ca h i vi trng lc ngoi khng c tnh trong ni nng. Nh vy ni nng s gm tng ng nng ca cc ht (vi gi thit khi tm ca h l ng yn) v th nng tng tc gia cc ht trong h:

U = E + Et =

a

(a)

+

1 wab 2 a b

(7.1)

( trong a ) l ng nng ca ht th a v wab l th nng tng tc gia hai ht a v b.

S hng ng nng c th vit li nh sau:

E =

a

(a)

= N

1 N

a

(a)

= N ,

(7.2)

y N l s ht trong h v l ng nng trung bnh ca phn t. 2. Nguyn l th nht nhit ng hc Theo 1.5 trong cc h nhit c hai hnh thc truyn nng lng: cng v nhit. S kt hp nh lut bo ton nng lng ca vt l vi cc hnh thc truyn nng lng ca h nhit dn ti nguyn l th nht (nguyn l I) nhit ng hc, pht biu nh sau: Bin thin ni nng ca h nhit trong mt qu trnh bng tng cng v nhit m h nhn trong qu trnh . Cng thc biu th nguyn l I l

U = A + Q,hay dng qu trnh v cng b

(7.3)

dU = A + Q = p dV +

m

C dT .

(7.4)

Trong cc cng thc (7.3) v (7.4) cn phi tnh n nhit tan chy v nhit ha hi nu qu trnh din bin qua cc trng thi . ng dng ca nguyn l I s cp n trong cc chng sau.

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Chng II

KH L TNG

2.1. KH L TNG

Chng ta s p dng cc khi nim trnh by trong chng I xt cc h nhit c th. H nhit n gin l cht kh v trong s ny n gin hn c l cht kh l tng. Cht kh l tng l cht kh tun theo cc gi thit sau: 1/- Cc phn t khng c kch thc, l nhng cht im. 2/- Cc phn t khng tng tc vi nhau, tr lc va chm trc tip. 3/- Va chm cc phn t l va chm n hi, tc l khng thay i tng nng lng. L thuyt m t cht kh bng cc gi thit trn gi l thuyt ng hc phn t cc cht kh. Chng ta bit rng ng knh ca phn t n gin thng c d = 0,3 0,4 nm. iu kin bnh thng trong khng kh khong cch gia hai phn t vo khong L = 3,5 nm. Nh vy kch thc d nh ng k so vi khong cch L. Mt khc, cng trong khng kh, th nng tng tc tnh trung bnh cho mt phn t rt nh so vi ng nng tnh tin trung bnh ca n. Tng qut li, c th ni rng cht kh quanh ta c xem l cht kh l tng khi p sut khng qu ln v nhit khng qu nh. p sut ln qu s lm cho khong cch L gim xung, khng cn ln so vi d, hn na cc phn t li gn nhau th tng tc khng cn yu. Nhit h thp qu s lm cho ng nng phn t khng cn ln so vi th nng. Cht kh photon l mt th d v mt cht kh l tng c trng: cc ht photon khng c kch thc (t ra l trong phm vi nng lng t ti hin nay) v gia chng gn nh khng c tng tc. Trong chng ny ta xt cc tnh cht ca cht kh l tng.

2.2. P SUT CA KH L TNG

Cht kh ng trong mt bnh cha s gy mt p sut ln thnh bnh. Ta hy tnh p sut ny. Trc ht hy gi thit cc phn t chuyn ng nhit u c cng mt vn tc v , p vung gc ln din tch S ca thnh bnh (Hnh 2.1). Do va chm l n hi nn mi phn t khi p ln thnh bnh s truyn cho n mt xung lng

P1 = m0v - (- m0v) = 2m0v.S phn t p ln din tch S trong khong thi gian t l

N = n0Svt.

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Trn thc t do cc phn t chuyn ng nhit hn lon trn ba phng c lp v mi phng c hai hng ngc nhau nn s phn t p ln S ch bng 1/6 con s trn

N =

1 n0 v S t . 6

T tnh c xung lng cc phn t truyn cho S

P = N P = 1

1 n0 m0 v 2 S t 3

By gi ta phi sa li gi thit cho rng cc phn t c cng vn tc. V cc phn t chuyn ng nhit vi vn tc khc nhau nn trong cng thc va thit lp phi thay v2 bng v 2 . Ta c

Hnh 2.1

P =

1 n0 m0 v 2 S t 3

Lc tc dng tng ng vi xung lng truyn ny l

F =

P 1 = n0 m0 v 2 S 3 t

T tnh c p sutp =

F 1 = n0 m0 v 2 . S 3

(2.1)

V = m0 v 2 / 2 nnp =

2 n0 . 3

(2.2)

H thc (2.2) c tn l phng trnh c bn ca thuyt ng hc phn t cc cht kh, cn gi l phng trnh Clapeiron-Mendeleev. N gii thch ngun gc ca p sut gy bi mt khi kh: chuyn ng nhit ca cc phn t. Phng trnh (2.2) kt hp vi h thc (I.3.2) chop =

2 3 n0 k BT = n0 k BT . 3 2p . k BT

T n0 = (2.3)

Cng thc ny biu th mt phn t ca cht kh l tng qua p sut v nhit . V mt phn t n0 l cng c nn t (2.3) thy rng cng mt nhit , p sut cng c tnh cht cng c.

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2.3. PHNG TRNH TRNG THI CA KH L TNG

Vi cht kh cc tham s trng thi in hnh l th tch V, p sut p v nhit T. Gia chng c mt mi lin h, tc l mt phng trnh trng thi. Ta c th thit lp phng trnh trng thi ny t cc kt qu 2.2 trn. Mt phn t bng n0 = N m NA = . V V m N k T. V A B

Thay vo biu thc ca p sut biu th qua nhit p = n0 k BT = Tip theo t R = kBNA = 8,31.103 J/kmolK,B

hng s ny c gi l hng s kh l tng, ta c: pV = m

RT .

(3.1)

y l phng trnh trng thi ca cht kh l tng. Nu k hiu = m/ l s kmol kh th phng trnh (3.1) vit li nh sau: pV = RT. pV = RT trong V l th tch ca 1 kmol kh. T phng trnh trng thi ca kh l tng c th rt ra mt s h qu: 1/- Trong qu trnh ng nhit (T = const) ta c pV = const, tc l p sut bin i t l nghch vi th tch (nh lut Boyle-Mariotte). 2/- Trong qu trnh ng p (p = const) ta c V = const.T, tc l th tch bin i t l thun vi nhit (nh lut Charles). 3/- Trong qu trnh ng tch (V = const) ta c p = const.T, tc l p sut bin i t l thun vi nhit (nh lut Gay-Lussac). 4/- T (3.3) c V = RT/p: biu thc cho thy cng mt nhit v p sut, th tch 1 kmol l nh nhau vi mi cht kh, khng ph thuc vo bn cht ca cht kh c th. Th d iu kin tiu chun (00C - 1,033 at) V = 22,4 m3/kmol hay 22,4 l/mol. (3.2) (3.3) Ni ring, khi xt cho 1 kmol kh ( = 1) th phng trnh trng thi c dng n gin sau

1 v 2 l s kmol ca mi cht kh, p1 v p2 l cc p sut ring phn ca chng, thp1V = 1 RT, p2V = 2 RT. Cng hai phng trnh li cho (p1 + p2)V = (1 + 2)RT.

5/- Xt mt hn hp hai (hay nhiu) cht kh trong mt bnh th tch V v c nhit T. Gi

(3.4)

14

Nhng = 1 + 2 l s kmol kh ca hn hp nn suy ra p = p1 + p2 phi l p sut tng hp ca hn hp kh. Nh vy p sut gy bi mt hn hp kh bng tng cc p sut ring phn do tng cht kh gy nn. y l nh lut Dalton. Cc h qu rt ra trn u l cc nh lut thc nghim tm ra vo th k 18-19.

2.4. PHN B PHN T THEO VN TC

Trong mt khi kh, cc phn t chuyn ng nhit c vn tc khc nhau. Ta hy xut pht t nhng tin c bn v chuyn ng nhit p dng cho cht kh l tng xc nh s phn b phn t theo vn tc. Bi ton t ra nh sau: Hy tm phn phn t dN/N c vn tc v nm trong khong v v + dv . Hin nhin i lng ny t l vi d 3v = dvx dv y dvz (th tch trong khng gian vn tc), ta c th vit dN = F (v ) d 3v . N (4.1)

Hm F (v ) c tn l hm phn b vn tc, n bng phn (hay xc sut ) phn t c vn tc nm trong khong mt n v vn tc chung quanh gi tr v . Trc ht ta nhn xt rng cc thnh phn vn tc vx, vy v vz l c lp nhau nnF (v )d 3v = F (vx ) dvx .F (v y ) dv y .F (vz ) dvz ,

t

F (v ) = F (v x ) F (v y ) F (vz ) .H thc ny ch c th tha mn khi F (v ) = F (v 2 ) , tc l2 2 F (v 2 ) = F (vx ) F (v y ) F (vz2 ) . 2 2 H thc ny, cng vi iu kin v 2 = vx + v y + vz2 cho thy hm F phi l mt hm m

F ( v 2 ) = C e v ,

2

(4.2)

trong C v l hai hng s dng. Kt qu l2 2 dN = C e v d 3v = C e v 4 v 2 dv . N

(4.3)

xc nh cc hng s C v ta phi s dng cc cng thc tnh tch phn sau:

In

0

e x x n dx , n nguyn 0.

2

Php tnh tch phn choI 2 n +1 = n! 2n +1

,

I 2n =

(2n 1)!! . 2n +1 n +1/ 2

p dng cho cng thc (4.3) ta c

15

1 = 4 C e v v 2 dv = 4 C I 2 = 4 C2

0

. 2 3/ 22

T C = 3/ 2

.

Tip theo hy tnh trung bnh ca bnh phng vn tcv2 =

F (v 2 ) v 2 d 3v = 4 C e v v 4 dv = 4 C I 4 = 4 C2

0

3 3 . = 3 5/ 2 2 2

Mt khc theo chng I th = m0 v 2 / 2 v =

3 k BT 3 k B T , ta c v 2 = , suy ra 2 m0

m0 m0 = , C = 2 k BT 2 k BT m0 dN = N 2 k BT 3/ 2

3/ 2

.

(4.4)

Kt qu l hm phn b phn t theo vn tc c dng sau e m0 v 2 / 2 k BT

m0 d v = 2 k BT 33/ 2

3/ 2

e m0v

2

/ 2 k BT

4 v 2 dv.

(4.5)

Cng thc c tn hm phn b Maxwell. C th vit li nh sau dN = f (v)dv, N m0 f (v ) = 2 k BT e m0v2

/ 2 k BT

4 v 2 .

(4.5)

Hm f(v) l hm phn b ca tr s vn tc. th ca hm ny nu trn Hnh 2.2. Gi tr cc i ca th ng vi vn tc c xc sut cc i, k hiu vp. D dng tnh c vn tc nyvp = 2 k BT . m0

(4.6)

Cng c th tnh c vn tc trung bnh:v = 8 k BT . m0

(4.7)

Hnh 2.2

Cui cng cn phi ni n khi nim vn tc ton phng trung bnh, k hiu vtf, nh ngha nh sau: vtf = v 2 . Theo trn ta cvtf = 3 k BT . m0

(4.8)

So snh ba vn tc ni trn: v p < v < vtf .

16

2.5. PHN B PHN T THEO CAO TRONG TRNG TRNG LC

Ta hy tnh hm phn b ca cc phn t trong mt iu kin khc, di nh hng ca tc dng ngoi. l phn b ca phn t kh trong trng trng lc. Gi th p(h) l p sut ca cht kh ti cao h so vi mt t. Hy xt mt khi khng kh hnh tr c din tch y l S, c chiu cao dh, y di nm ti cao h. Trng lng ca khi kh c th tnh theo hai cch nh sau:

S[p(h) - p(h + dh)] = gdm.V tri v v phi ca h thc c th bin i

S[p(h) - p(h + dh)] = - S gdm = gSdh.T

dp dh , dh

dp = g . dh

(5.1)

Hnh 2.3

n gin ta gi thit nhit l khng i (T = const) v cc phn t c cng khi lng m0, nh vy

mg dp p = m0 n0 g = m0 g = 0 p. dh k BT k BTPhp gii phng trnh vi phn ny cho m gh p (h) = p0 exp 0 , k BT

(5.2)

trong p0 l p sut ti mt t. Cng thc (5.2) cho thy cng ln cao p sut cng gim, v gim kh nhanh theo lut hm m. Ch rng wt = m0gh l th nng ca mt ht trong trng trng lc nn cng thc c th vit li theo nng lng nh sau: w p (h) = p0 exp t k BT .

(5.3)

T (5.2) cng rt ra c qui lut phn b mt phn t theo cao m gh n0 (h) = n0 (0).exp 0 . k BT

(5.4)

2.6. NI NNG CA KH L TNG

V cc phn t kh l tng khng tng tc vi nhau nn ni nng ca mt cht kh l tng l tng ng nng ca cc phn t.

17

cc cht kh m phn t c cu to t mt nguyn t nh cc cht kh tr, mi phn t thc t c th xem l mt cht im v chuyn ng ca phn t l chuyn ng tnh tin. Khi cc phn t c cu to t hai nguyn t tr ln th ta phi hiu chnh, xt n nh hng ca cu trc phn t m thc cht khng cn l mt cht im. C th l trong trng hp ny ngoi chuyn ng tnh tin, phn t cn c chuyn ng quay. Ta a ra khi nim s bc t do ca chuyn ng. S bc t do ca chuyn ng ca mt vt, k hiu i, l s ta c lp cn thit m t chuyn ng ca vt . Mi phn t n nguyn t th hin nh mt cht im, m mi cht im c 3 ta c lp (chng hn cc ta Descartes x, y, z) nn i = 3. Nu phn t gm hai nguyn t, nh H2, O2, CO, ... ta c 3 ta khi tm xC, yC, zC xc nh chuyn ng tnh tin ca khi tm phn t, ngoi ra cn c 2 ta gc 1, 2 biu th cc gc quay quanh hai trc vung gc vi trc dc phn t (php quay quanh trc dc phn t khng c ngha v khng c phn t vt cht no quay quanh trc ny). Nh vy phn t hai nguyn t c 3 bc t do chuyn ng tnh tin v 2 bc t do chuyn ng quay, v th i = 5. Cc phn t c cu to t ba nguyn t tr ln nh H2O, CO2, NH3, C2H4,... u c i = 6 gm 3 bc t do tnh tin (ng vi cc ta khi tm xC, yC, zC) v 3 bc t do quay (3 ta gc 1, 2, 3 ca 3 php quay quanh 3 trc vung gc nhau). V s phn b nng lng theo cc bc t do, c th xc nh nh sau. Nng lng tnh tin trung bnh

=

3 kB T 2

trn nguyn tc, c phn b u cho 3 bc t do tnh tin. Nh vy nng lng trung bnh ng vi mt bc t do l

(1) =

1 kB T . 2

(6.1)

Vi cc bc t do chuyn ng quay, do chuyn ng nhit l hn lon nn c th cho rng nng lng ng vi mi bc t do chuyn ng quay cng bng nng lng ca mt bc t do tnh tin v do cng biu th bng cng thc (6.1). Nh th nng lng c phn b u theo cc bc t do. l nh lut phn b u nng lng theo bc t do Maxwell. Nu phn t c i bc t do th nng lng trung bnh ca mt phn t chuyn ng nhit l

w =

i kB T . 2

(6.2)

T cng thc (6.2) c th thit lp biu thc ni nng ca cht kh l tng. Gi thit khi kh ch gm mt loi phn t v c N phn t, nh vy ni nng l v

U = N w .Thay cng thc (6.2) vo, ch rng N = mNA/, ta c

U =

i m RT . 2

(6.3)

l biu thc ni nng ca khi kh l tng. Nu h l mt hn hp kh, ta ch vic cng li cc biu thc ni nng ca cc cht kh thnh phn.

18

2.7. CNG V NHIT TRONG CC QU TRNH CA KH L TNG

Ta hy tnh cng v nhit trong mt s qu trnh cn bng ca kh l tng, lin h chng vi ni nng theo nguyn l I, v rt ra cc tnh cht ca cc qu trnh y. 1. Qu trnh ng tch Phng trnh ca qu trnh: V = const. Cng trong qu trnh ng tch v cng b: A = - pdV = 0, v qu trnh hu hn: A = 0. Nhit trong qu trnh v cng b l

Q =

m

CV dT ,

(7.1)

y nhit dung ca qu trnh ng tch c ghi CV. Mt khc, bin thin ni nng xc nh theo (6.3) l

dU =

i m R dT . 2

(7.2)

Theo nguyn l I: dU = A + Q, nhng A = 0 nn dU = Q, cn bng hai biu thc dU v Q va tm c ta rt ra

CV =

i R. 2

(7.3)

Ta thy nhit dung ng tch l mt hng s, ph thuc duy nht vo s bc t do phn t i. Trong mt qu trnh hu hn th

U = Q =2. Qu trnh ng p

i m R T . 2

(7.4)

Phng trnh ca qu trnh: p = const. Cng trong qu trnh ng p v cng b:

A = pdV =

m

R dT .

(7.5)

Nhit trong qu trnh v cng b l

Q =

m

C p dT ,

(7.6)

y nhit dung ca qu trnh ng p c ghi Cp. Bin thin ni nng vn l (7.2). Theo nguyn l I ta c Q = dU - A, thay vo y cc biu thc (7.6), (7.5) v (7.2), ta c

i C p = + 1 R . 2

(7.7)

Ta thy nhit dung ng p cng l mt hng s v ph thuc duy nht vo s bc t do phn t i. T (7.7) v (7.3) rt ra h thc sau gia nhit dung ng p v nhit dung ng tch 19

Cp CV = R.Biu thc cho thy nhit dung ng p ln hn nhit dung ng tch. T (7.5) v (7.6) d dng tnh c cng v nhit trong mt qu trnh hu hn

(7.8)

A =

m

R T ,

Q =

m

C p T .

(7.9)

3. Qu trnh ng nhit Phng trnh ca qu trnh: T = const hay pV = const. Cng trong qu trnh v cng b:

A = pdV =

m

RT

dV . V

(7.10)

V nhit khng thay i nn ni nng khng bin thin trong qu trnh ny: dU = 0, v th Q = - A. Trong qu trnh hu hn th

A =

m

RT ln

V2 m V RT ln 2 . , Q = A = V1 V1

(7.11)

4. Qu trnh on nhit Qu trnh on nhit l qu trnh trong h khng trao i nhit vi mi trng ngoi, tc l c lp nhit. C th to ra s c lp nhit khi t khi kh trong bnh cch nhit. Cng c th to ra c lp nhit bng cch dn nn nhanh h khng kp trao i nhit vi mi trng ngoi, nhng li phi chm ti mi thi im trng thi kp thit lp s cn bng. Ta c Q = 0, t dU = A. Thay (7.2) cho dU v (7.10) cho A ta c

m

CV dT =

m

RT

dV V

suy ra

CV

dT dV + R = 0. T V

Mt khc phng trnh trng thi ca kh l tng cho

dT dp dV . = + T p VKt hp hai biu thc trn dn n

dp dV + = 0, p Vtrong hng s Cp/CV c tn l ch s on nhit. Tch phn phng trnh ny cho

p V = const.

(7.12)

(7.12) l phng trnh on nhit. Ch rng v Cp > CV nn > 1, nh vy trn th (p, V) ng on nhit dc hn ng ng nhit. Kt hp (7.12) vi phng trnh trng thi cho ta cc dng khc sau y ca phng trnh on nhit

T V -1 = const, Tp (1-)/ = const.Trong qu trnh hu hn ta c

(7.12)

U = A =

i m R T . 2 20

(7.13)

2.8. QUNG NG T DO TRUNG BNH

Trong cht kh cc phn t chuyn ng hn lon v khng ngng. Nu quan nim cc phn t l nhng ht im th thc cht chng khng th va chm vi nhau, ch c th va chm ln thnh bnh truyn xung lng v t to nn p sut. Mt khc c nhiu hin tng m s gii thch cn phi xt n va chm gia cc phn t. V th trong ny ta cn hiu chnh li kch thc phn t, cho rng mi phn t c mt ng knh hiu dng d no , ng knh ny vn nh hn nhiu so vi khong cch gia cc phn t. Ta hy tnh s va chm trung bnh ca mt phn t. Hy tm gi thit phn t ang xt l chuyn ng vi vn tc bng vn tc trung bnh ca phn t v , cn li tt c cc phn t khc u ng yn. Khi s va chm trung bnh Z ca phn t xt trong mt n v thi gian bng s phn t c trong mt hnh tr bn knh y 2d v chiu cao v (Hnh 2.4)

Z = d 2 v n0 .

Hnh 2.4

Ngi ta chng minh c rng khi cc phn t u chuyn ng nhit th s va chm ni trn s tng ln 2 ln, tc lZ =

2 d 2 v n0 .

(8.1)

Qung ng t do trung bnh, k hiu , l qung ng m trn tnh mt cch trung bnh phn t khng b va chm vi phn t khc. Ta c = v / Z , tc l

=

1 = 2 d 2 n0

k BT . 2d2 p

(8.2)

Th d, cht kh oxy, ng knh hiu dng phn t l d = 0,29 nm, iu kin bnh thng T = 300 K, p = 1 at, c = 110 nm.

21

Chng III

KH THC

3.1. KH THC

Cht kh l tng m chng ta xt trong chng trc tun th cc gi thit l tng ha. Trn thc t c nhiu cht kh m nh hng ca kch thc v tng tc l khng th b qua, nht l trong iu kin p sut tng ln hoc nhit gim i ng k. Ta hy nu hai th d. Th d 1. iu kin tiu chun (00C 1,033 at) mt phn t khng kh l 2,69.1019 phn t/cm3. T xc nh c khong cch trung bnh gia hai phn t l 3,4 nm, kh ln hn so vi ng knh phn t 0,3 0,4 nm. Khi p sut tng ln n 10 at th khong cch gia hai phn t cn 1,1 nm. Kch thc phn t khng th b qua so vi khong cch ny. Th d 2. 1 mol cht kh N2 00C, c cc s liu o c nh sau v p sut v th tch:

p (at) V (lit) pV (at.lit)

1 22,4 22,4

100 0,24 24,0

300 0,085 25,5

500 0,0625 32,2

1000 0,046 46,0

Nu l kh l tng, tch pV phi l mt hng s v bng 22,4 at.lit. Bng s liu cho thy khi p sut tng ln th gi tr ca tch pV sai khc nhiu so vi kh l tng. Phn tch cho thy rng trong iu kin bnh thng, cht kh c th xem l kh l tng khi cc phn t ca chng gm mt n vi ba nguyn t. Khi phn t c cu to phc tp hn th gi thit v cht kh l tng khng cn ng ngay trong iu kin bnh thng. Nh vy ta cn phi xt cht kh bng phng php gn vi thc t hn. C th l phi tnh n kch thc ca phn t kh v tng tc gia chng. Mt cht kh c xt ti kch thc phn t v tng tc gia cc phn t, vi gi thit rng kch thc phn t nh hn ng k so vi khong cch gia cc phn t v tng tc gia cc phn t nh hn ng k so vi ng nng phn t c gi l kh thc. Ta s xt cc cht kh thc bng phng php hiu chnh cht kh l tng, tc l sa li hai gi thit v kch thc phn t v tng tc gia cc phn t.

3.2. PHNG TRNH TRNG THI CA KH THC

Phng php hiu chnh cho cht kh l tng trnh by trn thu c m hnh kh thc trc ht c dng rt phng trnh trng thi cho kh thc. Hy xt cho 1 kmol kh. Phng trnh trng thi ca kh l tng l

pV = RT.22

(2.1)

Trong cng thc ny V c hiu l khong khng gian t do ca khi kh. Vi kh l tng th khong khng gian t do trng vi th tch chim. Vi kh thc th khong khng gian t do phi bng th tch chim V tr i mt th tch ring hiu dng b no . C th hiu th tch ring b l tng th tch cc phn t trong 1 kmol khi ln cht nht. Phn tch cho thy rng th tch b ny bng 4 ln tng th tch hnh hc ca cc phn t: 1 2 b = 4 NA d3 = N A d 3 . 6 3 (2.2)

Tip theo hy xt hiu chnh do tng tc phn t. Th nng tng tc gia hai phn t trong php gn ng vi gi thit tc dng xuyn tm c dng nh sau

w(r ) =

A B + , > > 0, A & B > 0. r r

(2.3)

Th nng (2.3) c tn l th nng Van der Waals. th ca hm Van der Waals nu trn Hnh 3.1. Theo c hc th hnh chiu ca lc tc dng ln phng r bng

Fr =

dw(r ) . dr

(2.4)

T (2.4) suy ra trn on ng vi r < rc th nng gim theo khong cch nn lc Fr l y, cn trn on ng vi r > rc th nng tng dn nn lc Fr tr thnh ht, rc l khong cch cn bng lc gia hai phn t.

Hnh 3.1

Do chuyn ng nhit nn bnh thng khong cch gia hai phn t ln hn c ly rc, v vy lc tng tc phn t trong kh thc v c bn l lc ht. Lc ht Van der Waals lm cho tc dng ca cc phn t kh thc ln thnh bnh cha yu hn so vi trng hp kh l tng. Nh vy p sut do kh thc gy ra nh hn so vi p sut ca kh l tng. Ta c th vit

plt = pth + pitrong pi l phn p sut gim i do lc Van der Waals, gi l ni p sut phn t, cn pth l p sut o c ca kh thc, s k hiu n gin l p. Ni p sut pi c th tnh c theo lp lun sau: p sut pi t l vi mt phn t st cn thnh bnh n0 v ang b ko vo trong, cng t l vi mt phn t lp trong tip gip lp ni trn c vai tr ht cc phn t lp st cn thnh bnh vo trong lng khi kh, mt ny vn l n0. Nh vy2 pi ~ n0 ~

1 tc l V2

pi =

a , V2

(2.5)

a l hng s, ph thuc vo bn cht ca cht kh. Php tnh v mi quan h gia ni p sut v tng tc phn t cho biu thc sau v hng s a2 a = 2 N A

w(r ) r dr .2 0

(2.6)

By gi t phng trnh (2.1) thc hin cc php th V V - b, p p + a / V2 ta c

23

a p + 2 (V b ) = RT . V

(2.7)

y l phng trnh trng thi ca kh thc, vit cho 1 kmol kh, c tn l phng trnh Van der Waals (Van der Waals 1873). Mun vit phng trnh cho mt khi kh khi lng ty m, ta hy thay V = V/m, c

m2 a m m RT . p + 2 2 V b = V Phng trnh trng thi ca kh thc Van der Waals (2.7) c th vit li dng saup = RT a 2. V b V

(2.8)

(2.7)

Ta hy biu din th phng trnh (2.7) nh trn Hnh 3.2: ng vi mi gi tr T xc nh, ng cong biu th s ph thuc ca p vo V gi l ng ng nhit Van der Waals. Tp hp cc ng ng nhit y gi l h ng ng nhit Van der Waals. H ng ng nhit Van der Waals c cc c im sau: 1/- Khi T ln, ng Van der Waals c dng hyperbol, gn ging nh ng ng nhit ca kh l tng. 2/- Khi T nh, ng ng nhit c on Hnh 3.2 un khc, ng vi th tch tng th p sut tng, l iu khng xy ra trong thc nghim. Nhit T tng ln th on un khc thu ngn li. Tng n gi tr Tc th on un khc thu li thnh mt im, im ny k hiu l C. 3/- n C gi l im ti hn v Tc l nhit ti hn. Tc l nhit ranh gii gia vng c th ha lng v vng khng th ha lng. Tng ng vi im C, p sut pc gi l p sut ti hn v th tch Vc gi l th tch ti hn (cho 1 kmol kh). 4/- Vng nhit thp v c on un khc c gii hn bng mt parabol p, c nh l im C. 5/- im C l im un ca ng ng nhit ti hn, cc tham s ti hn c th tnh c t (2.7) hoc (2.7) v c cc biu thc nh sau

Vc = 3b,

pc =

a , 27 b 2

Tc =

8a . 27 bR

(2.9)

Sau y l th d v hng s hiu chnh ca mt s cht kh: Cht

CO23,64 0,043

O21,37 0,032

Ne0,213 0,017

H2O5,56 0,031

a (105 Nm4/kmol2) b (m3/kmol)

24

3.3. KIM TRA THC NGHIM

Nh vt l Andrews nm 1866 lm th nghim dn nn kh thc. Cc kt qu ca th nghim l kim chng l thuyt Van der Waals. Th nghim tin hnh nh sau Mt khi kh carbonic CO2 cha trong mt bnh c pitng v dng pitng dn nn kh trong iu kin ng nhit. H cc ng ng nhit thc nghim c nu trn Hnh 3.3 v gi l h ng ng nhit thc nghim Andrews. Ta hy so snh gia thc nghim l h ng ng nhit Andrews v l thuyt l h ng ng nhit Van der Waals.

Hnh 3.3

Hnh 3.4

1/- Khi T ln, c th l T > Tc, cc ng Van der Waals v Andrews trng nhau. 2/- Khi T < Tc, trong phm vi vng parabol p, on un khc Van der Waals c thay bng on nm ngang ca ng Andrews. Nh vy trn on ny thc nghim cho thy p sut khng thay i theo th tch. 3/- Cc ng ng nhit ti hn thc nghim v l thuyt cng trng nhau, k c gi tr cc tham s ti hn pc, Vc v Tc. Nh vy kt qu l thuyt ch sai khc thc nghim trn vng parabol p. Ta hy xt iu ny chi tit hn bng mt ng ng nhit Andrews khi T < Tc: ng ABDE trn Hnh 3.4. Qu trnh AB l qu trnh nn kh: th tch gim v p sut tng. Trn on BD cht kh carbonic bt u ha lng: cng nn th lng kh ha lng cng tng ln trong khi p sut khng thay i. p sut khng i ny chnh l p sut hi bo ha ca cht kh. n D cht kh ha lng hon ton. Qu trnh tip theo ng vi on DE: th tch gim t nhng p sut tng nhanh v cht lng kh nn.

Hnh 3.5

M hnh Van der Waals cho kh thc l kh ng n, kh ph hp thc nghim. Vi mt s cht kh ngi ta cn thy xut hin cc on qu trnh BB v DD (Hnh 3.5). Khi nn kh thay v i theo ng BD, qu trnh c th din bin theo on BB v l qu trnh chm ha lng.

25

Tng t khi dn kh t trng thi D, thay v i theo ng BD, qu trnh c th i theo on BB v l qu trnh chm ha hi. Cn qu trnh BD th khng xy ra trong thc t.

3.4. NI NNG CA KH THC

kh thc do cc phn t c tng tc vi nhau nn ni nng ca mt khi kh thc bng tng ng nng ca cc phn t v vi th nng tng tc gia cc chng

U =

(w ) a

a

+

(w )a

t a

,

(4.1)

trong (w)a l ng nng ca phn t th a, ( wt )a = (1/ 2) wab l th nng tng tc caba

phn t th a vi cc phn t cn li (tnh bnh qun), wab l th nng tng tc gia hai phn t a v b, tha s 1/2 hiu chnh vic trong tng mi phn t c mt hai ln. Ta hy phn tch hai s hng trong (4.1). S hng ng nng c th vit nh sau, xt cho 1 kmol kh:

(w ) a

a

= N A w = N A

i i k BT = R T . 2 2

(4.2)

S hng th nng, xt cho 1 kmol kh, c th nh gi bng tng nng lng ht lm gim p sut tc dng ln thnh bnh, tc l tng nng lng to nn ni p sut phn t:

(w )a

t a

=

V

pi dV .

Thay biu thc ca pi (2.5) vo ta c

(w )a

t a

=

a . V

(4.3)

Kt qu ta thu c biu thc ca ni nng kh thc (cho 1 kmol) nh sau:U = a i RT . 2 V

(4.4)

D dng i n cng thc biu th ni nng ca mt khi kh thc c khi lng ty U = m

U =

im m2 a RT 2 . 2 V

(4.5)

S c mt ca hng thc lin quan n th nng trong biu thc ca ni nng kh thc to nn mt hiu ng quan st c, gi l hiu ng Joule-Thompson. Ni dung ca hiu ng nh sau. Ly mt bnh cha kh c hai pitng P1 v P2 hai u. Khi kh gia chng c p sut p1, th tch V1 v nhit T1. trong bnh st pitng P2 c mt lp mng xp E cht kh thm chm khi dch chuyn pitng. Lp mng E gn c nh vi thnh bnh. By gi ta dch chuyn hai pitng sang phi, chm m bo qu trnh l cn bng, kt thc vo lc pitng P1 tin st mng E. Lc ny khi kh trong bnh c p sut p2, th tch V2 v nhit T2. Gi th qu trnh l on nhit, tc l h khng trao i nhit vi mi trng ngoi.26

Hin tng thay i nhit ca cht kh khi dn nn on nhit c gi l hiu ng JouleThompson. C th c trng qu trnh bin i nhit theo p sut bng i lng T/p, gi l h s bin i nhit theo p sut. Hiu ng Joule-Thompson c gi l dng hay m ty thuc vo du ca h s T/p. Vi cht kh Van der Waals tnh cT 1 2a = b. p C p RT

V Cp lun lun dng nn du ca T/p ty thuc vo du ca biu thc 2a/RT b. Hiu ng Joule-Thompson khng xy ra vi kh l tng v a = b = 0 nn T/p = 0.

27

Chng IV

CHT LNG

4.1. CHT LNG

V chuyn ng phn t, c trng c bn ca cht lng l th nng tng tc gia cc phn t c bng ng nng chuyn ng nhit: t ~ . Ch rng trong cht rn, th nng tng tc phn t ln hn ng k so vi ng nng chuyn ng nhit ca cc phn t nn chuyn ng nhit ca phn t l dao ng quanh v tr cn bng. Chuyn ng nhit cht lng l trung gian gia cht kh v cht rn: cc phn t dao ng quanh mt v tr cn bng no v sau mt s dao ng th bn thn v tr cn bng ny dch chuyn. Theo Frenkel, mi lin h gia thi gian phn t cht lng thc hin mt dao ng D v thi gian n dch chuyn v tr cn bng mt ln DC l

D = DC e / k TB

(1.1)

trong l nng lng ca phn t. Vi nc iu kin bnh thng th D ~ 10-13 s v DC ~ 10-11 s, tc l phn t thc hin khong 100 dao ng th dch chuyn mt ln.

4.2. CC HIN TNG B MT CA CHT LNG

1. Ni p sut Tng tc gia cc phn t c c ly ngn, v th mi phn t ch tng tc vi mt s phn t bn cnh n trong phm vi c ly tng tc. Nh vy nu ly phn t ang xt lm tm, cc phn t tng tc vi n s nm trong mt hnh cu c bn knh bng c ly tng tc. Hnh cu ny c tn l hnh cu tng tc. Khi xt gn b mt cht lng th v tr ca hnh cu tng tc c hai kh nng (Hnh 4.1). Phn t c gi l nm trong lng cht lng nu hnh cu tng tc nm gn trong lng cht lng, lc ny lc tc dng ca cc phn t trong hnh cu ln phn t tm s bng khng. Phn t gi l nm trn b mt nu hnh cu tng tc c mt phn nm ngoi mt cht lng. trng hp ny thc cht ta c mt chm cu tng tc, v trong s cc phn t cht lng tng tc vi phn t xt khng c

Hnh 4.1

phn nm ngoi khng kh (cht kh pha ngoi long hn nhiu so vi cht lng). Kt qu l tng hp lc tc dng ln phn t xt s khc 0 v l mt lc hng vo trong lng cht lng.

28

Nh th tt c cc phn t mt ngoi u chu mt lc hng vo trong lng cht lng. Lc ny to nn mt p sut (hng vo trong lng cht lng) c vai tr lm gim p sut cht lng tc dng ln thnh bnh, v gi l ni p sut phn t. Ni p sut phn t c biu thc ging nh kh thc:pi = a . V2

(2.1)

V bn cht th ni p sut phn t cht lng v kh thc l nh nhau, lp lun rt ra cng thc cng ging nhau. Tuy nhin v cc phn t cht lng c lc tng tc ln hn nhiu so vi kh thc nn ni p sut phn t trong trng hp ny cng ln hn nhiu. Ta c th thy ngay iu ny ch, vi cng mt cht v cng mt iu kin v p sut v nhit , th tch 1 kmol ca cht lng nh hn nhiu so vi th tch 1 kmol cht kh dng hi bo ha. Th d, nc 40C, a = 5,47.105 Nm4/kmol2, V = 0,018 m3/kmol, t pi = 17 210 at. p sut ny ht sc ln. Tuy nhin, n khng tc dng ln vt rn nhng vo cht lng v ni p sut phn t lun lun hng vo trong lng cht lng. 2. Sc cng mt ngoi Ni p sut pi lm cho mt ngoi c xu hng co li. Nh vy mt on ng AB trn b mt cht lng s chu tc dng ca cc lc vung gc vi AB v tip tuyn vi b mt (Hnh 4.2). Cc lc ny tc dng c hai pha ca AB nn cn bng. Tng cc lc tc dng ln on AB v mt pha c gi l sc cng mt ngoi trn on AB ca b mt cht lng. Lc ny s th hin nu AB l mt on dy ch mnh.

Hnh 4.2

Hnh 4.3

Thc nghim chng t sc cng mt ngoi t l vi chiu di ca on ng m lc ny tc dng, tc lF = l

(2.2)

trong h s t l c tn l h s cng mt ngoi. H s ny c n v l N/m. Sau y l th d v h s cng ca mt s cht lng: Cht ( 200C)H2O Hg

glycerin 0,065

ether 0,017

(N/m)

0,073

0,540

Nu c mt khung dy kim loi mnh v kn, trn c mt mng cht lng th trn mi on c chiu di l ca khung s chu mt sc cng mt ngoi lF = 2 l.

(2.3)

29

Cng thc c h s 2 do khung chu tc dng ca mng c hai pha b mt. Trn hnh 4.3 l mt khung nh th, cc mi tn ch phn lc ca sc cng. 3. Nng lng mt ngoi Sc cng mt ngoi biu th c d tr mt nng lng (th nng) trn b mt cht lng. Ta hy tnh nng lng ny. Ly mt khung dy hnh ch U bng dy kim loi mnh t nm ngang v tip xc vi mt cht lng (Hnh 4.4). Dng mt thanh kim loi mnh v thng gt nh trn khung li mt cht lng ln mt cht trong phm vi phn ng kn ca khung. Gi th l = AB l on thng ca thanh ta trn khung v vung gc vi hai cnh bin ca khung. Lc tc dng ln AB bi mng b mt cht lng chnh l lc biu th bng cng thc (2.3). Ta hy dch thanh kim loi sang phi mt on bng dx bng cch dng mt lc ngoi cn bng lc (2.3). Cng thc hin ldA = Fdx = ldx = dS

trong dS l din tch gia tng ca mt cht lng trn khung. Cng ny chuyn thnh th nng ca phn b mt cht lng tng thm. nh vy nng lng mt ngoi ca cht lngUm = S.

(2.4)

Cng thc (2.4) cho thy nng lng mt ngoi t l vi din tch. Cng thc cng dn n mt tn khc ca n v h s cng mt ngoi l J/m2.

Hnh 4.4 4. Gii thch mt vi hin tng mt ngoi

Hnh 4.5

Ta bit rng cc h vt l tn ti thc lun c nng lng nh nht c th. T nguyn tc c th dng cng thc (2.4) gii thch mt s hin tng mt ngoi ca cht lng. Mt git nc trong khng kh, mt git du trong nc c dng hnh cu v trong tt c cc khi c cng th tch th hnh cu c din tch mt ngoi nh nht v do nng lng mt ngoi nh nht. y ta phi b qua nh hng ca trng lc. Ta ly mt mng mng cht lng trn mt khung kn (Hnh 4.5). t vo mt cht lng mt vng dy ch ni kn, vng c hnh dng ty . By gi dng que nh chc thng mng cht lng trong vng ch. Vng ch s lp tc chuyn sang hnh trn. Gii thch nh sau: trong cc hnh c cng chu vi th hnh trn c din tch ln nht, nn phn mng cht lng cn li c din tch nh nht.

30

4.3. HIN TNG DNH T

Khi nh mt git cht lng ln mt phng lm bng cht rn, phn tip xc gia cht lng v cht rn s c dng nh trn Hnh 4.5. Pha trn ca h l khng kh. Ti mt on b ca ng bin im A, sc cng mt ngoi l FLK . Ch rng lc FLK c ph thuc vo bn cht ca cht kh vng CK. y thc cht cn hai lc na tc dng ln on bin ang xt l lc FRK xc nh bi (quan h gia) cht rn v cht kh, v lc FRL xc nh bi cht rn vi cht lng. Gc hp gia lc FLK v FRL gi l gc dnh t. Khi bin xc lp th chiu ca ba lc ny ln phng ngang cn bng nhau, tc lFRK = FRL + FLKcos.

(3.1)

S cn bng theo phng thng ng cn c trng lng ca git cht lng v phn lc ca mt bn tham gia. Nu k hiu RK, RL v LK l cc h s cng trn ng bin ca git cht lng trong mi quan h rn-kh, rn-lng v lng-kh ln lt th t (3.1) suy ra

RK = RL + LKcos.V cos lun 1 nn t (3.2) rt ra iu kin c th c c cn bng l |RK - RL| LK.

(3.2) (3.3)

Hnh 4.6a

Hnh 4.6b

Ta hy xt hai trng hp khng tha mn (3.3) hoc tha mn dng ng thc: 1/. Khi RK RL + LK: lc FRK ln hn hoc bng tng tr s hai lc kia, cht lng s chy loang v hn trn mt cht rn, ta c = 0, cht lng gi l dnh t hon ton (vi cht rn). 2/. Khi RL RK + LK: lc FRL ln hn hoc bng tng tr s hai lc cn li, mt tip xc cht lng-cht rn s thu v mt im (nu khng c lc trng trng), ta c = , cht lng gi l hon ton khng dnh t. Trong trng hp (3.3) c tha mn dng bt ng thc, tc l |RK - RL| < LK, th cht lng c gi l dnh t mt phn. C th hn: - Nu RK > RL th 0 < < /2: c dnh t (mt phn). - Nu RK RL th /2 < : khng dnh t (mt phn). Hin tng dnh t c ng dng trong vic lm giu qung. Ngi ta nghin qung c ln tp cht thnh cc ht nh v b vo mt cht lng sao cho cht lng lm t ht qung nhng khng lm t ht tp cht. Phng mt lung bt kh vo cht lng: cc ht tp cht s bm vo bt kh, ni ln v c vt i. 31

4.4. HIN TNG MAO DN

1. p sut ph di mt khum Hin tng dnh t lm cho b mt cht lng ng trong ng c dng mt mt khum, li ln hay lm xung ty thuc vo thnh bnh khng hoc c gy dnh t. Hnh 4.7 m t cc mt khum nh th, vi cc vect biu th phn lc ca sc cng mt ngoi (do thnh bnh to nn) t ln ng bin ca mt khum.

Hnh 4.7a

Hnh 4.7b

Hnh 4.8

Hnh 4.9

Phn lc F ca sc cng mt ngoi tc dng ln ton ng bin s l mt lc hng xung (nu khng dnh t) hoc hng ln (nu c dnh t). Lc ny gy nn mt p sut ln b mt cht lng trong ng, gi l p sut ph di mt khum. Ta hy tnh p sut ny. Gi th r l bn knh ca ng, nu r nh th mt khum l mt mt chm cu, bn knh R, c cos = r/R (Hnh 4.8):F' =

dF 'cos =

dF cos =

dl

r r r2 = 2 r = 2 R R R

trong dF l sc cng mt ngoi trn on b ca bin v dF l phn lc ca n. Gi pf l p sut ph, ta c pf = F/S, vi S = r 2, t pf =

2 . R

(4.1)

Ta qui c du cho cng thc ny nh sau: bn knh R hng xung mang du dng, hng ln mang du m, du ca p sut ph cng vy. 32

Ni chung th mt khum l mt mt cong, khng nht thit l chm cu. Ngi ta chng minh c rng trn mt mnh nh ca mt cong lun tn ti hai bn knh chnh, R1 v R2, c trng cho cong ca mt. Hai bn knh ny mt l cc i v mt cn li l cc tiu, thuc hai cung trn khng trng nhau v cng qua nh ca mnh mt cong. Khi p sut ph di mt khum l 1 1 pf = + . R2 R1

(4.2)

Du ca R1 v R2 vn xc nh nh qui c nu trn. Vi mt chm cu th R1 = R2 = R v cng thc (4.2) tr v cng thc (4.1). 2. Hin tng mao dn Ly mt ng trn bng thy tinh c bn knh nh v h hai u. Nhng ng theo phng thng ng vo mt chu nc, ta thy mc nc trong ng v bn ngoi ng khng bng nhau d vn bnh thng (Hnh 4.9). Nguyn nhn ca hin tng chnh l s dnh t: nu thnh bnh lm cht lng dnh t th mc cht lng trong ng dng ln, ngc li nu thnh bnh khng lm cht lng dnh t th mc cht lng h xung. S chnh lch chiu cao mc cht lng trong v ngoi mt ng bnh thng c gi l hin tng mao dn. Trn Hnh 4.9 c hai ng mao dn lm bng hai loi thy tinh khc nhau, mt ng c gy dnh t v mt ng khng gy dnh t i vi cht lng nm pha di. Ta hy tnh chiu cao ca ct cht lng mao dn. p sut ti hai im A v B bng nhau v u l p sut kh quyn: pA = pB. p sut ti hai im A v C cng bng nhau v cng cao v bnh thng: pA = pC. Nhng pC = pB + ph pf , trong ph l p sut do ct cht lng c chiu cao h = CB to nn. p sut ph pf hng ngc vi cc p sut cn li nn ng sau du m. Nh vy ph = pf. Nu l khi lng ring ca cht lng thB B

ph = gh.

Mt khcpf =

2 2 cos . = R r 2 cos . gr

Kt qu l thu c cng thc cho chiu cao ca ct cht lng mao dn nh sau:h =

(4.3)

Cng thc cng ng cho trng hp cht lng khng dnh t v lc ny (4.3) s biu th mc cht lng h xung. Cy ci ht nc bng cc ng mao dn trong thn cy: t cng thc (4.3) ta thy ht c nc ln cao th cy xanh phi to ra cc ng mao dn c bn knh r cng nh cng tt, nc phi c pha nha (do cy xanh t to ra) c h s cng tng ln, v tt nht l phi to dnh t hon ton (cos = 1). Nhiu cy xanh c chiu cao n 3040 m vn ht c nc ln tn ngn cy nh c ch ny.

33

Chng V

CHIU HNG CA QU TRNH NHIT

5.1. QU TRNH THUN NGHCH V KHNG THUN NGHCH

Trong chng ny chng ta s xt h v m trng thi c th cn bng hoc khng cn bng. Cc qu trnh thng gp trong thc t th a s l khng cn bng. Khi nghin cu cc qu trnh khng cn bng, vn quan trng hng u l lm r cc qui lut v chiu hng din bin ca chng. Trc ht ta nu mt khi nim rng hn: qu trnh thun nghch v khng thun nghch. Mt qu trnh c gi l thun nghch nu n c th din bin theo c chiu thun cng nh chiu ngc li, trong qu trnh ngc h i qua cc trng thi trung gian nh qu trnh thun v theo mt th t ngc li, sau khi tr v trng thi u th iu kin xung quanh khng thay i. Mt qu trnh l khng thun nghch nu vi phm iu kin t ra. Hy nu vi th d trong c hc v trong nhit hc. Th d 1: Con lc ton hc. Khi qu cu ca con lc chuyn ng t v tr cao nht mt pha sang v tr cao nht pha bn kia th n thc hin mt qu trnh thun nghch: chuyn ng ngc li tr v trng thi ban u l thc hin c. Mun tr v c trng thi ban u th phi loi tr hon ton ma st. Th d 2: Dn nn khi kh trong bnh c pitng. Phi nn chm ti mi thi im trng thi kp thit lp s cn bng. Phi loi tr ma st khng lm tng nhit do nguyn nhn ngoi lai. Nh th ta c mt qu trnh thun nghch: qu trnh ngc li s ging ht qu trnh thun, nhng theo th t ngc li. Nu dn nn khng chm th khi nn p sut gn pitng s ln hn v tr xa, cn khi dn p sut gn pitng s nh hn so vi v tr xa. iu ny lm cho qu trnh tr nn khng thun nghch. Ta thy mt qu trnh khng thun nghch c hc c nguyn nhn l ma st, trong khi qu trnh khng thun nghch nhit hc ngoi ma st cn c mt nguyn nhn na: s ph v cn bng ca trng thi trong qu trnh. D thy rng mt qu trnh cn bng (v khng hao ph nng lng) th thun nghch. iu ngc li khng ng: c nhng qu trnh thun nghch m khng cn bng. Th d, ly mt vng dy siu dn, cho mt thanh nam chm chuyn ng theo trc ca vng dy qua li quanh tm ca vng. Khi trong vng dy xut hin mt dng in cm ng xoay chiu. Qu trnh ny thun nghch nhng khng cn bng. Sau y l vi th d khc v cc qu trnh khng thun nghch ca h nhit: - S truyn nhit t vt c nhit cao sang vt c nhit thp, qu trnh ngc li khng th xy ra. - Dch chuyn mt bnh kh c ma st thnh bn ngoi, cng s bin thnh nhit lm tng nhit khi kh mt lng T.

34

5.2. NGUYN L TH HAI NHIT NG HC

Khi mt h l c lp v khng cn bng, th do chuyn ng nhit ca cc phn t, n s chuyn dn v trng thi cn bng. Qu trnh s din ra theo cch l cc i lng vt l s phn b dn dn ng u, cho n khi t c s cn bng. Qu trnh din ra nhanh cht kh, chm hn cht lng, vi cht rn th rt chm nhng vn xy ra. Chng hn, hy ly mt thanh ng v mt thanh nhm, c tit din ging nhau, mi phng pha u, cho tip xc v p li. Sau mt thi gian kh di ta s thy hai thanh dnh li vi nhau. Ti lp tip xc c hp kim ng-nhm: cc phn t ng khuch tn sang pha nhm, cc phn t nhm khuch tn sang pha ng. Sau y l cc th d khc: - Mt khi kh trong mt bnh c pitng gi p sut cao hn p sut kh quyn bn ngoi. Khi bung khng gi pitng, khi kh s dn t do n khi cn bng p sut. Trong qu trnh dn, cc phn t trong bnh phn b khng ng u, nhng sau mt thi gian th s ng u c thit lp. - Hai bnh cha hai cht kh khc nhau cho ni thng bng mt ng tit din nh. Cht kh trong mi bnh s khuch tn dn sang bnh kia, chng bao lu s c mt hn hp kh ng u khp mi ni. - Mt bnh nc nm tnh lng, ta th tht kho lo mt vin mc vo y bnh. D khng c chuyn ng i lu ca nc, mc vn tan dn n khi c mt dung dch ng u. Cc th d nu trn c chung mt c im: chiu hng ca qu trnh trong mt h c lp l h t lm cho mnh ng u dn dn, cho n khi ng u hon ton. Ngun gc ca chiu hng nh th chnh l chuyn ng nhit. Khng nh v mt chiu hng nh th chnh l ni dung ca nguyn l th hai (nguyn l II) ca nhit ng hc, pht biu nh sau: Qu trnh nhit ca mt h v m c lp lun din ra theo chiu tng ca mc phn b ng u. T tng y phi hiu c theo ngha: khi h t phn b ng u hon ton th mc phn b ng u t cc i v dng li. m t mc phn b ng u, ngi ta a ra khi nim trng thng k ca trng thi. Trng thng k ca mt trng thi (v m) l s cch phn b phn t ca trng thi . Theo c hc Newton, trng thi ca mt ht vt cht c m t bng ta r v xung lng p . Nh vy phn b ht theo trng thi l phn b trong khng gian pha, l hp nht ca khng gian ta v khng gian xung lng. Sau y ta hy xt phn b trong khng gian ta . Hy xt mt th d n gin, h c 4 phn t, k hiu a, b, c, d trong mt bnh chia tng tng lm hai ngn nh nhau (Hnh 5.1). C c thy 16 cch phn b 4 phn t vo hai ngn bnh. Tuy nhin ch c 5 trng thi v phn b phn t: trng thi v m ch phn bit c s phn t trong mi ngn, khng th nhn bit c phn t no ngn no. Nh vy: - trng thi 1: 4 phn t ngn tri, 0 phn t ngn phi c 1 cch phn b phn t. - trng thi 2: 3 - trng thi 3: 2 - trng thi 4: 1 - trng thi 5: 0 --------,1 ,2 ,3 ,4 -------- c 4 c 6 c 4 c 1

---------

. . . .

35

K hiu l trng thng k ca trng thi, ta c ln lt 5 trng thi nu trn: 1 = 1, 2 = 4, 3 = 6, 4 = 4, 5 = 1. T th d ny c th nu cc nhn xt sau: - Trng thi cng ng u hn th trng thng k cng ln, ln nht trng thi ng u hon ton. - Cc trng thng k nu trn ch mi tnh theo ta , k hiu r l T. Trng thng k tnh theo xung lng cng c cc tnh cht tng t, v k hiu l XL. Trng thng k ca trng thi s l = TXL. - Nu h c N ht th chnh xc hn c th xt bi ton chia hp thnh N ngn. - Cc kt qu chung khng thay i khi xt cho cc h c tng tc, k c cho cht rn. Hnh 5.1

Nh vy c th hiu trng thng k ca mt trng thi biu th mc phn b ng u cc ht ca trng thi . T nguyn l II c th pht biu: Qu trnh nhit ca mt h v m c lp lun din ra theo chiu tng ca trng thng k ca trng thi. Pht biu ny tng ng vi nguyn l II. Sau y ta tnh trng thng k ca trng thi phn b u N phn t trong bnh. Chia bnh thnh N ngn bng nhau. trng thi cn bng, mi ngn s c mt ht (Hnh 5.2). Gi th v l th tch b nht m mt phn t chim khi ln cht, gi n l mt , th tch ny hin nhin ph thuc nhit T. Th tch ca mt ngn l V1 = V/N, s trong mt ngn l m = V1/v = V/Nv. Ta c (S cch phn b 1 ht 1 ngn)N =

T = S cch phn b N ht vo N ngn= N! m N = N! (V/Nv) N . = N! (V/Nv) N XL.

Trong cng thc ny, v v XL ph thuc vo nhit , XL cn ph thuc s ht N.

Hnh 5.2 N = 12

5.3. ENTROPY

Khi h trng thi c trng thng k , entropy ca h trng thi y c nh ngha nh sauS = kB ln

(kB: hng s Boltzmann).B

(3.1)

Ta hy xt ngha ca khi nim ny. Trng thi km ng u nht, c th hiu l trng thi c trt t nht, s c = 1 v do S = 0. Trng thi cng ng u hn th entropy cng ln, v t cc i Smax trng thi cn bng. Entropy c tnh cht cng c: entropy ca mt 36

h gm bng tng entropy ca ca cc b phn nu tng tc gia chng khng ng k. Tht vy, nu h c hai phn, c cc trng thng k l 1 v 2, th trng thng k ca h l = 1 2, thay vo (3.1) s c S = S1 + S2. Theo phn tch trn ta thy khi nim entropy biu th mc hn lon trong mt h v m. Khi nim entropy v khi nim trng thng k c ngha tng ng nhau, khc nhau ch hai im: entropy c tnh cht cng c v entropy bng 0 h hon ton trt t. Hng s kB t trong cng thc (3.1) cho tin v ph hp vi lch s tm ra khi nim ny.B

T ngha ca khi nim entropy v nguyn l II ta c th pht biu: Qu trnh nhit ca mt h v m c lp lun din ra theo chiu tng ca entropy. l nguyn l tng entropy, tng ng vi nguyn l II. Trong s nhiu pht biu khc nhau ca nguyn l II, pht biu ny hin nay c xem l i din. Ta hy tnh entropy cho qu trnh cn bng ca kh l tng. T (3.1) v (2.1) c V N m S = k B ln N ! R ln V + f (T ) . XL = Nv

(3.2)

trong f(T) l phn cn li ca biu thc, ph thuc T v khng ph thuc V. T c th tnh c bin thin entropy cho mt qu trnh ng nhit (T = const) m th tch ca h thay i t V1 n V2:

S = S2 S1 =

m

R ln V2

m

R ln V1 =

m

R ln

V2 . V1

(3.3)

i chiu (3.3) vi cng thc (II.7.11) ta c

S =

Q . T

(3.4)

trong Q l lng nhit nhn vo ca qu trnh. Nu qu trnh khng ng nhit nhng vn l cn bng th ta xt mt on nh sao cho nhit trn l khng i. p dng (3.4) chodS =

QT(2)

.

(3.5)

V do vi qu trnh hu hn:

S =

(1)

QT

.

(3.5)

Ngi ta chng minh c rng cng thc (3.5) v do (3.5) ng cho qu trnh thun nghch h nhit bt k, khng nht thit l cht kh l tng. Vi qu trnh khng thun nghch th entropy c sn sinh thm nn thay cho (3.5) l mt bt ng thcdS0tn >

QT

.

(3.6)

Tip theo ta tnh bin thin entropy cho mt qu trnh cn bng ty ca kh l tng. T (3.5) v p dng nguyn l I ta cdS =

1 m dT dV + R ( dU + pdV ) = CV . T T V

V CV l hng s nn tch phn h thc trn cho 37

S =

T2 V m + R ln 2 . CV ln T1 V1

(3.7)

Mt ng dng khc na ca khi nim entropy l gii thch chiu hng truyn nhit. Gi th c hai vt c nhit T1 v T2 khc nhau v tip xc vi nhau, lp thnh mt h c lp, c mt lng nhit Q > 0 truyn t vt th nht sang vt th hai. Bin thin entropy ca qu trnh ldS = dS1 + dS 2 =

QT1

+

QT2

1 1 = Q . T1 T2

Entropy ca h phi tng trong qu trnh ny nn dS > 0, suy ra T1 > T2. Vy: Nu hai vt c nhit khc nhau v tip xc vi nhau, lp thnh mt h c lp th nhit ch truyn t vt nng sang vt lnh. Pht biu ny cng biu th chiu hng truyn nhit v tng ng vi nguyn l II.

5.4. MY NHIT

1. My nhit My nhit l mt dng c bin i nhit thnh cng. Hai b phn quan trng ca my nhit l ngun nng c nhit cao T1 v mt bung cha kh c pitng nhn nhit bin thnh cng hu ch nh dn nn. Cht kh trong bung c gi l tc nhn, n hot ng theo chu trnh (qu trnh kn): tc nhn nhn nhit, dn pitng sinh cng, h nhit tr li trng thi ban u, nhn nhit tip...Tht s th cc chu trnh vn hnh ca cc my nhit u l cc qu trnh khng thun nghch, song n gin c th xem chng l thun nghch. Gi th trong mt chu trnh cht kh tc nhn nhn mt lng nhit Q, sinh mt cng hu ch A, hiu sut ca my nhit l i lng sau y

=

A . Q

(4.1)

Hiu sut cho bit t l (thng tnh bng %) nng lng nhit cung cp bin thnh cng hu ch. Nu my nhit ch gm c ngun nng v bung kh tc nhn th khng hot ng c. Tht vy, bin thin entropy ca h sau mt chu trnh l

S = S1 + S M =

Q Q + 0 = , T1 T1

(4.2)

trong S1 l bin thin entropy ca ngun nng (lng nhit nhn sau mt chu trnh l Q v ngun ta nhit), bin thin entropy ca khi kh SM = 0 v sau mt chu trnh tc nhn tr li trng thi ban u. V ton b my nhit l kn nn entropy phi tng sau mi chu trnh, gi tr m ca S l biu hin qu trnh khng th din ra. my hot ng c cn phi c thm mt b phn na: ngun lnh, c nhit T2 (< T1), tc nhn thi bt nhit (Hnh 5.3). Ngun lnh thng l mi trng khng kh bn ngoi. Gi th nhit ta ra cho ngun lnh trong mi chu trnh l Q (Q > 0) th bin thn entropy sau mi chu trnh by gi s l 38

S = S1 + S M + S2 =

Q Q' Q Q' . + 0 + = + T1 T2 T1 T2

(4.3)

Trong biu thc (4.2) mi quan h gia cc i lng T1, T2, Q, Q phi m bo sao cho S 0 th my nhit mi hot ng c. Khi c ngun lnh th cng hu ch bng A = Q Q, thay vo cng thc (4.1) s cQ' . Q

=1

(4.4)

Cng thc ny trc ht cho ta mt nhn xt sau. V Q phi khc khng nn A phi nh hn Q tc hiu sut phi nh hn 1. iu ny c ngha l nhit khng th bin hon ton thnh cng. Nhn xt ny tng ng vi nguyn l II v chnh l mt dng pht biu ca nguyn l II. Trc y tng c nhng nh ch to cc ng c vnh cu loi hai, tc l loi ng c nhn nhit v bin hon ton thnh cng hu ch. Theo nguyn l II, nh chng t bng cc cng thc (4.2), (4.3) v (4.4), iu ny khng th thc hin.

Hnh 5.3 2. Chu trnh Carnot

Hnh 5.4

Trong s cc chu trnh hot ng ca my nhit th chu trnh Carnot c ngha c bit hn c. y l mt chu trnh gm hai qu trnh ng nhit v hai qu trnh on nhit k tip xen k nhau (Hnh 5.4). Qu trnh t trng thi (1) n trng thi (2) l qu trnh ng nhit, bung kh tip xc vi ngun nng nhit T1, th tch tng ln y pitng ra v p sut gim i. n trng thi (2), nh mt c ch ring, khi kh tc nhn dn on nhit n trng thi (3), th tch vn tng v p sut gim tip, kt qu l nhit gim n T2 bng nhit mi trng (tc l ngun lnh). Do tc dng ca ngun lnh, khi kh co li ng nhit theo qu trnh t (3) n (4) v ko pitng vo. Qu trnh tip theo t (4) n (1) l nn on nhit, nh c ch ring nu trn, a cht kh tc nhn tr v nhit bng nhit ngun nng v li tip xc vi ngun nng. Ta tnh hiu sut ca chu trnh Carnot. Cc phng trnh ca 4 qu trnh thnh phn l (1) (2): p1V1 = p2V2, (2) (3): p2 V2 = p3V3 , (3) (4): p3V3 = p4V4, (4) (1): p4 V4 = p1V1 . Hai phng trnh on nhit nu vit theo bin s T v V th: (4) (1): T1V1 1 = T2 V4 1 , (2) (3): T1V2 1 = T2 V3 1 . (4.5) (4.5)

39

Trong ton chu trnh, nhit nhn vo din ra trong qu trnh (1) (2):Q = Q((1) (2)) = m

RT1 ln

V2 . V1 V3 . V4

(4.6a)

Nhit ta ra l nhit ca qu trnh (3) (4):Q ' = | Q((3) (4)) | = m

RT2 ln

(4.6b)

T (4.6a,b) s dng (4.5) v (4.5) ta i n biu thc sau y cho hiu sut ca chu trnh Carnot

C = 1

T2 . T1

(4.7)

Ta thy hiu sut ca chu trnh Carnot ch ph thuc vo nhit ngun nng v ngun lnh, khng ph thuc vo cht tc nhn. Nu my nhit hot ng theo chu trnh Carnot trn Hnh 5.4 nhng theo chiu ngc kim ng h th l my lm lnh. Cng thc hiu sut vn l (4.7), khng thay i. Ni chung, ng c hot ng theo chu trnh thun kim ng h l my bin i nhit thnh cng, theo chu trnh ngc kim ng h l my lm lnh. Ta c nh l sau (nh l Carnot): Gia hai ngun nng nhit T1 v ngun lnh nhit T2 cho trc, ng c hot ng theo chu trnh Carnot s cho hiu sut cc i. C nhiu cc chng minh nh l ny, sau y l mt chng minh n gin. Gi th c mt chu trnh (thun nghch) khc m my X to hiu sut cao hn hiu sut C (4.7). Ta hy ni my ny vi my chy theo chu trnh Carnot M sao cho cng A sinh ra t my X chuyn cho my M lm lnh (Hnh 5.5a). Gi thit Q1 l nhit nhn vo, Q1 l nhit ta ra my X, Q2 l nhit ly t ngun lnh v Q2 l nhit ta ra cho ngun nng (cc lng nhit qui c u dng du dng). Ta c

= A/Q1 , C = A/Q2 .

Hnh 5.5 a/.

b/.

V > C (theo gi thit) nn Q2 > Q1. Mt khc A = Q1 - Q1 = Q2 - Q2 nn Q2 - Q1 = Q2 Q1 Q > 0. Vi kt qu ny h s tng ng vi my lm lnh nh trn Hnh 5.5b. y l my rt nhit lin tc t ngun lnh chuyn sang ngun nng khng hao ph nng lng. My nh th khng th c theo nguyn l II. Vy C. nh l c chng minh cho chu trnh dng ty nhng thun nghch.

Nu my hot ng theo chu trnh khng thun nghch th do hao ph nng lng v do khng cn bng nn hiu sut phi gim i so vi chu trnh thun nghch tng ng. nh l Carnot c chng minh.

40

5.5. CC HM TH NHIT NG LC

Hm th nhit ng lc l hm trng thi sao cho o hm ca n theo tham s trng thi cho ta cc i lng nhit ng lc, cng l nhng tham s trng thi. Cc hm th c vai tr quan trng c bit trong vic xt bin i ca trng thi, nht l trong cc chuyn pha (Chng VI). Trc ht ta xt chng trong trng thi cn bng v cho h c s ht c nh. Theo nguyn l I th bin thin ni nng trong mt qu trnh c dng sau dU = Q + A. Theo Ch.I th A = - pdV, theo (3.5) th Q = TdS, nh vydU = TdS pdV.

(5.1)

H thc ny biu th vi phn ca U lin quan n vi phn ca hai tham s trng thi S v V. Theo qui tc vi phn th t (5.1) ta c U T = , S V U p = , V S

(5.2)

trong ch s di nhn mnh i lng ghi km l khng i khi ly o hm theo i lng kia. Theo nh ngha nu trn th ni nng U l mt hm th nhit ng lc, c hai bin s l S v V: U = U(S, V), c hai o hm ring biu th hai i lng nhit ng lc l (5.2). C th thay i bin s xc nh cc hm th khc. Nu dng bin s T v V th bin i nh sau:dU = TdS - pdV = d(TS) - SdT- pdV.

Sau t F = U TS, i lng ny gi l hm th (nhit ng lc) Gibbs, thdF = - SdT pdV.

(5.3)

Nh vy F l hm th c hai bin s T v V: F = F(T, V), v c cc o hm sau F S = , T V F p = . V T

(5.4)

Lm tng t nh trn, ta c th thu c hai hm th na nh sau: Hm th Helmholtz = F + pV = U TS + pV, = (T, p):d = - SdT + Vdp,

S = . , V = T p p T W W T = . , V = S p p S

(5.5)

Hm th enthalpy W = U + pV, W = W(S, p): dW = TdS + Vdp, (5.6)

Nu xt cho h c s ht thay i th cc hm th ni trn c thm bin s na: s ht N (gi thit h ch c mt loi ht): U = U(S, V, N). Khi cc biu thc vi phn v o hm c dng sau dU = TdS pdV + dN, U T = , S V , N U U p = , = . V S , N N S ,V (5.7)

41

i lng c tn l th ha, n bng nng lng tng ln ca h khi thm vo mt ht (vi iu kin gi nguyn entropy v th tch). Vic thm bin s N vo cc hm th F, v W thc hin tng t. By gi ta xt qu trnh khng cn bng, cho h c s ht c nh. Lc ny cc tham s trng thi phi hiu theo ngha ly trung bnh trn ton h. Theo (3.5) v (3.6) ta c

Q TdS,trong du = ng vi qu trnh cn bng, du < ng vi qu trnh khng cn bng. Nh vy dU TdS pdV hay dU - TdS + pdV 0. (5.8) Nu qu trnh c S = const, V = const th suy ra dU 0, tc l khi dch chuyn v trng thi cn bng th ni nng U gim. Ta i n kt lun sau: Nu S = const, V = const th U = Umin ti trng thi cn bng. Bng cch tng t ta chng minh c: Nu T = const, V = const th F = Fmin ti trng thi cn bng. Nu T = const, p = const th = min ti trng thi cn bng. Nu S = const, p = const th W = Wmin ti trng thi cn bng.

42

Chng VI

CN BNG PHA V CHUYN PHA

6.1. CC PHA CA H V M

Trong cc chng II, III v IV ta kho st cc h v m c cu trc ng nht, tc l c vt cht phn b ng u ti mi im trong h. Mt ngoi l l chng III khi xt nn kh thc: trong h cn bng c th tn ti ng thi cht kh v cht lng vi cng p sut v nhit . Mt b phn trong h c cc tnh cht vt l ng nht (trn ton b phn y) c gi l mt pha. H v m c th gm nhiu b phn c cu trc khc nhau nhng cn bng vi nhau, tc l nhiu pha cn bng vi nhau. chng III khi xt th d v nn kh carbonic, ta thy hai pha kh v lng tn ti c trong pitng khi dn nn kh nhit di nhit ti hn. Trong thc t hng ngy, chng ta thng gp cc h nhiu pha nh: kh-lng, kh-rn, lngrn, kh-lng-rn. Hai nhm hin tng c bn xy ra i vi quan h gia cc pha ca h l cn bng cc pha v chuyn di pha.

6.2. CN BNG PHA

1. Cn bng hai pha Khi trong h c hai pha th s cn bng ca chng trc ht biu th cn bng nhit v p sut T1 = T2, p1 = p2. (2.1)

Theo 5.5, cc hm th nhit ng lc t cc tiu ti cn bng. Nu chn hm th c cc bin s T, p v Ni (s ht loi i) th l hm th nhit ng lc Helmholtz :

= (T, p, Ni): d = - SdT + Vdp + i i dNi.

(2.2)

i lng i c tn l th ha (ca ht loi i), n biu th nng lng thm vo h khi c thm mt ht (loi i). V hm th Helmholtz t cc tiu ti cn bng nn d = 0, cng vi (2.1) dn n i i dNi = 0. p dng cho h c hai pha th 1 dN1 + 2 dN2 = 0. Nu gi N1 v N2 l s ht ca mi pha th tng ca chng l khng i N1 + N2 = N = const, tc dN = dN1 + dN2 = 0. Kt hp hai kt qu trn cho

1(T, p) = 2 (T, p).

(2.3)

43

y l iu kin th ba ca cn bng cho hai pha. Trong (2.3) ta vit r hai bin s ca ca th ha l T v p. Trn th (T, p) biu thc (2.3) biu th mt ng cong, gi l ng cong cn bng pha (Hnh 6.1). Trn ng cong ta c (2.1). Ngoi ng cong v hai pha l cc trng thi ca hai pha.

Hnh 6.1 2. Cn bng ba pha

Hnh 6.2

Tng t nh trn ta c th thit lp cc h thc biu th cn bng ba pha: T1 = T2 = T3, p1 = p2 = p3,

1(T, p) = 2 (T, p) = 3 (T, p).

(2.4)

V h thc cn bng th ha l hai phng trnh nn nghim ca chng xc nh mt im (Tb, pb) trn th (T, p) gi l im ba (im M trn Hnh 6.2). Trn th ny c ba pha vi cc k hiu: R, L, K. im ba l giao im ca ba ng cong cn bng tng cp pha (th d: rn-lng, rn-kh v lng-kh). 3. Cn bng nhiu pha By gi xt trng hp tng qut cn bng ca nhiu pha. Gi thit h c r pha, k hiu bng cc ch s i = 1, 2, ..., r, v c n cht thnh phn, k hiu k = 1, 2, ..., n. Gi N i( k ) l s ht ca cht th k pha i th nng t i ca n s lCi( k ) = N i( k ) . Ni(k )k

Cc lng ny l cc tham s trng thi. V

k

Ci( k ) = 1 nn ch c (n - 1)r tham s nng l

c lp. Ngoi ra cn hai tham s c lp na l T v p, nh vy tng s tham s c lp l (n 1)r + 2. iu kin cn bng pha ca cc th ha lk 1k = 2 = ... = rk .

(2.5)

Cc h thc (2.5) biu th (r - 1)n phng trnh. h phng trnh (2.5) c nghim th s tham s phi bng hoc ln hn s phng trnh, tc l (n - 1)r + 2 (r - 1)n. T rt ra bt ng thc sau y, gi l qui tc Gibbs

r n + 2. 44

(2.6)

Qui tc Gibbs biu th mi quan h gia s pha v s cht c th cn bng vi nhau khi lp thnh mt h. Th d, nu h gm mt cht, n = 1, th qui tc Gibbs l r 3; nu c hai cht, n = 2, th r 4.

6.3. CHUYN PHA

By gi chng ta xt s chuyn pha, tc l s chuyn trng thi ca h t pha n qua pha kia. C hai loi chuyn pha, l chuyn pha loi mt v chuyn pha loi hai. 1. Chuyn pha loi mt Chuyn pha loi mt l chuyn pha trong c cc i lng qung tnh nh th tch, ni nng, entropy,... bin i gin on. V cc i lng nh th u l o hm bc nht ca cc hm th nhit ng lc nn cng c th ni chuyn pha loi mt l chuyn pha trong o hm bc nht ca cc hm th nhit ng lc l gin on, cn bn thn cc hm th th vn lin tc. in hnh nht ca chuyn pha loi mt l s bay hi ca cht lng, tc l chuyn pha lngkh. Xt mt hm th nhit ng lc, chng hn, hm th Gibbs F = F(T, V), hy c nh th tch v xt bin i hm ny theo nhit c hai pha (Hnh 6.3). pha I, ta c F = F1(T), ng vi ng AA, pha II, ta c F = F2(T), ng vi ng BB, hai ng gp nhau ti C.

Hnh 6.3

Hnh 6.4

Gi th ban u qu trnh din bin t A theo hm F = F1(T), n C qu trnh din bin tip l CB' theo hm F = F2(T) m khng phi l CA v ng CB' c nng lng thp hn. Nh vy ti im C xy ra chuyn pha. Qu trnh t B n C cng xy ra chuyn pha ti C v sau din bin tip theo CA, m khng phi l CB. S gy khc ca ng AB ti C chng t o hm bc nht ca F theo T b gin on. Theo (V.5.4) th o hm ny, ly vi du ngc li, chnh l entropy. Ta c

F F = S1 , = S2 . T Tc 0 T Tc + 0Nu ( F / T )T 0 > ( F / T )T + 0 th S2 > S1, tc l entropy ca khi vt cht pha kh s ln hnc c

so vi pha lng. Theo Chng V th S = Q/T, Q l nhit ha hi, cn gi l n nhit, thng tnh cho 1 kmol (nhit ha hi cng k hiu l V).

45

By gi ta thit lp phng trnh ca ng chuyn pha p = p(T), hay dng vi phn dp/dT. Trn Hnh 6.4 ta gi thit chuyn pha ti (p, T) ng vi (1) (2): V1 V2, ti (p, T) ng vi (1) (2): V1 V2 , vi p = p + dp v T = T + dT. Xem (1) (2) (2) (1) (1) l mt chu trnh th cng v nhit nhn vo ca ton chu trnh y l

A = - p(V2 V1) - pdV2 p(V1 V2) pdV1 == - p(V2 V1) - pdV2 (p + dp)(V1 + dV1 - V2 - dV2) (p + dp)(- dV1) = = dp(V2 V1), Q = T(S2 S1) + TdS2 + T ( S1 S 2 ) + T dS1 = = + T(S2 S1) + TdS2 + (T + dT)(S1 + dS1 - S2 - dS2) + (T + dT)(- dS1) = = dT(S2 S1) = dT(Q/T). V sau mt chu trnh th bin thin ni nng bng dU = 0 nn A + Q = 0 tc l dp(V2 V1) = dT(Q/T), do dp Q . = dT T (V2 V1 ) (3.1)

y l phng trnh Clapeiron-Clausius. Trong phng trnh ny nu bit trc V1, V2 v Q (ph thuc vo T) th gii n ta s thu c ng p = p(T). 2. Chuyn pha loi hai Chuyn pha loi hai l chuyn pha c cc hm th nhit ng lc lin tc cng vi cc o hm bc nht ca chng, nhng cc o hm bc hai th gin on. Trong s ny in hnh nht l nhit dung 2 F Q dS = T = T 2 , CV = dT V dT V T V

2 Q dS Cp = = T = T 2 , ... dT p dT p T p Do o hm bc nht lin tc nn nhng i lng nh n nhit Q, bin i th tch V2 V1,... u bng 0. Th d c trng nht v chuyn pha loi hai l chuyn pha t cht dn in thng sang cht siu dn. C mt s cht bnh thng dn in khng tt lm, nhng khi h nhit xung di mt gi tr ti hn no th chuyn thnh siu dn, dn in khng c in tr. Phn tch cho thy rng ngun gc ca chuyn pha loi hai l do bin i tnh cht i xng bn trong ca h.

46

Chng VII

CC QU TRNH KHNG CN BNG

7.1. QU TRNH KHNG CN BNG

Qu trnh khng cn bng l qu trnh trong c tham gia cc trng thi khng cn bng. Mn nhit ng hc nghin cu cc trng thi v cc qu trnh khng cn bng gi l nhit ng hc khng cn bng. Mt qu trnh nh th khng th biu din bng ng cong trn cc th m trc ta l cc tham s trng thi, v trng thi khng cn bng c nhng tham s trng thi khng c gi tr xc nh. Th d, mt khi kh khng cn bng th th tch vn xc nh, nhng nhit v p sut c th khng xc nh, chng nhn cc gi tr khc nhau cc khu vc khc nhau trong h. Trong mt qu trnh b, bin thin entropy ca h c th phn lm hai phn dS = deS + diS, (1.1) vi deS l entropy t mi trng chuyn vo, deS = Q/T, diS l entropy ni sinh, theo nguyn l II th diS 0. Thay hai kt qu ny vo (1.1) cho dS - Q/T 0, tc dS

QT

.

(1.2)

H thc ny nu 3 Ch.V. Mt h c lp th c trng ca qu trnh khng cn bng l lm h chuyn dn v cn bng. Tc dng ca mi trng ngoi ln h c th lm cho n cng mt cn bng mnh, nhng din bin trong h vn l xu hng chuyn v cn bng. un nng mt bnh nc m la ch t t y bnh l mt th d. Nh vy c trng c bn ca qu trnh khng cn bng l tn ti cc dng vn chuyn ca cc i lng khng cn bng. Chng hn, khi nhit khng cn bng th nng lng nhit vn chuyn, khi mt ht khng cn bng th bn thn cc ht vn chuyn,... Dng vn chuyn c o bng lng vn chuyn ca i lng vt l (ang b mt cn bng) qua mt n v din tch vung gc trong mt n v thi gian. Cc i lng ny c k hiu chung l Ji, trong ch s i nhn mt s gi tr no . Nguyn nhn to nn dng vn chuyn l do mt cn bng, hay chnh lch, ca i lng vt l. Cc chnh lch ny c tn l lc. Ta hy biu th X1, X2,... l cc lc m cc dng Ji ph thuc vo, ta c Ji = fi (X1, X2,... ). (1.3) Ngi ta chng minh c rng s thnh phn ca dng lun bng s thnh phn ca lc. Nu mc mt cn bng khng ln lm th cc lc X1, X2,... l nh, ta c th khai trin dng theo chng, ch gi li cc hng thc bc nht (cc s hng bc khng khng c nu gi thit rng khi khng c chnh lch th khng c dng). Ji =

Lk

ik

Xk .

(1.4)

Nh vy dng biu th tuyn tnh qua cc lc. Phn nhit ng hc nghin cu cc qu trnh khng cn bng trong cc dng tha mn h thc (1.2) gi l nhit ng hc tuyn tnh. 47

Sau y chng ta s xt mt s qu trnh ca nhit ng hc tuyn tnh.7.2. KHUCH TN

S chnh lch u tin m chng ta xt ti l chnh lch v mt ht n, hay khi lng ring . Gi thit gia hai im A v B cch nhau mt on AB = x, c khi lng ring 1 v 2 khc nhau v 1 > 2 (Hnh 7.1). Hin tng vn chuyn khi lng gi l khuch tn. Ta hy tnh khi lng khuch tn, xt cho cht kh hoc cht lng.. Gi M l khi lng vn chuyn qua S trong khong thi gian t. C th nh gi c rng M t l vi chnh lch khi lng ring = 2 - 1 = - (1 - 2), t l vi S v t, t l ngc vi x, tc l

M = D

S t x

(2.1)

Hnh 7.1

trong D l mt h s dng, gi l h s khuch tn, du tr c trc cng thc v < 0, t s /x gi l gradien khi lng ring. Gradien l bin i trn mt n v di, cng thc chung l d/dx, hoc tng qut l d / dr . C th tnh h s khuch tn nh sau. Hy chn x bng hai ln qung ng t do trung bnh khi vt qua S phn t va chm t nht mt ln. tin hy coi trong h ch c mt loi phn t. S phn t vn chuyn qua S trong t l

N =

1 v S t (n1 n2 ) 6

y l vn tc trung bnh, tha s 1/6 c mt do chia u cho 6 hng. Mt khc n1 n2 = - n = - (n/x)x = - (n/x)2 . K hiu mo l khi lng mt phn t th M = moN v mon = . T tnh c khi lng khuch tn

M =

1 v S t. 3 x

(2.2)

So snh (2.1) v (2.2) cho biu thc sau y ca h s khuch tn D = 1 v. 3 (2.3)

Dng khi lng thng k hiu l JM, theo nh ngha th JM = M/(St), t (2.1) rt ra JM = D

d , hay J M = D , dng vect: J M = D , dx x

(2.4)

trong l gradien ba chiu ca khi lng ring : = , , . x y z

48

5.3. NI MA ST

Xt khi cht kh hoc cht lng c cu to mt loi phn t v chy theo phng ngang trn Hnh 7.2, vn tc phn t ti A l u1, ti B l u2, vi u2 > u1. Xu hng cn bng vn tc chy dn ti vn tc lp A tng ln, vn tc lp B gim i. Trn lp tip xc gia hai lp vi din tch S xut hin mt lc gi l lc ni ma st F. Hin nhin F t l vi u v S, t l ngc vi x, nn

F =

u S . x

(3.1)

H s t l l h s ni ma st, cn gi l h s nht. Ta tnh h s ny. S phn t vn chuyn qua S trong t l 1 n v S t. 6

N =

Hnh 7.2

T xung lng vn chuyn qua S trong t l P = n v S t mo (u1 u2 ) / 6 . Nhng u1 - u2 = - (u / x) 2 nn 1 3

P = v Lc F = P/t:

u S t . x u S . x

(3.2)

F = v

1 3

(3.3)

i chiu biu thc ny vi (3.1) cho kt qu sau v h s ni ma st

=

1 v. 3

(3.4)

7.4. TRUYN NHIT

Xt s truyn nhit gia hai im c nhit T1 v T2, vi T1 > T2. S chnh lch nhit dn ti vn chuyn nng lng nhit. Lp lun ging nh cc tit trn dn n biu thc sau y cho lng nhit truyn Q =

T S t x

(4.1)

trong l h s truyn nhit. H s truyn nhit c tnh nh sau cho cht kh. S ht truyn t A n B qua S trong t l 49

N12 =

1 n v S t. 6 1 i n v1S t. k BT1 6 2

T nng lng chuyn t A n B qua S trong t E12 = N121 ==

8k BT1 i i k BT1 n v1S t = n S t.k BT1 m0 12 123/ i kB 2 n S t T13/ 2 . = 3 2 m0

Hnh 7.3

Nng lng vn chuyn E21 t B n A xc nh bng cng thc tng t trong hon v cc ch s 1 v 2. Hiu nng lng vn chuyn s l3/ i kB 2 n S t (T13/ 2 T23/ 2 ) . 3 2 m0

E = E12 - E21 =V T13/ 2 T23/ 2 = nn

T13/ 2 T23/ 2 3T 1/ 2 T 2 , x = 2 x x

3/ 3/ i kB 2 i kB 2 3T 1/ 2 T T E = n S t n T 1/ 2 S t . 2 = 2 x x 3 2 m0 2 m0

Nhng theo (II.8.2) th n = 1/( 2 d 2 ) , d l ng knh phn t, kt qu l

E = =

3/ i kB 2 T T 1/ 2 S t , 3/ 2 2 x 2 m0 d

(4.2)

3/ 3/ i kB 2 ik i kB 2 n T 1/ 2 = B n v = T 1/ 2 . 3/ 2 2 4 2 m0 2 m0 d

(4.3)

7.5. NHIT NG HC XA CN BNG

c im chung ca cc trng thi xa cn bng: 1. Sinh entropy mnh, 2. Dng Ji ph thuc vo lc Xk mt cch phi tuyn. Nhit ng hc xa cn bng cn gi l nhit ng hc phi tuyn. Khi cc dng mnh v cnh tranh nhau c th lm xut hin cc trng thi dng (khng cn bng) c cc cu trc xc nh. Cc cu trc ny th hin bng cc hnh thi i xng no 50

, ni cch khc, h tn ti trong trng thi v m dng vi mt trt t xc nh. V cc cu trc ny tn ti khi h xa cn bng nn gi l cc cu trc tiu hao. Ngi ta cn gi y l hin tng t t chc ca cc h s ln. Nh vy khi h khng cn bng mnh th li c th xut hin trt t. Theo l lun nhit ng hc th trng thi c trt t, entropy phi gim thiu. Phn tch cho thy qu tht cc trng thi dng c xut hin cu trc tiu hao th entropy S c cc tiu a phng. Mt trong cc th d in hnh ca cu trc tiu hao l hin tng Benard, m t trn cc hnh 7.4 v 7.5. Ly mt bnh thy tinh hnh tr c y rng, ng mt t nc (Hnh 7.4). un nng bnh bng ngn la phn b tht u t y. Nh th y bnh gi nhit T1 cn mt trn ca nc c nhit T2, chnh lch nhit l T = T1 - T2. Khi T nh hn mt gi tr (T)c no th dn nhit t y bnh thc hin bng truyn nhit thng thng (da trn va chm phn t). Khi T vt qua gi tr ti hn (T)c th h mt cn bng qu mnh, thc hin dn nhit phi c thm qu trnh i lu. Trn Hnh 7.4 cc dng i lu trong bnh nc c ch ra bng cc ng cong c mi tn. S cnh tranh ca cc dng i lu lm cho mt nc c cu trc nh b mt t ong (Hnh 7.5).

Hnh 7.4

Hnh 7.5

Mt cu trc nh th tn ti c chng no m bo c iu kin T > (T)c. Khi cc iu kin m bo kiu nh th khng cn th cu trc t tan r. Mt th d khc l phn ng ha hc Belousov-Zhabotinsky. Phn ng ny phc tp, nhng vai tr trung tm l ba qu trnh ha hc sau: CH2(COOH)2 + 6Ce4+ + 2H2O 2CO2 + HCOOH + 6Ce3+ + 6H+, 10Ce3+

(5.1a) (5.1b) (5.1c)

+ 2HBrO3 + 10H

+

10Ce

4+

+ Br2 + 6H2O ,

CH2(COOH)2 + Br2 CHBr(COOH)2 .

Phn ng (5.1b) l t xc tc. Chn liu lng sao cho phn ng (5.1a) v (5.1b) din ra vi cng tc , nn lng Ce4+ sinh ra trong (5.1b) b li lng ion y mt i trong (5.1a). Nh vy lng ion Ce3+ gi khng i trong h. Axit bromomalonic CHBr(COOH)2 to thnh trong phn ng (5.1c) kt hp vi Ce3+ s cn tr phn ng (5.1b), dn ti gim lng Ce4+. Cht cn tr khng bn v phn tch thnh CO2 v axit dibromoacetic. Khi phn ng t xc tc (5.1b) li tip tc. Kt qu l din ra qu trnh bin i thun v ngc gia hai loi ion cerium: Ce4+ Ce3+. (5.2) Trong nc cc ion Ce4+ c mu lam cn cc ion Ce3+ mu . Khi cn xa cn bng th cc phn ng (5.1) lm cho nc trong bnh bin i mu dn t lam sang v sau ngc li.

51

l mt cu trc tiu hao, bin i theo thi gian, vi chu k khong 4 pht. Cu trc ny s mt dn khi h tin ti cn bng. Ngi ta cho rng cc vnh bi ca Sao Th l mt cu trc tiu hao theo quan im ca nhit ng hc khng cn bng: nhit bc x t Mt Tri l ngun nng lng to nn qu trnh khng cn bng y. Theo trn th cu trc tiu hao hnh thnh trn c s ba iu kin: 1/. H m, lun trao i vi mi trng ngoi. 2/. Trng thi xa cn bng, quan h dng v lc l phi tuyn. 3/. Sn sinh entropy ng vi mt cc tiu a phng. Theo l lun nh th, khi cc phn t hu c hp li to thnh mt t bo th chnh l hnh thnh mt cu trc tiu hao. Cc t bo hp li to thnh mt c th sng cng l hnh thnh mt cu trc tiu hao. l cch thc m nhit ng hc xa cn bng tham vng gii thch s hnh thnh v pht trin ca s sng. Ni chung cc h s ln m ba iu kin nu trn tha mn th u c kh nng hnh thnh cc cu trc tiu hao.

52

TI LIU THAM KHO

1. m Trung n, Phm Vit Knh: Vt l phn t v nhit ng hc. NXB i hc v THCN, H Ni 1974. 2. Lng Duyn Bnh: Vt l i cng Tp I: C-Nhit. NXB Gio dc, H Ni 2007. 3. A. Kikoin, I. Kikoin: Molecular Physics. Mir Publishers, Moscow 1978. 4. .. , .. , .. : . . . , 1978. 5. B. Linder: Thermodynamics. Wiley-Interscience Publ., New York 2004. 6. W. Nolting: Statistische Physik. Springer Verl., Berlin 2002. 7. P. Glansdorff, I. Prigogine: Thermodynamic Theory of Structure, Stability and Fluctuations. Wiley-Interscience Publ., London 1971.

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