1 HA L I (Physical Chemistry I) 2 Ni dung mn hc(Course Content)
Chng 1. Nguyn l th nht ca nhit ng ha hc Chng 2. Nguyn l th 2 ca
nhit ng ha hc Chng 3. Cn bng ha hc Chng 4. C s l thuyt cn bng pha
Chng 5. Cn bng pha ca h mt cu t Chng 6. Cn bng pha dung dch lng-hi
Chng 7. Cn bng pha dung dch lng-rn 3 Chng 1. Nguyn l th nht ca nhit
ng hc (NH) 1) i tng v phng php nghin cu 2) Khi nim c bn 3) Nguyn l
I ca nhit ng4) nh lut Hess 5) nh lut Kirchhoff 1. i tng v phng php
nghin cu a) Nhit ng lc hc (NLH): nghin cu lin h gia nhit, cng v cc
dng nng lng khc nhauCc quy lut xy ra trong h. C s ch yu ca NLH:
Nguyn l I: Mt trng hp ca nh lut bo ton v bin ha nng lng Nguyn l II:
Xt kh nng, chiu hng v gii hn ca cc qu trnh 4 b) NHH nghin cu ch yu
nhng vn sau: Nhit ha hc: Hiu ng nhit ca p..h.h, qu trnh ha l nh ha
tan, hp ph, chuyn pha (nng chy, ha hi, thng hoa) Cn bng ha hc: Kh
nng, chiu hng v gii hn ca p. Cn bng pha: iu kin cn bng pha, qui lut
ca cc qu trnh chuyn pha 5 1. i tng v phng php nghin cu (tt) 2. Nhng
khi nim c bn a. H nhit ngb. Trng thi v qu trnh. Hm trng thi v hm qu
trnh c. Ni nng Nhit v Cng. Quy c v du ca Nhit v Cng 6 a. H nhit ng:
L phn vt cht m chng ta nghin cu, nhng vt cht khc nm ngoi phm vi
nghin cu c gi l mi trng xung quanh. 7 H nhit ng Mi trng ngoi V tr H
kh H2 v O2 Mi trng 2. Nhng khi nim c bn (tt) H h 8 H knH c lp Cc h:
2. Nhng khi nim c bn (tt) 9 H on nhit: Khng trao i nhit vi mi trng
(Q=0). Lu : c th trao i cng A V d: Phch nc, Bnh gi nng caf H ng th:
tnh cht ca h khng thay i hoc thay i lin tc t im ny n im khc v trong
h khng c b mt phn chia.V d: Cc nc ng, chai c-ca-c-la, mt hn hp khng
kh 10 H d th: h c b mt phn chia. Tp hp nhng phn ng th ca h ging
nhau v mi tnh cht v c ngn cch khi cc phn khc ca h bng b mt phn chia
gi l pha (hay tng). V d: mt cc nc ang tanVy:- H ng th: 1 pha - H d
th: 2 pha tr ln b. Trng thi v qu trnh. Hm trng thi v hm qu trnh
Trng thi ca mt h: xc nh bi mt tp hp cc thng s l ha nh T, P, V, C,
Nhng i lng ny gi l cc thng s trng thi. 11 2. Nhng khi nim c bn (tt)
12 Thng s trng thi TS cng - Khng ph thuc vo lng cht - Khng c tnh
cng tnh - T,P, t khi, C TS khuch - Ph thuc vo lng cht - C tnh cng
tnh - V,m,P, nng lng b. Trng thi v qu trnh. Hm trng thi v hm qu
trnh (tt) Khithayibtkmtthngstrngthi(TT) no th h s chuyn t TT ny
sang TT khc, ta ni h thchinmtqutrnh(QT)hayqutrnhnhit ng (QTN) V d:
Qu trnh un nng h t T1 T2
Qu trnh gin n kh t V1V2 13 Qu trnhc im ng (Chu trnh)TT u TT cui
MTT u TT cui Cn bngCc TS ca h c nh hoc bin thin v cng chm. Thun
nghch(QT 1 2) (QT2 1) Khng TN(QT 1 2) (QT2 1) 14 b. Trng thi v qu
trnh. Hm trng thi v hm qu trnh (tt) Ch : QT no c ma st u khng thun
nghch Hm trng thi (HTT): l hm m bin thin ca n ch ph thuc vo TT u v
TT cui ca h, khng ph thuc vo cch tin hnh qu trnh. Vd: Nhit Q v cng
A l cc thng s nhit ng xut hin trong QT v c trng cho QT. Chng c
gilcchmqutrnh.Gatrcachngkhng nhngphthucTTu,TTcuimcnphthuc vo ng i
ca qu trnh. 15 b. Trng thi v qu trnh. Hm trng thi v hm qu trnh (tt)
16 K hiu ton cho hm trng thi v hm qu trnh: Mt bin thin v cng b ca
hm trng thi x s ghi l dx (dT, dp, dV, dS, dG, dH) Bin thin hu hn ca
n l:i vi chu trnh (Qa trnh ng):Mt i lng v cng nh ca hm qu trnh k
hiu x (Q,A) Bin thin trong qu trnh ca n:1212x x dx x = = A}} }= = =
A110 dZ dZ Z} }o = o =2121Q Q ; A A17 c. Ni nng Nhit v Cng. Quy c v
du ca Nhit v Cng Ni nng: tng NL bn trong h, bao gm nng lng chuyn ng
tnh tin, quay, dao ng, th nng tng tc gia cc phn t, nng lng cc mc e,
nng lng ht nhn U = Etnh tin +Equay+Edao ng+Eht,y + Ee + Enhn 18 Ni
nng U l mt hm trng thi, bin thin ca n ch ph thuc trng thi u v trng
thi cui ca h, khng ph thuc ng i ca qu trnh i vi bin i hu hn: i vi
chu trnh:1212U U dU U = = A}01211= = = A}U U dU Uc. Ni nng Nhit v
Cng. Quy c v du ca Nhit v Cng (tt) 19 Nhit v Cng: Hai cch khc nhau
chuyn nng lng: - Nuschuynchclinquanvistngcng chuyn ng phn t trong h
nhn nng lng th s chuyn c thc hin di dng nhit (Q)
Vd:vtlnhtipxcvtnng,ccphntchuynng nhanhcavtnngvachmviccphntchuynng
chmcavtlnhvtruyn1phnngnng,lmtngs chuyn ng cc phn t vt lnh. c. Ni
nng Nhit v Cng. Quy c v du ca Nhit v Cng (tt) 20 -
Nuschuynclinquannschuyndch ca h di tc dng ca lc no , th s chuyn c
thc hin di dng cng (A) Vd:khginntrongxi-lanhlmpittngchuyn
ng,khtruynnnglngchopittngdi dng cng c. Ni nng Nhit v Cng. Quy c v
du ca Nhit v Cng (tt) Quy c v du ca Nhit v Cng: H nhn nhit (quy c
+) s sinh cng (quy c +) H nhng nhit (-) s nhn cng (-) 21 H
Q>0A>0 c. Ni nng Nhit v Cng. Quy c v du ca Nhit v Cng (tt) 3.
Nguyn l I ca nhit ng Ni dung: NguynlI(NL1)lnhlutbotonv
chuynhannglngpdngvocchclin quan n s trao i nhit v cng vi mi trng bn
ngoi 22 Xt mt h nhn nhit lng Q v sinh cng A. Theo quy c v du Q v A
u + 23 3. Nguyn l I ca nhit ng (tt) Trng hp 1: 24 3. Nguyn l I ca
nhit ng (tt) TT u QT 1 QT 2QT 3 QT 4 Q =A (3.1) Trng hp 2:25 3.
Nguyn l I ca nhit ng (tt) TT 1TT 2 Q = U + A (3.2) 1 2U U U = AQ =
dU +A (3.3) 26 Nhn xt: - T (3.1) Q = A, nu A > 0 th Q > 0,h
mun sinh cng th phi nhn nhit. Nu1ngchotngtunhonsinhcngmkhngcn
nhnnhitlngcvnhculoi1thcthkhngnh: Khng th c ng c vnh cu loi 1. cng l
mt cch pht biu ca NL 1 -T (3.2) Q = U + A, nu Q = 0 v A = 0 th U =
0, U = Const. Ni nng h c lp c bo ton. cng l 1 cch pht biu ca NL 1
3. Nguyn l I ca nhit ng (tt) 27 Pht trin biu thc tng qut ca NL
1
(d cho hm trng thi U, cho hm qu trnh Q, A) A l cng chng li lc
ngoi: A= PdV + A cng c ch cng gin n NL 1 tr thnh: Q = dU + PdV + A
(3.4) Trong nhit ng hc thng xt cc h khng chu tc ng ca in trng, t
trng. Khi ch c cng gin n pdV. NL 1 thng l: Q = dU + PdV (3.5) 3.
Nguyn l I ca nhit ng (tt) Q = dU +A (3.3) Pht trin biu thc tng qut
ca NL 1 28 - T (3.5), suy ra trong qu trnh ng tch (V = Const): pdV
= 0, Qv = dU Qa trnh hu hn: Suy ra: Qv = Uv - Trong qu trnh ng p (P
= Const): Qp = dU + PdV + VdP = dU + d(PV) = d(U+PV) i lng H = U +
PV c gi l entanpi. U, P, V l cc hm trng thi, H l hm trng thi. H: hm
entanpi hay cn gi l hm nhit. Ta c: Qp = dH Suy ra: Qp = Hp 29 - Mi
quan h gia Qp v Qv QP = HP = (U+PV)P = UP + PVP
Qv = Uv , U ch ph thuc T, Suy ra UP = UV
Vy QP = QV + PV i vi p..h.h, nu l cht rn v lng, bin thin th tch
V trong qu trnh p. l khng ng k, ta c: QP = QV i vi p. c cht kh tham
gia, v cc kh c coi l l tng. PT trng thi PV = nRT, PV = nRT Vy QP =
QV + nRT, n l bin thin s mol kh trong qu trnh p.. Ta tnh c nhit ca
p. trong qu trnh ng p khi bit nhit ca p. trong qu trnh ng tch v ngc
li. 30 Nhit dung (C): - nh ngha: Nhit dung l nhit lng cn thit nng
nhit ca h ln 1 . (Ni cch khc nhit dung l nhit lng cn thit lm nng h
ln 1 ) n v ca C l Cal.K-1 hoc Jthc .K-1 Hai iu kin hay gp trong t l
ng p v ng tch: - QT ng tch V=Const - QT ng p P=Const VVdTQC|.|
\|=oPPdTQC|.|
\|=odTQCo=Mt s ng dng ca NL 1 31 - Quan h CP v CV :
Suy ra: PV=nRT, p dng cho 1 mol, PV=RT, ta c d(PV) = d(RT) =RdT
U ch ph thuc nhit Vy CP CV = R (H thc mayer) CP>CV , H ng p,
nhit cp cho h khng nhng lm tng ni nng ca h m cn sinh cng gin n. Cn
QT ng tch V = Const, nhit cp cho h ch dng tng ni nng. ) ( dH
QdTdHdTQCPP PP=|.|
\|=|.|
\|= oo) () ( ) (V VV VVP P PPdU QdTdUdTQCdTPV ddTdUdTPV U
ddTdHC=|.|
\|=|.|
\|=|.|
\|+|.|
\|=+=|.|
\|=ooMt s ng dng ca NL 1 (tt) 32 -S ph thuc ca nhit dung vo nhit
: Nhit dung thng thng l mt hm ca nhit : CP = a+bT+cT2. Hoc CP=
a+bT+cT-2 a,b,chng s kinh nghim V d: Hi nc,
CP=36,8-7,9.10-3T+9,2.10-6T2 (J.mol-1.K-1) Q a 5 mol hi nc t 100C
ln 200C, P=Const ? kJ Q dT T Qmol n dT C n QdTQCP PTTP PPP586 , 17
) 10 . 2 , 9 10 . 9 , 7 8 , 36 ( 5) (4733732 6 321= + ==|.|
\|=}} oMt s ng dng ca NL 1 (tt) 33 Nhit dung trung bnh: nhit
dung trong mt khong nhit T1 n T2 , k hiu C: }}= = ==A==2121) (1) (1
21 21 2TTTTCdTT TCT T C CdT QdTQCTQT TQCo(Nhit dung thc) Mt s ng
dng ca NL 1 (tt) Nhit dung theo thuyt ng hc cht kh: 34 Mt s ng dng
ca NL 1 (tt) RidTdUCRTiUVV22=|.|
\|==i: bc t do,i=3 cho kh n nguyn t nh He, Ar, i=5 cho kh lng
nguyn t R C R C C R CP V P V2523= = =R C R CP V2725= = Tnh ton A, Q
v U trong cc QT: NL 1: 35 Mt s ng dng ca NL 1 (tt) ) (2121Const P
pdV A PdV ACdT QdTQCA U QA dU QVVTT= = == =+ A =+ =}}ooo oTH1) Qa
trnh ng tch (V=Const) Cng gin n: AV = 0 (dV=0) ) ( ) (1 221Const C
T T C QdT C U QV V VTTV V V= == A =}36 TH2) Qa trnh ng p (P=Const)
Cng gin n: Nhit ng p: Ni nng:V P V V P PdV AVVPA = = = }) (1 221) (
) () (1 221Const C T C T T C H QdT C H QH PV U QP V V P U V P U QP
P P P PTTP P PPP= A = = A == A =A = + A =A + A + A = A + A =}P PA Q
U = AMt s ng dng ca NL 1 (tt) 37 TH3) Qa trnh ng nhit T=Const Ni
nng ch ph thuc nhit , dU=0, UT=0, Q=U+A, QT=AT Cng gin n ng nhit xc
nh bi t l th tch cui v th tch u hoc p sut u v cui. 212 2 1 112ln)
(ln) 1 (2121PPRT AV P V PConst T Const RT PVVVRT Amol dVVRTPdV
ATTVVVVT=== = === =} }Mt s ng dng ca NL 1 (tt) 38 TH4) Qa trnh on
nhit (Q=0, Q=0) QT tin hnh trong iu kin c lp v nhit, Q=0 NL1,
Q=U+A, An=-U, An=?P, V, T u bin i, khng th tnh A theo PT trng thi
PV=RT ! Thit lp biu thc phng trnh pot-xng: Const P TConst V PConst
V T=== 11...Mt s ng dng ca NL 1 (tt) 39 - Thit lp:
Const V T =1.) ( . ; ..ln ) 1 ( ln) ( ln ) 1 ( ln0 ln ) 1 ( ln0
) 1 (0) (0) (011RT PV v Const P T Const V PConst V TConst V TConst
d V d T dV d T dCCVdVCCTdTVdVCC CTdTdVVT C CdT CR C C dVVRTdT CRT
PV dT C dU O PdV dU A du QVPVPVV PV PVV P VV= = === += += += =
+=+=+= = += = = + = + =} } } o oMt s ng dng ca NL 1 (tt) 40 - Tnh
An Ch : Cc i lng A, U, Q tnh trn cho 1 mol kh. (((
||.|
\|= = ====== = } } / ) 1 (12 12 1 2 11 1 2 21112211212111) ( )
(1) . (1. .1. .1.PP RTT T C T TRAConst V PV P V PAV Const V Const
VConst AdVVConstPdV AV nnnnMt s ng dng ca NL 1 (tt) 41 Chng minh L
Hess: P.: aA+bB=cC+dD T P..H.H: thu nhit (Q>0) hay ta nhit (Q 0)
A cng dn n }=21PdV A71 Qu trnh ng tch , dV=0 A=0 Qv A= AU C phh
A> 0 Qv A = AU Qv A hiu ng nhit ng tch Khng c p ho hc A=0 Qv= AU
Qv-nhit ng tch
Hiu ng nhit ca phn ng k ng tch bng bin thin ni nng ca h. 72 Khng
c phhA=0QP = AH , Qp l nhit ng p Qu trnh ng p P= const A = P
(V2-V1) QP A = (U2 U1) + P(V2 V1) Qp A = (U2 + PV2) (U1 + PV1)
Entanpi H = U + PV Qp A = H2 H1= AH C phh A>0 Qp A = AH(Qp-A )
gi l hiu ng nhit ng p Hiu ng nhit ca phn ng k ng p bng bin thin
entanpi ca h. 73 Enthalpy l hm trng thi , thng s dung . a s cc phn
ng ho hc thng c tin hnh k p sut khng i nn hiu ng nhit ca phn ng c
xc nh bng AH ca phn ng. 74 75 }= A =o=21TTpppdT . C . n HdTdHdTQC}=
A =o=21TTvvvdT . C . n UdTdUdTQCNhit dung ng p , nhit dung ng tch
76 2. Hiu ng nhit ca cc qu trnh ha hc v phng trnh nhit haa. Hiu ng
nhit ca cc qu trnh ha hc b. Phng trnh nhit ha c. Nhit to thnh v
nhit t chy 77 Hiu ng nhit ca cc qu trnh ha hc Llng nhit Q m h thu
vo hay pht ra trong qa trnh ha hc .Qu trnh ng pQp = AH P to nhitP
thu nhit AH = (H)sn phm - (H)cht u 78 Quan h gia AH v AU trong qu
trnh ng p ,ng nhit AH = AU + P.AV Phn ng ch c cht rn, cht lng AV ~
0 nn AH ~ AU Phn ng c cht khP.AV = An.R.T (xem kh l kh l tng) AH =
AU + An.R.TAn = (s mol kh)sp - (s mol kh)c
tnh trong phng trnh phn ng 79 Quan h gia Qv v Qp 1(V,P, T,U1
,H1) Cht u 3(V,P,T,U3) Sn phm , Qv 2(V,P,T,U2 ,H2) Sn phm , Qp Qp=
H2-H1= U2 - U1+An.R.T = U3-U1+ An.R.TQp-Qv = An.R.TP=const V=const
80 So snh nhit phn ng Qp v Qv 2CO(k)+ O2 (k) = 2CO2(k) QP = AH =
-566 kJ Qv = AU = ? Qp-Qv = An.R.T PAV = An.R.T=
(-1).(8,314).298.10-3 = -2,5kJQv = Qp - PAV= -563.5 kJ 81 Phng trnh
nhit ha hc AH c tnh vi gi thit : p xy ra hon ton , cht u v sn phm
theo lng hp thc v trng thi vt l tng ng vi p.Nhit c th bin i trong
qu trnh p nhng mi tnh ton c thc hin khi nhit cui bng nhit u. 0298H
A2C(gr) + 2O2(k) = 2CO2(k) ;AH0298= -787,02kJ y l p to nhit. 82
Trong iu kin bnh thng (nhit thp), phn ng ta nhit (AH < 0) l phn
ng c kh nng t xy ra . Hiu ng nhit tiu chun :AH0T p sut chun 1 atm
Nhit T tu , thng chn 250C Cht u v sn phm trng thi chun v cng nhit .
83 Nhit to thnh lhiu ng nhit ca phn ng to thnh 1 mol cht t cc n cht
ng vi trng thi t do bn nht trong nhng iu kin cho v p sut v nhit .
iu kin chun, nhit to thnh tiu chun k hiu (AH0298)tt
Cc n cht bn k chun nh: Cl2(k); H2(k); O2(k); N2(k); Br2(lng);
I2(r); C(gr) (AH0298)tt ca cc n cht bn k chun = 0 (AH0298 )tt ca
ion H+.nH2O =0 84 -1131Na2CO3(s)49C6H6(l)-92HCl(g)
-127AgCl(s)-278C2H5OH(l)95.4N2H4(g)
-1260C6H12O6(s)-201CH3OH(g)-46NH3(g)
-207NO3-(aq)-85C2H6(g)-286H2O(l) -240Na+(aq)52C2H4(g)-394CO2(g)
106Ag+(aq)-74.8CH4(g)-111CO(g) Nhit to thnh tiu chun, AHf (kJ/mol)
Phn ng phn hu l phn ng nghch ca p to thnh. 85 NHIT TO THNH TIU CHUN
86 Nhit t chy lhiu ng nhit ca phn ng t chy 1 mol cht bng kh oxy to
thnh cc oxyt cao bn k phn ng.C2H4(k)+ 3O2(k) 2CO2(k) + 2H2O(l)
(AH0298)c =-1411 kJ/mol 0. 298 dcH A0. 298 ttH A0. 298 ttH AK hiu
-nhit t chy tiu chun (AH0298)c
n v -[kcal/mol] hay [kJ/mol] 87 Nhit t chy tiu chun ca kh CO2 v
nc lng c qui c bng khng Nhit to thnh nguyn t l hiu ng nhit ca p to
thnh 1mol hp cht trng thi kh t cc nguyn t cng trng thi kh. Hiu ng
nhit ca qu trnh ho tan l hiu ng nhit ca qu trnh ho tan 1mol cht tan
nn chu nh hng ln n lng v bn cht ca dung mi. Khi lng dung mi ln th
AHht khng ph thuc vo dung mi. 88 3. nh lut Hess v cc h qu a. nh lut
Hess b. Cc h qu 89 nh lut Hess Hiu ng nhit ca phn ng ha hc iu kin
ng p hoc ng tch ch ph thuc vo bn cht v trng thi ca cc cht u v sn
phm cui ch khng ph thuc vo ng i ca qu trnh, ngha l khng ph thuc vo
s lng v c im ca cc cht giai on trung gian. 90 AB , AH1 = HB -HA
AH1 = HB Hc+Hc-HA CAH1 = AH3 + AH2
Trong cng mt iu kin , hiu ng nhit ca mt phn ng bng tng hiu ng
nhit ca cc phn ng trung gian. AH1 AH2AH3
91 3H2 (k) +N2(k)= H2 (k) +N2H4(k) AH01 = 95,4 kJ N2H4(k)+ H2(k)
= 2NH3(k) AH02= -187,6 kJ 3H2(k)+N2(k)= 2NH3 (k) AH03 = -92,2kJ 92
The Solution Process for NaCl AHsoln = Step 1 + Step 2 = 788 784 =
4 kJ/mol 6.7 93 H qu 1- nh lut Lavoisier La Place A BAHthun = HB-HA
BAAHnghch = HA-HB AHthun = - AHnghch AHto thnh = - AHphn hu 94 H qa
2. Chu trnh Born - Haber NaCl (r) Na+(k)+ Cl-(k) Na(r) + 1/2Cl2(k)
Na (k) + Cl (k) Um= ????? -(AH0298)tt NaCl (AH0298)th Na
1/2(AH0298)pl Cl2 I1Na FCl Um= -(AH0298)tt NaCl + (AH0298)th Na+1/2
(AH0298)plCl2 + I1 Na+FCl 95 H qu 3 A A = Attcdttsp puH H H A A =
Attcdttsp puH H H96 aA + bB=cC + dD ;AHp AH = [cAHtt(C) + dAHtt(D)]
[aAHtt(A) +bAHtt (B)] CH4 (g) + 2 O2 (g) CO2 (g)+2 H2O (l) AH =
(AHtt CO2 + 2 AHtt H2O)- (AHtt CH4 +2 AHtt O2) = (-394+2 x
-286)(-75+0) =(-1066) (-75) =-891 kJ/mol 97 H qu 4 A A = Attcdttsp
puH H H A A = Adcspdccd puH H HC(gr) + O2(k) = CO(k) AH0298 =?
C(gr) + O2(k) = CO2(k),(AH0298)c = - 393.5kJ/mol CO2(k) = CO(k) +
O2(k), - (AH0298)c = +283.0kJ/mol C(gr) + O2(k) = CO(k)AH0298 =
(AH0298)cC-(AH0298)c CO=-110,5kJ 98 aA + bB=cC + dD ;AHp AHp =
[aAHc(A) + bAHc(B)] - [cAHc(C) + dAHc(D)] 99 H qu 5- AHp = (Elk)c
-(Elk)sp AHp = (Elk)t -(Elk)tt H-H(k)+Cl-Cl(k)2H-Cl(k) 2H(k) ElkH2
+ 2Cl(k) ElkCl2 -2ElkHCl AH0298 = [Elk(H2 )+Elk(Cl2)] [2Elk(HCl)]
AH0298 =? Cc cht trong phn ng trng thi kh 100 101 Calculate AH (in
kilojoules) for: C2H4 (g) + H2O (g)C2H5OH (g) AH = D (Bonds Broken)
D (Bonds Formed) =(H-O+C=C)-(H-C+C-O+C-C) = O H HHHHH+CCHOHHH
HHCC102 T2 aA + bB cC+dDT2 T1 aA + bBcC+dD T1 }21TTpAC . aAH0 T2 dT
. bC21TTpB}dT . aC21TTpA}AH0 T 1 dT . cC21TTpC}dT . dC21TTpD}103 aA
+ bB cC+dD dT ) C . b C . a ( ) C . d C . c ( H HB A21D C 1 2p pTTP
p0T0T+ + + A = A}dT ) C . b C . a ( ) C . d C . c ( H HB A21D C 1
2p pTTP p0T0T+ + + A = A}4. S ph thuc ca hiu ng nhit vo nhit Phng
trnh Kirchhoff 104 Phng trnh Kirchhoff , khi ACp = f(T) ( )1 2 2 1T
T C H Hp A + A = A) T T ( C H dT . C H H1 2 p0TTTp0T0T1211 2 A + A
= A + A = A}AH thay i theo nhit khng nhiu
