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8/13/2019 BaiGiangVLDC VUI - 4ChuongCuoi
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BAI GING VT LY I CNG 1
1
CHNGIV. C NNG V TRNG LC TH
4.1.CNG V CNG SUT
4.1.1. Cng
Gis c F
khng i, im t ca
n chuyn di MM ' S
Cng A do F
sinh ra:
A = F.S.cos hayA F.SDthy F.Cos l hnh chiu SF ca F trn MM
sA F .S (4.1)Cng l i lng v hng, gi trththuc gc Nu im t ca lc F
chuyn di trn ng cong C D.
Cng ca lc F
trong chuyn di v cng nhdS
l: dA F.dS
Cng A ca F
trong chuyn di CD l:
CDCD
SdFdAA
(4.2)
(4.2) l biu thc tnh cng tng qut.
4.1.2. Cng sut.
Cng sut c trng cho sc mnh ca cc my (P).
Gistrong t, mt lc sinh cng A.
Ta c:tb
AP
t
(4.3)(4.3) l cng sut trung bnh ca lc trong t.
tnh cng sut ti tng thi im, ta cho t 0, khi A
t
dn ti mtgii hn l cng sut tc thi P
t
AP lim
t 0 hay dAP dt (4.4)
Cng sut c gi trbng o hm ca cng theo thi gian.
M
Hnh 4-1 S
F M
F
M
M
F
dS
C D
Hnh 4-2
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BAI GING VT LY I CNG 1
2
Ta cng c th vit cng sut di dng mt biu thc khc:
Do:dS
dA F.dS P F. F.vdt
(4.5)
Vy, cng sut bng tch phn v hng ca lc tc dng vi vectvn tc
ca chuyn di.
4.2. NNG LNG V NH LUT BO TON NNG LNG
nh ngha:Nng lng l i lng c trng cho mc vn ng ca
vt cht.
Ta bit cng l i lng c trng cho qu trnh trao i nng lng gia
vt ny v vt khc. Ni cch khc: khi mt hthc hin cng th nng lng ca n
sbin i.Gistrong qu trnh no hbin i ttrng thi 1 sang trng thi2; qu trnh ny hnhn tngoi cng A, ta c bin thin nng lng:
W2- W1= A (4.6)
bin thin nng lng ca mt h trong qu trnh no , c gi tr
bng cng m hnhn c tbn ngoi trong qu trnh .
- Nu hnhn cng t ngoi A > 0 th nng lng htng.
- Nu hsinh cng cho bn ngoi A < 0 th nng lng hgim.
-Nu hc lp, ta c A = 0; W2 = W1= const
Vy, nng lng khng tmt i m cng khng tsinh ra, nng lng ch
chuyn thny sang hkhc - nh lut bo ton nng lng:
Ch :
+ Khng thc mt hsinh cng mi mi m khng nhn thm nng lng t
mt ngun bn ngoi.
+ ng cvnh cu l mt hsinh cng mi mi m khng nhn nng lng t
ngun bn ngoi.
4.3. NG NNG VA CHM
4.3.1. ng nng
Xt cht im khi lng m chu tc dng F
v chuyn di tvtr 1 2.
Ta c cng ca lc F
l: A F.dS2
1
mdv
F ma m.dt
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BAI GING VT LY I CNG 1
3
dv dS v
A m. dS m. dv m.v.dv m.ddt dt
2 2 2 2 2
1 1 1 1 2
mv mvmv
A d A
2 2 22
2 1
1
2 2 2 (4.7)
T(4.7) biu thc ng nng: dmv
W 22
Khi (4.7) thnh: W2- W1= A (4.8)
nh l ng nng:bin thin ng nng ca mt cht im trong mt
qung ng no , c gi trbng cng ca ngoi lc tc dng ln cht im
sinh ra trong qung ng .
ng nng ca vt rn quay:d
IW
22
ng nng ca vt va quay va tnh tin:d
I mvW
2 22 2
4.3.2. Va chm xuyn tm
Xt hai qucu c khi lng l m1v m2. Trc khi va chm c: v1
, v2
cng
phng; sau va chm chng c vectvn tc 'v1
, 'v2
cng phng nhban u.Ta githit h(m1+ m2) c lp. Ta hy tnh
'v1v 'v
2:
Theo nh lut bo ton ng lng, ng lng ca h trc v sau va
chm:
' 'm v m v m v m v1 1 2 2 1 1 2 2
(4.9)
Ta phi tm thm mt phng trnh na i vi 'v1 v 'v2 , mun vy ta xc
nh iu kin khi va chm gn vi hai bi ton cth
a. Va chm n hi.
Do ng nng ca hbo ton, ta c:
m v' m v' m v m v 2 2 2 21 1 2 2 1 1 2 22 2 2 2
(4.10)
T (4.9) v (4.10) ta suy ra:
1
v
dS
F
2
Hnh 4-3
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BAI GING VT LY I CNG 1
5
MN t(M) t(N)A W W (4.15)V d:
- Trong trng trng u, cht im c cao h: tW ( h) m.g.h C- Trong in trng, thnng ca in tch q0ti vtr cch q mt on r:
tqq
W (r) k C.r
0
Tnh cht:
- Hiu thnng gia hai vtrhon ton xc nh.
- Trng lc v thnng c hthc:
MN t( M ) t (N )MN
A FdS W W
(4.16)
Nu cht im dch chuyn theo mt vng kn: 0SdF
(4.17)
ngha ca thnng:L dng nng lng c trng cho tng tc.
V d:
- Dng thnng ca cht im trong trng trng ca qut l nng lng
c trng cho tng tc gia qut vi cht im.
- Thnng ca in tch q0trong in trng Culong ca in tch q l th
nng tng tc gia q v q0.
4.4.3. nh lut bo ton cnng trong trng trng
Khi cht im c khi lng m chuyn ng tM N trong trng lc
th:
Ta c: MN t(M) t(N)A W W
Theo nh l vng nng:
MN (M) (N)A W W t(M) t(N) (N) (M)W W W W Hay: W+ Wt)N= (W+ Wt)M= const.
Suy ra: WC= W+ Wt= const (4.18)
Tng ng nng ca cht im trong trng lc thc gi l cnng ca
cht im
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BAI GING VT LY I CNG 1
6
nh lut bo ton cnng:Khi cht im chuyn ng trong trng lc
thth cnng ca cht im trong trng lc thl i lng bo ton.
Hqu:v W = W+ Wt= const trong qu trnh chuyn ng ca cht
im nu Wtng th Wtgim v ngc li.
Ch :Khi cht im chuyn ng trong trng lc thcn chu tc dng
ca mt lc khc F
( th dms
F
)cnng khng bo ton.
4.5. TRNG HP DN
4.5.1.nh lut niutn vlc hp dn vtr. ng dng
Hai cht im khi lng m v m t cch nhau mt khong r sht nhau
bng nhng lc c phng l ng thng ni hai cht im , c cng t
lthun vi hai khi lng m v m v tlnghch vi bnh phng khong cchr.
Biu thc:2
m.m'F F' G.
r (4.19)
Trong : G = 6,67.10-11Nm
kg
2
2gi l hng shp dn
ng dng:a. Sthay i gia tc trng trng theo cao.
Xt vt c khi lng m trn mt t:M.m
P G.R
0 2 (4.20)
vi: P0= mg0vM
g G.R
0 2gi l gia tc trng trng trn mt t.
Ti im cch mt t mt cao h, ta c :
M.mP G. m.g
R h 2 (4.21)
Gia tc trng trng cao h:M
g G.R h
2 (4.22)
Do :R
g g .
R h
2
0
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BAI GING VT LY I CNG 1
7
Ta thy:R h
R h Rh
R
2 2
2
11
1
Ta chxt vt cao h
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BAI GING VT LY I CNG 1
8
Khi lng ca Mt tri l:. .R'
M 'T .G
2 3
2
4 (4.27)Thay sv tnh ton, ta c : M = 2.1030(kg)
4.5.2. Trng hp dn
* Khi nim trng hp dn.
- Xung quanh mt vt c khi lng tn ti mt trng hp dn m bt k vt c
khi lng t trong khng gian ca trng u chu tc dng ca lc hp dn.
-V d: Trng hp dn ca tri t chnh l trng trng ca n.
* Bo ton momen ng lng trong trng hp dn.
Xt cht im khi lng m chuyn ng trong trng hp dn ca cht
im M cnh ti O. Chn O lm gc to.
Ta c:dL
/ o(F)dt
M
m F
lun hng vo tm O
/ o(F) 0M v dL
dt 0
L Const * Kt lun: Cht im m chuyn ng trong trng hp dn ca mt cht im
M th momen ng lng ca m c bo ton.
* Hqu: Cht im m chuyn ng trn quo phng, mt phng quo
ca cht im vung gc vi vect L .
* Tnh cht thca trng hp dn.
C lc F
tc dng ln cht im m chuyn ng trong trng hp dn ca
M tA n B. Cng ca F
trong chuyn di dS
: dA F.PQ F.PQ.CosNu ta vQH OP th PQ. cos = PH dA F.PHv PQ
l chuyn di vi phn nn ta t OP r
OH OQ r dr
(m)
F
L
O
(M)
v
Hnh 4-4
B
rA
rB
r +dr
r
F
P
H
Q
A
O
Hnh 4-5
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BAI GING VT LY I CNG 1
9
Khi PH OH OP r dr r dr mM
dA Fdr G dr r
2
Cng ca lc F
di chuyn m tA n B l:
B B
A A
r r
AB
B Ar r
Mm Mm MmA Fdr G dr G G
r r r 2
AB
A B
Mm MmA G G
r r
(4.28)T(17) ta thy trng hp dn ca cht im M l mt trng lc th.
Thnng ca cht im m ti vtr cch O mt khong cch r l:
t r
mMW G C
r
* Bo ton cnng trong trng hp dn.
- Cnng ca cht im m chuyn ng trong trng hp dn c bo ton:
W = W+ W
t=
mv mMG Const
r
2
2
(4.29)
- Hqu: khi r tng, thnng tng th ng nng gim v ngc li.
4.5.3. Chuyn ng trong trng hp dn ca tri t
Nu tmt im A no trong trng hp dn ca Tri t, ta bn i
vin n c khi lng m vi vn tc u l v0 th l thuyt v thc nghim
chng ttutheo trsca v0 c thxy ra cc trng hp sau:
a)Vin n ri trvTri t.b)Vin n bay vng quanh Tri t theo quo kn hnh elip.
c)Vin n bay ngy cng xa Tri t.
*Trsvn tc ban u v0 cn thit bn vin n bay vng quanh Tri t
theo quo kn hnh elip gi l vn tc vtrcp I.
* Trsvn tc ban u v0 cn thit bn vin n bay ngy cng xa Tri t
gi l vn tc vtrcp I.
* Tnh vn tc vtrcp I.
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BAI GING VT LY I CNG 1
10
Gismt vin n bay cch mt t khng xa coi bn knh quo
bng bn knh R ca Tri t:
Vn tc vIca vin n lin hvi gia tc hng tm
II
v
a g v g RR 2
0 0 0 (4.30)
viM
g GR
0 2
tnh vI= 7,9 km/s 8km/s
Vy, nu v < 8 km/s n ri vTri t
v > 8 km/s n chuyn ng quanh Tri t theo quo elip.
* Tnh vn tc vtrcp I I .
Gisn xut pht tA cch tm Tri t mt khong bng bn knh
R ca Tri t vi vn tc v0v bay ngy cng xa n .Bo ton cnng p dng cho n:
mv Mm mv MmG G
R
2 2
0
2 2
Domv 2 0
2nn
mv mM GMG v
R R 20
0
2
2 m
GMg
R
0 2
v g R0 0
2 . Gi trti thiu ca v0l vn tc vtrcp II.
K hiu IIv g R02 = 11,2km/s
CU HI N TP V BI TP CHNG 4
1. Khi no ni lc thc hin cng. Vit biu thc cng ca lc trong trnghp
tng qut. Nu ngha ca cc trng hp: A> 0, A < 0, A = 0.
2.Phn bit cng v cng sut. n v ca cng v cng sut?
3.Khi nim v nng lng, nh lut bo ton nng lng v ngha ca n.
Nu cc thnh phn ca c nng. Nu ngha ca ng nng v th nng.
4.Khi nim v trng lc th? Tnh cht ca trng lc th, p dng chotrng
lc th ca qu t?
5. Chng minh nh l ng nng v nh l th nng. ng nng ca mt cht
im c c xc nh sai khc mt hng s cng khng? Ti sao?
6. Chng minh nh lut bo ton c nng trong trng trng.
7. Ti sao ni th nng c trng cho s tng tc gia cc vt?
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BAI GING VT LY I CNG 1
11
8. Thit lp nh lut bo ton c nng. Xt trng hp h gm cht im v qu
t.
9. Mt t khi lng 10 tn ang chy trn on ng phng ngang vi vn
tc khngi bng 36km/h. Sau khi tt my v hm phanh, t chy chm dn
v dng li. H s ma st ca mt ng l 0,3 v lc hm ca phanh bng82.103N. Ly gia tc trng trng g = 9,8m/s2. Hy xc nh cng ca lc ma st
v on ng t i c t khi tt my n khi dng li. p
s: Ams 20,9 . 106J ; s 355m
10. Mt xe chuyn ng t nh xung chn ca mt
phng nghing DC v dng li sau khi i c mt
on ng nm ngang CB (hnh v). Cho bit AB = s =
2,5m; AC = l = 1,5m; DA = h = 0,5m. H s ma st k
trn cc on DCv CB l nh nhau. Ly gia tc trng
trng g = 9,8m/s2. Hy xc nh h s ma st v gia tc
ca xe trn cc on DC v CB.
p s: 0,2; 1,24m/s2; -1,96m/s2.
11.Mt t khi lng mt tn, khi tt my chuyn ng xung dc th c vn
tc khng i v = 54km/h. nghing ca dc l 4%. Hi ng ct phi c
cng sut bao nhiu n ln c dc trn cng vi vn tc 54km/h. S:
11,8kW12. Mt chic xe khi lng 20000kg chuyn ng chm dn u di tc dng
ca lc ma st bng 6000N. Sau mt thi gian xe dng li. Vn tc ban u ca
xe l 54km/h. Tnh:
a) Cng ca lc ma st;
b) Qung ng m xe i c ktlc c lc ma st tc dng cho ti khi xe
dng hn.
S: 6| 2,25.10msA J ; 375S m
13.Tnh cng cn thit cho mt on tu khi lng m = 8.105kg:
a) Tng tc tv1= 36km/h n v2= 54km/hthng
b) Dng li nu vn tc ban u l 72km/h.
S: a) A = 5.107J; b) A = -1,6.108J
14.Mt vin n khi lng m = 10kg ang bay vi vn tc v=100m/s th gp
mt bn gdy v cm su vo bn gmt on s = 4cm. Tm:
a) Lc cn trung bnh ca bn gln vin n;
b) Vn tc vin n sau khi ra khi bn gchdy d = 2cm.
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BAI GING VT LY I CNG 1
12
S: a) 1250cF N
; b) v = 71m/s
15.Mt vt khi lng m = 10kg trt tnh mt mt phng nghing cao 20m
xung. Khi ti chn dc vt c vn tc 15m/s. Tnh cng ca ma st.
S: A = -835J
16.Mt t khi lng 20 tn ang chuyn ng vi vn tc khng i trn on
ng phng nm ngang th phanh gp. Cho bit t dng li sau khi i thm
c 45m. Lc hm ca phanh xe bng 10800N. H s ma st gia bnh xe v
mt ng bng 0,2. Ly gia tc trng trng g = 9,8m/s 2. Hy xc nh:
a.Cng cn ca cc lctc dng ln t.
b.Vn tc ca t trc khi hm phanh.
p s:a) A = -2,25.106J. b) v =15m/s.
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BAI GING VT LY I CNG 1
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CHNGV. DAO NG V SNG C5.1. Dao ng c hc iu ho, dao ng tt dn, dao ng cng bc5.1.1. Dao ng c hc iu ho.
a. Phng trnh ng lc hc dng vi phn ca vt dao ng iu ho(dh)
Lc tc dng ln vt dao ng iu ho: f = -k.x (5.1)
Trong : f l tc dng ln vt chuyn ng: Hng v v tr cn bng,
cng t l vi lch khiv tr cn bng ca vt; x l ta ca vt dao
ng (vi gc O ti v tr cn bng) ; k l h s t l. Gi f l lc phc hi.
p dng nh lut th II Niutn:f = max =mx
''
thay vo biu thc (5.1) ta c:
x'' +o2 x=0 (5.2)
(5.2) l phng trnh ng lc hc dng vi phn
Trong phng trnh trn o2 = k/m l hng s ph thuc vo c tnh ca h
dao ng.V d v biu thc xc nh k v oca mt s h.
+ Con lc lxo: k l h s n hi ca l xo v o2=k/m
+ Con lc n: k = mg/l v o2 = k/m = g/l (g l gia tc trng trng)
+ Con lc vt l: k = m2.gl/I v o2= mgl/I (vi I l m men qun tnh)
b.Phng trnh dh.
Nghim ca phng trnh 5.1 c dng l
X = A sin(ot+) ( 5.3)
x l li ca vt dao ng , l hm iu ho i vi thi gian
c. Cc c trng cho dh.
Chu k T= 2/o
Tn s f= 1/T
Vn tc: v= Aocos(t+)
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BAI GING VT LY I CNG 1
14
Gia tc: a= -o2 Asin(t+)
Nng lng (dng c nng): ng nng hoc th nng bin thin iu ho
cng tn s vi li v vn tc, nhng nng lng ton phn ca dao ng
khng thay i theo thi gian: 2o
22 mA
2
1kA
2
1W
= const
5.1.2. Dao ng c hc tt dn.
a.Phng trnh ng lc hc dng vi phn.
Lc tng hp tc dng ln vt dao ng trong thc t cn phi k n lc
ma st hoc lc cn tc dng ln vt trong qu trnh dao ng. Xt con lc l xo
trong mi trng nht, ngoi thnh phn lc phc hi (-kx) cn c thm thnh
phn lc cn chuyn ng )dt
dxr( . Do lc tng hp l:
dt
dxrkxf ( r l h s nht) (5.4)
p dng nh lut th II Niutn f = max =mx'' , thay vo biu thc (5.4) ta
c:2
2
d x m
dt
dx dxkx r mx kx r
dt dt
2 22
o2 2
d x r d dx: 0 ; 2 2 x 0 (5.5)
dt m dt
r dx k xHay x
m dt m dt
Phng trnh (5.5)l phng trnh ng lc hc dng vi phn bc hai i
vi vt dao ng.
b.Phng trnh dao ng ca vt.
Nghim ca phng trnh (5.4) c dng:
)tcos(eaxt
o (5.6)
Trong :m
k;;
m2
ro
22
o
x l li ca vt dao ng, l mt hm i vi thi gian.
Trong biu thc nghim (5.6), tnh cht ca bin dao ng c th
hin biu thc: toeaa (5.7)
Ngha l bin dao ng gim dn i vi thi gian tun theo quy lut
hm m, iu ni ln s tt dn ca dao ng v do gi l dao ng tt dn.
c. Tnh cht tt dn ca dao ng.
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BAI GING VT LY I CNG 1
15
thy r c tnh cht tt dn ca dao ng c, ta biu din phng
trnh (5.7) bng ng cong lin nt trn th nm gii hn gia hai ng
cong nt t trn hnh 5-1.
Tnh cht tt dn ca dao ng c c trng bng mt i lng vt l
gi l gim lng lga, k hiu bng
TT
ea
ealn
a
aln
)Tt(o
to
2
1
(5.8)
trong :
a1l bin dao ng ti thi im t
a2l bin dao ng ti thi im t+ T
5.1.3. Dao ng c hc cng bc.a. Phng trnh ng lc hc dng vi phn.
Vt dao ng cng bc bao gm cc lc
tc dng ln vt l:
- Lc ma st (hoc lc cn): dxr
dt ,
- Lc phc hi: - kx,
- Lc kch thch tun hon tc dng ln vt trong qu trnh dao ng :
cosoF t .
Do lc tng hp l:
tcosFdt
dxrkxf o (5.9)
(l tn s gc ca lc cng bc)
Phng trnh ng lc hc dng vi phn: p dng nh lut th II Niutn f
= max =mx'' , thay vo biu thc trn ta c:
2
2
2
2
d xcos m cos
dt
d x: cos (5.10)
dt
o o
o
dx dxkx r F t mx kx r F t
dt dt
Fr dx k Hay x t
m dt m m
t: 2ooo
m
k;2
m
r;f
m
F thay vo phng trnh (5.10) ta c:
x
t
ao
-ao
to ea
toea
Hnh 5-1
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BAI GING VT LY I CNG 1
16
a
om
of
ma
=
2
2
1
Hnh 5-2
22
2
d x 2 cos (5.11)
dt o o
dxx f t
dt
Phng trnh (5.11) l phng trnh ng lc hc dng vi phn bc hai i
vi vt dao ng cng bc.
b. Phng trnh dao ng ca vt.
Nghim ca phng trnh (5.11) c dng: )tcos(ax (5.12)
Trong : 2tg;4)(
fa
22o
22222o
o
(5.13)
x l li ca vt dao ng, l mt hm i vi thi gian.
c. Tnh cht cng hng ca dao ng.
Ngita chng minh c bin dao ng cc i khi tn s gc ca yu
t cng bcl: 22om
2 , th bin dao ng cng bc cc i l:
22o
omax
2
fa
(5.14)
Hin tng bin dao ng cng bc
t gi tr cc i c gi l hin tng cnghng.
th v s cng hng: S phthuc ca
bin a vo tn s gc
ng vi h s tt dn gi l h ng
cong cng hng.
Vi cng nh th a cng ln. Khi = 0 th
amax s ln nht ng vi hin tng cng hng
nhn.
5.2. Sng c hc
5.2.1. Khi nim v cc c trng ca sng.
a. Khi nim:
Qu trnh lan truyn dao ng c trong mi trng vt cht n hi c
gi l sng c.
Phn loi sng c theo tnh cht mt u sng:
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+ Mt u sng l mt phng th sng c c gi l sng phng.
+ Mt u sng l mt cu th sng c c gi l sng cu.
Phn loi sng c theo phng truyn sng:
+ Sng ngang: Phng daong vung gc vi phng truyn sng.
+ Sng dc: Phng dao ng trng vi phng truyn sng.
b. Cc c trng.
Vn tc(v): Vn tc truyn dao ng c c gi l vn tc truyn sng.
Bin sng(a): Bin ca sng ti mt im bng bin dao n g
ca im khi c sng truyn qua.
Tn s sng(f): Tn s ca sng bng tn s dao ng dao ng ca mt
im khi c sng truyn qua.
Bc sng(): Khong cch gn nhau nht gia hai im dao ng cng
pha
Mi lin h gia cc i lng:f
vv.T;
T
2f2
5.2.2. Phng trnh truyn sng v tnh cht tun hon.
a. Phng trnh truyn sng.Gi s sng phng c bin a
khng thay, itn s , vn tc v.
Ti thi im t ngun sng Mo c phng trnh: tinsa)t(u x (phng
trnh dao ng ca im Mo).
Nu gi thi im sng truyn ti im M vo thi im t, th khi sng
im Mo l thi im (t-x/v), dao ng ca im M ging nh im Monhng
xy ra chm hn mt khong thi gian (t- x/v), do vy biu thc ca sng l:
)x2
ft2sin(a)v
xt(insa)t(u x
(5.15)
Trong : )t(u x l li dao ng ca im c tax ti thi im t khi c sng truyn qua.
b. Tnh cht tun hon ca sng theo thi gian v khng gian.
Tun hon trong khng gian: Ta c th chng minh c rng)t(u x = )t(u x , khi chu k khng gian l , vi f2
(phng truyn sng) xMo
u (phng dao ng)
M
Hnh 5-3
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Tun hon theo thi gian: Tng t nh trn ta cng chng minh c
)t(u x = )Tt(u x , vi
2
T
CU HI N TP CHNG 5
1. Thnh lp phng trnh ng lc hc dng vi phn v phng trnh dao ng iuho ca con lc vt l?
2. Phn bit dao ng ttdn vi dao ng iu ho? S tt dn ph thuc yu t no?
Vit phng trnh ng lc hc dng vi phn v phng trnh dao ng tt dn?
3. Th no l dao ng cng bc? Phn tch qu trnh qu ca dao ng cng
bc? Phng trnh ng lc hc dng vi phn v phng trnh dao ng cng bc?
4. Th no l hin tng cng hng? iu kin ca hin tng cng hng l g? Nu
cc ng dng ca hin tng cng hng?5.Nu nhng im ging nhau v khc nhau gia cc dao ng: Dao ng iu ho;
dao ngduy tr; dao ng cng bc; dao ng cng hng?
6. Th no l sng c hc? iu kin hnh thnh sng c hc? Phn loi sng?
7. Vit phng trnh truyn sng c? Tnh cht ca sng?
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CHNG VI. THUYT TNG I
Cui th k th XVIIIcc th nghim cho thy cc kt qu ca php bin
i Ga-li-l khng cn ng na, in hnh l th nghim o vn tc nh sng
ca Michelson v Morlay (nm 1887). Ngha l c hc Niu-tn vi quan nim v
khng gian v thi gianc th hin trong php bin i Ga-li-l khng th p
dng cho cc vt chuyn ng vi vn tc c nh sng. B tc nyca c hc c
in l tin cho mt ngnh Chc mi ra i l Chc Tng i tnh
hay Thuyt tng i Anhxtanh m csca n l hai tin caAnhxtanh.
Sau y ta xt mt cch s lc v c bn mt s ni dung chnh ca thuyt
tng i hp.
6.1. Nhng tin ca thuyt tng i hp Einstein
xy dng nn thuyt tng i ca mnh, nm 1905 Einstein a ra
hai nguyn l sau:
6.2.1. Nguyn l tng i
Mi nh lut Vt l u nh nhau trong cc h quy chiu qun tnh.
6.2.2. Nguyn l v s bt bin ca vn tc nh sng
Vn tc nh sng trong chn khng u bng nhau i vi mi h qun
tnh. N c gi tr bng c = 3. 108 m/s v l gi tr vn tc cc i trong tnhin.
y cn phn bit vi nguyn l tng i Galille trong c hc c in.
Theo nguyn l ny ch cc nh lut c hc l bt bin khi chuyn t mt h
qun tnh ny sang mt h qun tnh khc. iu c ngha l phngtrnh m
t mt nh lut c hc no , biu din qua ta v thi gian, s gi nguyn
dng trong tt c cc h qun tnh. Nh vy, nguyn l tng i Einstein m
rng nguyn l Galille t cc hin tng c hc sang cc hin tng Vt l nichung.
Trong c hc c in Newton, tng tc c m t da vo th nng
tng tc. l mt hm ca cc ta nhng ht tng tc. T suy ra cc
lc tng tc gia mt cht im no vi cc cht im cn li, ti mi thi
im, ch ph thuc vo v trca cc cht im ti cng thi im . S tng
tc s nh hng ngay tc thi n cc cht im khc ti cng thi im. Nh
vy, tng tc c truyn i tc thi. Nu chia khong cch gia hai cht im
cho thi gian truyn tng tc t ( t = 0), v l truyn tc thi) ta s thu c vn
tc truyn tng tc. T suy ra rng trong c hc c in vn tc truyn
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tng tc ln v hn.
Tuy nhin, thc nghim chng t, trong t nhin khng tn ti nhng
tng tc tc thi. Nu ti mt cht im no ca h cht im c xy ra mt
s thay i no , th s thay i ny ch nh hng ti mt cht im khc ca
h sau mt khong thi gian t no ( t > 0). Nh vy, vn tc truyn tng tcc gi tr hu hn. Theo thuyt tng i ca Einstein vn tc truyn tng tc
l nh nhau trong tt c cc h qun tnh. N l mt hng s ph bin. Thc
nghim chng t vn tc khng i ny l cc i v bng vn tc truyn nh
sng trong chn khng (c = 3.108m/s). Trong thc t hng ngy chng ta thng
gp cc vn tc rt nh so vi vn tc nh sng (v
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Gi s ta xe lin h vi x v t theo phng trnh:
x' = f (xt) (6.1)
tm dng ca phng trnh f (x,t) chng ta vit phng trnh chuyn
ng ca cc gc ta O v O' trong hai h K v K'. i vi h K, gc O'
chuyn ng vi vn tcV. Ta c:
x - Vt = 0 (6.2)
trong x l ta ca gc O' xt vi h K. Cn i vi h K' gc O' l
ng yn. Ta xe ca n trong h K' bao gi cng bng khng. Ta c: x' = 0.
Mun cho phng trnh (6.1) p dng ng cho h K', ngha l khi thay x' =
0 vo (6.1) ta phi thu c (6.2), th f (x,t) ch c th khc (x - Vt) mt s nhn
no :
x' = (x - Vt) (6. 3)
i vi h K' gc O chuyn ng vi vn tc - V. Nhng i vi h K gc O
l ng yn. Lp lun tng t nh trn ta c:
x = (x' + Vt') (6.4)
trong l h s nhn.
Theo tin th nht ca Einstein mi h qun tnh u tng ng nhau,ngha l t (6.3) c th suy ra (6.4) v ngc li bng cch thay th V-V, x'
x, t t Ta rt ra c: = .
Theo tin th hai, ta c trong h K v K': nu x = ct th x' = ct', thay cc
biu thc ny vo trong (6.3) v (6.4) ta thu c:
2
2
1
1 v
c
(6.5)
Nh vy ta c:2 2
2 2
'' ,
1 1
x Vt x Vtx x
v v
c c
,
(Cng thc vi thi gian c dng tng t )
V h K' chuyn ng dc theo trc x nn r rng l y = y' v z = z'. Tm
li, ta thu c cng thc bin i Lorentz nh sau:
-i vi php bin i ta v thi gian t h K sang h K'
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2
2 2
2 2
' , y' y, z' z, t'
1 1
Vt x
x Vt cxv v
c c
(6.6)
-i vi php bin i ta v thi gian t hK' sang h K: (dng tng t)
Cc cng thc (6.6) c gi l php bin i Lorentz. Qua ta thy c
mi lin h mt thit gia khng gian v thi gian.
T cc kt qu trn ta nhn thy rng khi c hay khi V/c 0 th cc
cng thc (6.6) s chuyn thnh:
x' = x - Vt ; y' = y ; z' = z ; t = t ; x = x' + Vt'; y = y', z = z', t = t'
Ngha l chuyn thnh cc cng thc ca php bin i Galille. iu kin c
tng ng vi quan nim tng tc tc thi, iu kin th hai V/c 0tng ng vi s gn ng c in.
Khi V > c, trong cc cng thc trn cc ta x, t tr nn o, iu
chng t khng th c cc chuyn ng vi vn tc ln hn vn tc nh sng c.
Cng khng thdng h quy chiu chuyn ng vi vn tc bng vn tc nh
sng, v khi mu s trong cc cng thc (6.6) s bng khng.
6.3. Cc h qu ca php bin i Lorentz
6.3.1. Khi nim v tnh ng thi v quan h nhn qu
Gi s rng trong h qun tnh K c hai hin tng (hoc cn gi l bin
c) ; hin tng A1(x1y1z1t1) v hin tng A2(x2y2z2t2) vi x2# x1chng ta hy
tm khong thi gian t2- t1gia hai hin tng trong h K', chuyn ng vi
vn tc V dc theo trc x. T cc cng thc bin i Lorentz ta thu c:
2 1 2 12
2 1
2
( )
' '
1
Vt t x x
ct t
Vc
(6.7)
T suy ra rng cc hin tng xy ra ng thi trong h K (t2= t1) s
khng ng thi h K' v (t2t1) # 0. Ch c mt trng hp ngoi l l khi c
hai bin c xy ra ng thi ti nhng im c cng gi tr ca x (ta y c th
khc nhau).
Nh vy, khi nim ng thi ch l mt khi nim tng i, hai bin c c
th ng thi trong mt h quy chiu ny ni chung c th khng ng thi trong mt h quy chiu khc.
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Biu thc (6.7) cng chng t rng i vi cc bin c ng thi trong h
K, du ca (t2t1) c xc nh bi du ca biu thc (x2x1)v. Do , trong
cc h qun tnh khc nhau (vi cc gi tr khc nhau ca V), hiu t2 t1s
khng nhng khc nhau v ln m cn khc nhau v du. iu c ngha l
th t ca cc bin c A1 v A2c th bt k (A1c th xy ra trc A2hocngc li).
Tuy nhng iu va trnh by trn khng c xt cho cc bin c c lin
h nhn qu vi nhau. Lin h nhn qu l mt lin h gia nguyn nhn v kt
qu. Nguyn nhn bao gi cng xy ra trc kt qu, quyt nh s ra i ca
kt qu. Th t ca cc bin c c quan h nhn qu bao gi cng c bo
m trong mih qun tnh. Nguyn nhn xy ra trc, kt qu xy ra sau.
6.3.2. S co ngn LorentzBygi da vo cc cng thc (6.6)chng ta so snh di ca mt vt v
khong thi gian ca mt qu trnh trong hai h K v K'. Gi s c mt thanh
ng yn trong h K' t dc theo trc x', di ca n trong h K' bng
l0= x2x1
Gi l l di ca n o trong h K. Mun vy, ta phi xc nh v tr cc
u ca thanh trong h K ti cng thi im. T php bin i Lorentz ta vit
c:
2 2 1 12 2
2 12 2
2 2
' , '
1 1
V Vx t x t
c cx xV V
c c
Tr v vi v hai biu thc v lu t2= t1ta c:
2
2 12 1 0 22
2
' ' 1
1
x x Vx x l l
cVc
(6.8)
Vy:di (dc theo phng chuynng) ca thanh trong hquy chiu
mthanh chuyn ng ngn hn di ca thanh trong h m thanh ng yn.
Ni mt cch khc, khi vt chuyn ng, kch thc ca n b co ngn theo
phng chuyn ng.
Nh vy, kch thc ca mt vt s khc nhau ty thuc vo ch ta quan st
n trong h ng yn hay chuyn ng. iu ni ln tnh cht ca khnggian trong cc h quy chiu thay i. Ni mt cch khc, khng gian c tnh
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cht tng i, n ph thuc vo chuyn ng. Trng hp vn tc ca.chuyn
ng nh (V
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Khng th m t chuyn ng ca cht im vi vn tc ln c. m t
chuyn ng, cn phi c phng trnh khc tng qut hn. Theo thuyt tng
i, phng trnh c dng: ( )d
F mvdt
(6.10)
trong khi lng m ca cht im bng:
0
2
21
mm
V
c
(6.11)
m l khi lng ca cht im trong h m n chuyn ng vi vn tc v
c gi l khi lng tngi;m0 l khi lng cng ca cht im do
trong h m n ng yn (v = 0) c gi l khi lng ngh.
Ta thy rng theo thuyt tng i, khi lng ca mt vt khng cn lmt hng s na; n tng khi vt chuyn ng; gi tr nh nht ca n ng vi
khi vt ng yn. Cng c th ni rng: khi lng c tnh tng i; n ph
thuc h quy chiu.
Phng trnh (6.11) bt bin i vi php bin i Lorentz v trong trng
hp v
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0
3/22
2
21
m vdvdm
vc
c
So snh hai biu thc trn ta rt ra c: dW = c2dm hay: W = mc2
Hthc ny thng c gi l hthc Einstein.
T h thc Einstein ta tm c nng lng ngh: l nng lng lc vt
ng yn: m = m0 W= m0c2
Lc vt chuyn ng, vt c thm ng nng W
CU HI L THUYT
1.Nu gii hn ng dng ca c hc Newton.
2.Pht biu hai tin Anhxtanh
3.Vit cng thc ca php bin i Lorentz.
4.Gii thch s co ngn ca di v s gin ca thi gian5.Phn tch tnh tng i ca s ng thi gia cc bin ckhng c quan hnhn qu vi nhau.
6. Da vo php bin i Lorentz, chng t trt t k tip v thi gian gia ccbin c c quan h nhn qu vi nhau vn c tn trng.
7. Chng t c hc Newton l trng hp gii hn ca thuyt tng iAnhxtanh khi v
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CHNG VII. C HC CHT LU
7.1. TNH HC CHT LU
Theo nh ngha, cht lu l cht c th chy c. Trn ngha cht lu
bao gm c cht kh v cht lng. V mt c hc c th coi cht lu l mt mi
trng lin tc c to thnh bi v s cc cht im lin kt vi nhau bng cc
ni lc tng tc. Khc vi cht rn, cht lu c nhng tnh cht c th sau y:
- Cht lu khng c hnh dng nht nh. Hnh dng cht lu l hnh dng
ca bnh cha chng.
- Cht lu bao gm cc cht d nn nh cc cht kh v c cc cht kh nn
nh cht lng.
- Khi mt cht lu chuyn ng th n chuyn ng thnh lp. Cc lp can chuyn ng vi cc vn tc khc nhau nn gia cc lp cht lu xut hin
cc lc tng tc gi l lc ni ma st hay lc nht. Lc nht lm cn tr chuyn
ng ca cc lp cht lu chuyn ng nhanh v h tr cho cc lp cht lu
chuyn ng chm hn.
- Cht lu c gi l l tng nu n hon ton khng chu nn v trong
n khng tn ti cc lc nht.
Trong chng ny chng ta nghin cu cht lu trng thi cn bng(khng chuyn ng) do phn c hc cht lu ny c gi l tnh hc cht
lu. Cng c nghin cu s cn bng ca cht lu l cc nh lut Niutn I v II.
7.1.1. Khi lng ring v p sut ca cht lu
a. Khi lng ring:
Theo nh ngha , khi lng ring ca bt k mt cht no chnh l khi
lng ca mt n v th tch ca cht . Trng hp cht lu l ng nht th
khi lng ring r ca n c xc nh bi h thc:
M
V (kg/m3)
Trong V l th tch ca khi cht lu v m l khi lng ca cht lu
cha trong th tch V.
Tuy nhin , trong thc t ta cng hay gp cc trng hp mt cht lu
thay i t v tr ny n v tr khc: trn cao khng kh long hn so vi mt
t, nc bin nhng ni su c mt cao hn gn b mt. Ni chung, khilng ring ca mt cht lu ph thuc vo p sut v nhit ca n.
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b. p sut ca cht lu:
Khi mt cht lu trng thi cn bng th n tc dng mt lc thng gc
vi bt k mt mt no tip xc vi n. Chng ta d dng cm nhn iu ny khi
nhng chn vo nc hay khi ngm mnh trong nc bin.Chng ta hy tng
tng c mt tit din dS nm trong lng mt cht lu. Cht lu hai bn titdin dS tc dng ln tit din nhng lc bng nhau v ln nhng ngc
chiu nhau (nu khng nh vy th tit din dS s chuyn ng v cht lu s
khng trng thi cn bng). Gi dF l lc tc dng vung gc vi tit din dS.
Ngi ta nh ngha p sut p ti mt im l lc tc dng theo phng
thng gc ln mt n v din tch cha im , tc l : p = dF/dS
c. S phn b p sut trong cht lu nh lut Pascal :
Do cht lu chu lc ht ca qu t, tc l n c trng lng nn nhchng ta s thy s phn b p sut trong cht lu ti cc im c cao khc
nhau s khc nhau.
Xt mt khi cht lu tng tng c dng hnh tr c tit din y l S v
chiu cao l dz (hnh 7-1).
Hnh 7- 1
Th tch ca khi cht lu l Sdz, cn khi lng ca n l dm = dV =
Sdz. Trng lng ca khi cht lu trn c ln: dP = dm.g = .g.Sdz, lc ny
hng thng ng xung pha di nn hnh chiu ca lc ny trn trc Oz l:
- .g.Sdz
Mt y ca hnh tr ca khi cht lu chu mt p lc hng thng ng
ln pha trn: pS, cn mt trn ca khi cht lu ang xt chu mt p lc
(p+dp)S v hng thng ng xung pha di nn hnh chiu ca lc ny trn
trc Oz l:
-(p+dp)S
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Mt bn ca hnh tr chu cc p lc vung gc nn khi chiu ln trc Oz
th hnh chiu ca cc lc ny bng 0 v chng c phng vung gc vi Oz.
V khi cht lu trng thi cn bng theo Oz nn theo nh lut I Niutn:
pS - (p+dp)S - r gSdz = 0 (7.1)
Chia hai v phng trnh trn cho S ta i n phng trnh:
dP = - .g.dz (7.2)
Ly tch phn hai v phng trnh (7.2) ta i n biu thc:
P2P1= - .g.z (7.3)
trong p1, p2l p sut trong lng cht lu ti cc im c cao tng
ng l z1v z2cn z = z2z1l hiu cao tng ng gia hai im trn.
(7.3) chng t rng hiu p sut trong mt khi cht lu bng trng lng
ca ct cht lu c tit din bng mt n v din tch v c cao bng hiu
cao gia hai im y.
Thng th ngi ta mun bit p sut trong lng cht lu mt su no
so vi psut b mt ca cht lu (b mt cht lu tip xc vi khng kh gi
l mt thong). Mun vy, ta vit li (7.3) di mt dng thch hp hn. Ta chn
gc O ca h trc ta nm mt thong v trc Oz c chiu dng hng
thng ng xung pha di. Chn im 1 l mt im bt k nm trong lng chtlu su z v k hiu p sut ti l p. im 2 chn l im nm ngay mt
thong v gi p sut ti im 2 l po (ch s 0 ch rng su ti bng 0)
a) b)
Hnh 7- 2
(7.3) c vit li nh sau:
pop = - gz hay p = po+ gz (7.4)
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(7.4) chng t rng p sut su z ln hn p sut mt thong mt
lngl gz. Lu rng p sut hai im bt k l nh nhau min l chng
cng cao. Dng ca bnh cha khng nh hng g n kt lun trn. l
nguyn tc ca bnh thng nhau nh thy hnh 7.2 a.
T (7.4) chng ta thy rng nu bng cch no talm tng p sut po mt thong (chng hn bng cch dng piston to ra lc nn) th p sut p trong
lng cht lng ti bt k su no cng s tng ln mt lng ng bng by
nhiu. iu ny chnh l ni dung ca nh lut Pascal c pht biu nm 1653:
Trong mt cht lu khng chu nn, bt k mt tng p sut no cng
c truyn nguyn vn cho ton khi cht lu k c thnh ca bnh cha.
Hnh 7-2. b trnh by nguyn tc lm vic ca mt ci kch thy lc da trn
nh lut Pascal.Mt piston vi tit din nhS1tc dng mt lc F1ln bmt mt cht lng
l du nht. p sut tc dng P1=F1/S1c truyn qua ng ni sang mt piston
c tit din S2ln hn nhiu. V rng p sut hai piston l nh nhau, tc l:
1 2
1 2
F Fp
S S (7.5)
Nh vy kch thy lc c tc dng nhn lc ln S2/ S1ln v do c th
lm nng bng t ln cao mc d ta chtc dng mt lc nhF1ln kch.
7.1.2. Cng thc phong v biu
Trong phn ny chng ta nghin cu s cn bng ca kh quyn trn tri
t. Chn trc Oz hng thng ng ln pha trn. Phng trnh cn bng ca kh
quyn l:
dpg
dz (7.6)
Ta c thly tch phn phng trnh ny tm sphthuc ca p theo z,
tuy nhin cn lu rng trong trng hp ny khi lng ring ca khng kh
phthuc vo cao z v khng kh l mi trng chu nn.
Mt khc, v cc i lng p sut p, mt v nhit T ca mt cht
kh trng thi cn bng tun theo phng trnh trng thi (xem phn nhit hc):
p RT
(7.7)
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trong l khi lng ng vi mt kilomol ca khng kh cn R l
hng skh l tng: R = 8,31.103J/Kmol.oK. Tphng trnh trng thi thay
vo (7.7) ri sau thay vo (7.6) ta c:
dp gp
dz RT
hay
dp gdz
p RT
(7.8)
c th tch phn phng trnh trn tm sph thuc ca p sut theo
cao th ta phi coi nhit T ca kh quyn l mt hng s v khng ph
thuc vo cao z. Githit ny chng nhng cao khng qu ln (c1-
2km). Vi iu kin ny c thxem T l hng sv khi ly tch phn t0 n z
phng trnh (7.8), ta c:
0
gz
RTp p e
(7.9)
trong pol p sut kh quyn ngay mt t, tc p sut ti z = 0, cn p l p
sut kh quyn ti cao z.
V mt ca kh quyn tlvi p sut p, nn ta cng rrng suy ra biu
thc tng ti vi mt . Cc cng thc c gi l cc cng thc phong
vbiu.Cng thc phong v biu gip ta c thxc nh cao ca cc im c
a hnh phc tp nh cao ca ni khi o p sut v nhit ti nh ni v ti
mt bin (vi cao z = 0). Cng thc phong v biu chng khi nhit ti
nh ni v ti mt bin l nh nhau, nu khng ta cn phi hiu chnh.
7.2. NG HC CHT LU
Trong phn ny, chng ta nghin cu cht lu trng thi chuyn ng.
Chuyn ng ca cht lu ni chung l kh phc tp, tuy nhin trong mt s
trng hp th chuyn ng ca cht lu c thc biu din bng cc m hnh
l tng mt cch n gin. C hai cch nghin cu chuyn ng ca cht lu:
- Theo di chuyn ng ca tng cht im (ht) ca cht lu v nghincu qu o, vn tc, gia tc ca chng nh cch ta lm i vi chuyn ng
ca cht im cc chng trc.
- Ly mt vtr M xc nhtrong khi cht lu v xt cc cht im khc
nhau di chuyn qua vtr ti nhng thi im khc nhau. Ti mi thi im t,
vn tc ca ht cht lu ti mt im M trong khi cht lu l mt hm ca im
M v thi gian t, tc l: ( , )v v M t
Nu v chphthuc vo vtr ca M m khng phthuc vo thi gian tth ta ni rng chuyn ng ca cht lu lchuyn ng dng. Trong phn ny,
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chng ta chnghin cu chuyn ng ca cht lu trng thi dng v sdng
cch nghin cu thhai nghin cu chuyn ng ca cht lu.
7.2.1. Phng trnh lin tc ca cht lu trng thi dng
a. Mt vi nh ngha:
Ta gi nhng ng cong m tip tuyn ti mi im ca n trng vi
phng ca vn tc ht cht lu ti im l nhng ng dng. trng thi
dng th cc ng dng trng vi qu o chuyn ng ca cc ht cht lu.
Hnh (7-3) trnh by cc ng dng. Chiu ca ng dng c qui c
chn l chiu ca vn tc. Cn lu l khng nn ng nht khi nim ng
dng vi qu o chuyn ng ca cc ht cht lu. Qu o l ng m mt ht
cht lu xc nh vch ra trong sut thi gian chuyn ng ca n.
Hnh 7-3
Tri li, ng dng c trng cho chuyn ng ca mt tp hp v hn
cc ht cht lu m ti thi im kho st nm trn ng .
Tp hp nhng ng dng ta ln mt ng cong kn lm thnh mt
ng dng. Cht lu chuyn ng trong ng dng ta nh l chuyn ng trong
mt ng thc.
Hnh 7-4
b. Phng trnh lin tc:
Xt mt ng dng c cc tit din ngang rt nh. Khi ta c thxem
rng vn tc ti mi im trn tit din l nh nhau v hng theo trc ca ng
dng. Lng cht lu chuyn qua mt tit din ngang ca ng dng sau mt
khong thi gian dt l:
dm = (S.v.dt) trong l mt (khi lng ring) ca cht lu, v l
vn tc ca ht cht lu ti tit din ngang, S l din tch ca tit din .
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Do tnh khng chu nn ca cht lu nn trng thi dng th lng cht
lu dm chuyn qua bt ktit din ngang no trong cng mt thi gian dt phi
khng i, tc l :
dm = v1S1dt = v2S2dt = const hay: v1S1= v2S2 (7.10)
T(7.10) ta suy ra: 21 21
Sv v
S . Hthc dng ny cho ta mt hqul: nu
S1< S2th v1 > v2; cn nu S1> S2th v1 < v2.
Vy ta c th kt lun: ni ng dng c tit din ln th cht lu chy
chm cn ni ng dng c tit din nhth cht lu chy nhanh.
Phng trnh (7.10) gi lphng trnh lin tcca cht lu trng thi
dng.
7.2.2. Phng trnh Bernoulli (Bec-nu-li)
Xt chuyn ng ca cht lu l tng trng thi dng. Ta ly trong
cht lu mt ng dng hp v xt lng cht lu cha trong mt on ng dng
bgii hn gia hai tit din MN v CD (Hnh 7-5). Gissau mt khong thi
gian dt lng cht lu tit din MN dch chuyn n M1N1cn lng cht lu
tit din CD dch chuyn n C1D1.
Hnh 7-5Ta hy tnh cng dA m cc lc p sut gy ra trong chuyn ng . p
sut tc dng vo mt bn ca ng dng lun lun vung gc vi dch chuyn
nn khng sinh cng do cng dch chuyn chdo cc lc p sut p1v p2ti
cc tit din MN v CD gy ra m thi.
Lc tc dng ln tit din MN c din tch S1l p1S1cn lc tc dng ln
tit din CD c din tch S2l (p2S2). Du "-" v p2c chiu chng li chuyn
ng ca dng cht lu. Do cng dA bng:
1 1 1 2 2 1 1 2dA p S MM p S CC dA dA (7.11)
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nhng1 1 1S MM dV cn 1 1 2S CC dV dV2, trong dV1v dV2l thtch ca cc
khi cht lu cha trong cc on ng dng MNM1N1v CDC1D1.
Nu ta gi dm1v 1l khi lng v mt cht lu cha trong dV1v
dm2v 2 l khi lng v mt cht lu cha trong dV2th:
11
1
dmdV
v 22
2
dmdV
do thay vo (7.11) ta tnh c cng dA ca cc lc p sut gy ra s
chuyn ng ca dng cht lu trong khong thi gian dt l:
1 21 2
1 2
dm dmdA p p
V chuyn ng ca cht lu m ta xt l chuyn ng dng nn khilng cht lu cha trong on ng dng M1N1CD l khng thay i v tnh
lut bo ton khi lng ta suy ra dm1 = dm2= dm.
1 2
1 2
( )p p
dA dm
Theo nh lut bo ton nng lng, cng ny phi bng tng dE ca
nng lng ton phn ca khi cht lu m ta ang xt.
V chuyn ng l dng nn nng lng ca khi cht lu cha trongon ng dng M1N1CD l khng i nn tng nng lng dE phi bng hiu
nng lng ca khi cht lu dm cha trong thtch dV1v dV2.Nu ta gi e l
nng lng ton phn ng vi mt n vkhi lngca cht lu th :
dE = (e2- e1)dm
Cho dA = dE, ta c:
1 2
2 1
1 2( )
p p
dm e e dm
hay1 2
1 2
1 2
p p
e e
V on ng dng m ta ang xt c thly ty nn ta c thrt ra kt
lun:
11
1
=consp
e t
(7.12)
Vy theo (7.12): "Dc theo mt ng dng, trng thi dng th tng
11
1
pe ca mt cht lu l tng l mt hng s". Phng trnh (7.12) gi l
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phng trnh Bec-nu-li dng tng qut. l phng trnh c bn m t
chuyn ng ca cht lu trng thi dng.
Ta hy p dng l phng trnh Bec-nu-li tng qut (7.12) vo trong
trng hp ring quan trng m ta thng hay gp: cht lu khng chu nn v
chuyn ng trong trng trng. Ni chung, nng lng ton phn e ng vi mtn vkhi lng phthuc vo trng thi bnn ca cht lu. Tuy nhin, trong
trng hp ca cht lu khng chu nn th nng lng ton phn e ch l c
nng ca mt n vkhi lng ca cht lu, tc l chgm ng nng v th
nng m thi:
21
2e v gz
(Lu l khi lng y bng n v nn ng nng v thnng cdng nh trn).
Thay biu thc ca e trn vo (7.12) v bin i ta c :
2 2
2 12 1 1 2
( ) ( ) g(z ) g2 2
v vp p z h (7.13)
Trong cng thc (7.13) ta gi cc p sut: p sut tnhp v p sut ng2
2
v (gi l p sut ng l v do cht lu chuyn ng m c). Tng hai p sut
ny gi l p sut ton phn
Vy ta c th pht biu nh lut Bec-nu-li nh sau: Hiu p sut ton
phn gia hai im trong mt ng dng ca cht lu khng chu nn trng
thi dng bng trng lng ca ct cht lu c tit din bng mt n vdin
tch v c cao bng hiu cao gia hai im y".
7.2.3. Cc hquca phng trnh Bec-nu-li
Phng trnh Bec-nu-li lphng trnh c bnm tchuyn ng ca chtlu. Tphng trnh ta c thrt ra mt shququan trng sau y :
a. Hin tng Ven-tu-ri :
Ta xt schy ca cht lu khng chu nn qua mt ng nm ngang c tit
din thay i.Hnh v7-6 trnh by mt ci ng nm ngang c tit din khng
ngu. Trong ng ta c mt ng dng, chng hn nm trc ca ng. Cc ht
cht lu trn ng dng ny c cng cao z nn t(VII.4c) khi ta cho h = 0 ta i
n phng trnh :
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Hnh 7-6
2 22 1
2 1 co2 2
v vp p nst (7.14)
Tphng trnh ny, ta suy ra ni no ht cht lu c vn tc ln th ti ni
p sut nhv ngc li ni no ht cht lu c vn tc nhth ni c p
sut ln.
Mt khc kt hp vi phng trnh lin tc: S1v1= S2v2= = const , ta
suy ra vn tc ht cht lu ti ni no ln th tit din ca ng ni li nhcn
ni no vn tc ht cht lu nhth ni tit din ng li ln. hnh 7-6, titit din AB vn tc ht cht lu nhhn vn tc ca n ti tit din CD.
Vy tm li thai kt lun va nu trn , ta c thgp li thnh kt lun
chung nh sau:Khi mt cht lu khng chu nn chy dc theo mt ng dn nm
ngang th ni no c tit din ln th ti ni p sut ln nhng vn tc ht
cht lu li nh, ngc li ni no ng c tit din nh th p sut ti nh
nhng vn tc ht cht lu li ln. Hin tng trn c gi l hin tng Ven-
tu-ri.b. nh l Trixenl (Torricelli):
Ta p dng nh l Becnuli kho st sthot ra ca cht lng qua mt
lnhy bnh (hnh 7-7).
Hnh 7-7
Gista c mt ci bnh y phng v rng, y bnh c mt lT nh.
Bnh cha cht lng m mc nc AB c cao h. Do mt thong AB rng v l
T nhnn khi nc thot ra ngoi bnh ta vn c thxem mc nc AB l khngi ngha l vn tc ca cht lng mt thong AB c thcoi nh bng khng.
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9. Trn bn c t mt bnh nc, thnh bnh c mt l nh nm cch y bnh
mt on h1 v cch mc nc mt on h2. Mc nc trong bnh c gi
khng i. Hi tia nc ri xung mt bn cch l mt on L bng bao nhiu
(theo phng nm ngang)? Gii bi ton trong hai trng hp:
a) h1= 25cm v h2= 16cm; P S: cmL 4016.252 b) h1= 16cm v h2= 25cm. P S: cmL 4025.162
10. Gia y mt gu nc hnh tr b thng mt l nh. Mc nc trong gu
cch y gu H = 30cm. Hi nc chy qua l vi vn tc bng bao nhiu trong
cc trng hp sau:
a) Gu nc ng yn;
b) Gu c nng ln u:
c)Gu chuyn ng vi gia tc 1,2m/s2ln trn ri xung di.
d) Gu chuyn ng theo phng nm ngang vi gia tc 1,2m/s2.
P S: a, b) v = 2,42 m/s; c) v = 2,57 m/s v v = 2,27 m/s; d) v = 2,42 m/s
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PHN TH HAI. VT L PHN T V NHIT HC
Nhit hc nghin cu cc hin tng lin quan n nhng qu trnh xy ra
bn trong vt nh vt nng chy,vt bay hi,vt nng ln khi ma st nhng
hin tng ny lin quan n mt dng chuyn ng khc ca vt cht l
chuyn ng nhit. Chuyn ng nhit l i tng nghin cu ca nhit hc.
nghin cu chuyn ng nhit ngi ta dng hai phng php:
1. Phng php thng k(ng dng trong vt l phn t): Da vo cu tophn t ca cc cht v s chuyn ng hn lon ca chng; Ngi ta s dng
cc quy lut ca xc sut thng k tnh gi tr trung bnh ca cc i lng
trn c s nghin cu ccqu trnh xy ra cho tng phn t. Phng php ny
cho ta bit mt cch su sc bn cht ca hin tng. Tuy nhin trong mt s
trng hp vic ng dng phng php ny tng i phc tp.
2. Phng php nhit ng (ng dng trong phn nhit ng lc hc):
Nghin cu qu trnh trao i v bin ho nng lng da trn hai nguyn l cbn c rt ra t thc nghim gi l nguyn l th nht v nguyn l th hai
nhit ng hc. Phng php nhit ng hc khng gii thch c su sc bn
cht ca hin tngnhng n li c phm vi ng dng su rng hn v n gin
hn phng php thng k
CHNG VIII. CC NH LUT THC NGHIM V CHT KH
8.1. MT S KHA I NIM C BN.
8.1.1. Thng s trng thi.
Trng thi ca mt h hon ton c xc nh nu bit c cc c tnh
ca h. V d: nng hay lnh, c hay long, b nn t hay nhiu, ... Mi c tnh
nh vy u c c trng bng mt i lng vt l, bao gm: nhit , th
tch, khi lng, p sut, .... Cc i lng c trng nh vy c gi l cc
thng s trng thi ca h. Vi h l cht kh, cc thng s trng thi c trng l
p sut, th tch v nhit
a. p sut.
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i lng vt l c xc nh bng lc tc dng vung gc ln mt n v
din tch:
Gi Fnl lc tc dng vung gc ln din tch S th p sut l:
nF
PS
(8.1)
Trong h SI n v p sut l N/m2, haypascal (Pa). Ngi ta cn dng cc
n v: Atmophe k thut, Milimet thu ngn (cn gi l tor)
1 at = 736 milimet thu ngn (mmHg) = 9,81.104N/m2 = 736Pa
b. Nhit .
Nhit ca mt vt cho ta cm gic v mc nng lnhca vt . C
th nu nhit ca vt A ln hn nhit ca vt B th ta ni vt A nnghn vt B, hay vt B lnhhn vt A . Tuy nhin, iu ch mang tnh tng
i, v cm gic nng, lnh ph thuc vo tng ngi v tng trng hp c th
(ngha l mang tnh ch quan). Tnh cht nng, lnh m ta cm nhn c vt
lin quan n nng lng chuyn ng nhit ca cc phn t. V th, nhit
c nh ngha mt cch chnh xc nh sau:
Nhit l i lng vt l, c trng cho tnh cht v m ca vt (hay h
vt), th hin mc nhanh, chm ca chuyn ng hn lon ca cc phn t
ca vt (hay h vt) .
Nhit lin quan n nng lng chuyn ng nhit ca cc phn t.Tuy
nhin khng th dng nng lng o nhit v khng th o trc tip nng
lng chuyn ng nhit, hn na nng lng ny li rt nh. Do d ngi ta o
nhit bng n v l .
Tu theo cch chia ngi ta s dng cc nhit giai khc nhau:
n v nhit trong nhit giai Celsius (Xenxiuyt) (bch phn) l
o
C n v nhit trong nhit giai (Kelvin)Kenvin (tuyt i): oK
Trong nhit giai Xenxiuyt th 0oC th P 0; cn trong nhit giai Kenvin th
0oK th P = 0.
Gi T l nhit tuyt i, th n lin h vi bch phn t qua biu thc:
T = t + 273,15
(Khi khng cn chnh xc cao v tnh ton n gin ta ly: T = t + 273)
8.2. CC NH LUT THC NGHIM V CHT KH .
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8.3.2. Phng trnh trng thi kh l tng.
Trng thi ca mt khi kh l tng c m t bi cc thng s: nhit
T, p sut P v th tch V.
Xt qu trnh bin i trng thi ca mt khi lng kh t trng thi (1)
sang trng thi (2) thng qua mt trng thi trung gian (*) nh s din binhnh 8-1:
nh lut Bil - Marit vit cho qu trnh th nht: P1V1=P*V*
Ta rt ra c:*11*
V
VPP (8.1)
nh lut Gay-luytxac vit cho qu trnh th hai:
1
2
2
2
*
*
T
P
T
P
T
P suy ra
2
2
**
11
T
P
TV
VP
(8.2)
Thay V* = V2 vo biu thc (8.2) v chuyn cc i lng c cng ch s
sang cng mt v, ta c:
1 1 2 2
1 2
PV PV PVconst
T T T
(8.3)
Gi P,V,T l cc thng s trng thi ca mt kmol cht kh. p dng h
thc (8.3) cho 1 kmol cht kh :
o
oo
T
VP
T
PV.................
T
VP
T
VP
2
22
1
11
Vi: Po Vo To l thng s trng thi ca mt kmol kh iu kin tiu chun:
Po=1,033 at = 1,013.105N/m2 , Vo = 22,4m
3 , T = 273o K.
Khi ta c: k RPV
T
(8.4)
Trng thi 1P1V1T1
Trng thi *P* V* T*
Trng thi 2P2 V2 T2
u trnh n nhi t u trnh n tch
Hnh 8-1
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(8.4) l phng trnh trng thi vit cho mt kmol kh l tng ( phng
trnh Merdeleev-Clapeyron)
Trong : K.kmol/J10.31,8273
4,22.01.013,1R
35
, l hng s kh l tng;
Vkl th tch ca 1kmol khXt khi lng m kh bt k c cc thng s trng thi P,V,T. Trong lng
kh m(kg) , p sut v nhit ging nhau i vi mi kmol, do vy T = Tk; P
= Pkv th tch km
V
V
Thay vo phng trnh trng thi (8.4) ta c: RVmT
P
Vit li thnh: RTmPV
(8.5)
(8.5) l phng trnh trng thi kh l tng p dng cho khi lng kh bt k.
CU HI N TP V BI TP CHNG 8
1. Trnh by cc khi nim c bn v: p sut, nhit v nhit giai.
2.Nu c im ca kh l tng? Trnh by phng trnh trng thi kh l tng.
3.Nu ni dung nh lut, biu thc ca cc nh lut thc nghim v cht kh?4.Mt lng kh xy m = 500gam, ng trong bnh c dung tch bng 2lt, nhit
27O C. Tnh p sut ca kh cn li trong bnh khi mt na lng kh
thot ra khi bnh v nhit nng ln 87O C. Cho bit xy c =32kg/kmol.
P S: 6 22 11,7.10 /P N m
5.40g kh O2chim th tch 3l p sut 10at.
a.Tnh nhit ca kh. P S: T1= 292,5Kb.Cho khi kh gin n ti th tch 4l. Hi nhit ca khi kh sau
khi gin n. P S: T2= 390K
6. C 10g kh ng trong mt bnh, p sut 107Pa. Ngi ta ly bnh ra mt
lng kh cho ti khi p sut ca kh cn li trong bnh bng 2,5.10 6Pa. Coi nhit
kh khng i. Tm lng kh ly ra. P S: 7,5kg
7.C 8g kh xy hn hp vi 22g kh ccbonnc (CO2). Xc nh khi lng ca
1 kilmol hn hp . P S: 40 /kg kmol
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8. Kh n l mt hn hp gm mt phn khi lng h yr v tm phn khi
lng xy. Hy xc nh khi lng ring ca kh n iu kin thng.
P S: 12 /g mol
CHNG IX: VT LY PHN T
9.1. THUYT NG HOC PHN T
9.1.1 N i dung
Thuyt ng hc phn t l mt trong nhng thuyt Vt L ra i smnht. N k tha nhng quan im c i v cu to vt cht v nhng kt qu
ca cuc u tranh ko di nhiu th k gia cc t tng i lp nhau v bn
cht ca nhit, cng vi s quan st bng thc nghim, ngi ta a ra thuyt
ng hc phn t. Ni dung c bn ca thuyt c th tm tt bng cc quan im
sau:
Cc cht kh c cu trc gin on gm s ln cc phn t.
Cc phn t lun trng thi chuyn ng hn lon v khng ngng. Kch thc ring ca cc phn t rt nh b so vi khong cch gia
chng, coi phn t nh mt cht im chuyn ng.
Cc phn t khng tng tc ln nhau tr lc chng va chm vo nhauhoc va chm vo thnh bnh. Quy lut va chm l hon ton n hi tun
theo cc nh lut c hc ca Niutn.
9.1.2. Phngtrnh thuyt ng hc phn t
Cc phn t kh chuyn ng hn lon khng ngng va chm vo thnhbnh hoc vo b mt S bt k nm trong khi kh, to nn p sut. Chuyn ng
ca cc phn t cng nhanh, tc ng nng cng ln, th p vo bnh vi p lc
cng ln, gy ra p sut cng ln. Ngoi ra, mt cc phn t kh cng ln th
kh nng va chm vi thnh bnh cng cao, suy ra p sut cng ln. Nh vy p
sut ca khi kh c lin quan n ng nng ca cc phn t kh v mt kh.
H thc lin h gia p sut, mt v ng nng ca cc phn t kh, gi l
phng trnh c bn ca thuyt ng hc phn t
a. Thit lp phng trnh v.t
S thnhbnh
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Xt bnh cha kh c mt phn t l no, cc
phn t chuyn ng hn lon vi vn tc trung bnh
l v, khi ccphn t p vo thnh bnh th gy nn
p sut i vi thnh bnh v cng l p sut ca
cht kh bn trong bnh cha (hnh 9-1) .Gi F l lc tc dng vung gc vo din tch s ca thnh bnh
Theo biu thc nh ngha v p sut:s
FP
Trong F l cng lc tng hp ca n cc phn t tc dng vung
gc ln din tch S trong khong thi gian t. Ta c F = n.f ( f l cng lc
do mt phn t tc dng vo s thnh bnh, n l s ht c kh nng n va chm
vo s )
Tnh n ?
S ht c kh nng n va chm vo s trong thi gian t s nm trong
th tch V ca hnh tr y l s v ng cao l v.t. Do vy t.v.SV . S
phn t N c trong th tch V l: N = no. V = no . s.v. t
Do tnh cht hn lon ca cc phn t nn theo hng vung gc s ch
c 1/6 s ht trong tng s ni trn miti va chm vo thnh bnh, tc l:
n =6
t.v.s.n
6
N o (ht)
Tnh f?
Theo nh l v ng lng th bin thin ng lng ca mt ht phn
t va chm vo thnh bnh trong t l: k = f.t m k = 2mv
2 f
k mv
t t
Suy ra: 2. . 2 .
.6 3
o on s v t nm v
F n f mv st
Tnh p sut P?
2 2
0
. 2 .
3 3 2o
F m v m vP n n
s
Trong : 2 2 2 21 21
( ... )nv v v v
n
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Do d:2.
2
m v= Wc gi l ng nng trung bnh ca cc phn t kh
Kt qu l:2
.W3
oP n (9.1)
(9.1) l phng trnh c bn ca thuyt ng hc phn t kh. Phngtrnh cho thy mi quan h gia p sut (i lng v m, c trng cho tc dng
tp th ca cc phn t) vi mt v ng nng trung bnh ca cc phn t kh
(cc i lng vi m, c trng cho phn t v chuyn ng ca phn t).
Phng trnh (9.1) ch r c ch vi m ca p sut cht kh tc dng l n
thnh bnh v phn nh mt cch tng minh cc quan im c bn ca Thuyt
ng Hc Phn T. Phng trnh (9.1) c tnh thng k. Cc i lng trong
(9.1) l cc i lng thng k. Ta ch c th ni ti p sut v ng nng trungbnh ca mt tp hp rt ln cc phn t; khng th ni ti p sut v ng nng
ca mt hoc mt s t phn t c.
b. H qu:
Gii thch cc nh lut cht kh bng thuyt ng hc phn t. Ta c th tmli c cc mi quan h ny t phng trnh c bn ca thuyt ng hc phn
t
Biu thc ng nng tnh tin trung bnh ph thuc vo nhit . Ta chngminh di y:
Xt 1 Kmol cht kh l tng, ta c PV = RT. Suy ra: RT
PV
.
Mt khc t phng trnh (9.1):
2W
3 o
RTP P n
V
W=3 3
2 2o A
RT RT
n V N
Suy ra c: W=3
2kT (vi k l hng s Bnzman)
(9.2)
Tr s ca k l: k =3
23
26
8,31.10 / .1,38.10 /
16,023.10
oR J Kmol K J KN
Kmol
Tnh vn tc cn qun phng:
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21 3 3W . ; W2 2 2 A
RTm v KT
N
RT3
mN
RT3v
A
2 (m: khi lng 1
phn t v NA.m = )
RT
v
32
gi
2
v vn tc cn qun phng.
Tnh mt phn t: Ta c 2 .W3
oP n , vi
3W
2KT
Suy ra: .kTo o
PP n n
kT
Mi cht kh c mt phn t bng nhau di cng p sut v nhit .
Xt trong KTC: P = 1,013.105N/m, To = 273oK , mt phn t kh l:
no=25
25
25
o
o 10.687,2do273.do/J10.38,1
m/N10.013,1
KT
P
phn t/m3
9.2. NI NNG KHI LY TNG
9.2.1. nh lut phn b nng lng theo bc t do
a. Bc t do ca phn t.
Khi nim: Thng s c lp cn thit xc nh v tr ca phn t trongkhng gian.
V d: xc nh v tr ca phn t trong khng gian ta cn phi bit cc to x, y, z. Cc ta c gi l bc t do.
Nu phn t ch chuyn ng tnh tin th s bc t do bng 3, cn nu phnt va tnh tin, va quay th s bc t do bng 5. Theo , phn t n nguyn
t c i = 3; phn t 2 nguyn t c i = 5; phn t 3 nguyn t c i = 6.
b. nh lut phn b nng lng theo s bc t do
Ni dung:Nng lng ca phn t kh c phn b u theo cc bc t do. Biu thc: Trong chuyn ng tnh tin cc phn t c s bc t do bng 3,ng nng trung bnh chuyn ng hn n cc phn t tng ng bng
3W
2KT . T c th suy ra nng lng (ng nng) tng ng vi mi bc
t do bng KT
2
1.
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Ta c th nhn xt mt cch tng qut: nu phn t c s bc t do l i th
nng lng ca phn t s l:1
2KTi .
9.2.2. Ni nng ca kh l tng.
a. Khi nim ni nng
Nng lng ca h gm ng nng ng vi chuyn ng c hng ca c
h, th nng ca c h v phn nng lng ng vi chuyn ng bn trong ca h
tc l ni nng U ca h: W = W+ Wt+ U.
Trong nhit ng hc ta gi thit rng chuyn ng c hng ca h
khng ng k v h khng t trong mt trng lc no, do nng lng ca
h ng bng ni nng ca h: W = U
i vi h l khi kh l tng, ni nng l tng nng lng chuyn ng
nhit ca cc phn t cu to nn h, l nng lng do chuyn ng hn lon ca
cc phn t to nn v chnh l ng nng ca cc phn t. Nng lng ny
ph thuc vo nhit ca cc phn t vt cht v c gi l nhit nng.
Nh vy, ni nng ca mt khi kh l tng bng tng ng nng ca
cc phn t: U = W
a. Ni nng ca 1 kmol kh l tng.
Trong 1kmol kh l tng gm NA = 6,023.1026phn t, mi phn t c
ng nng KT2
i.
Tng cng ng nng ca cc phn t c trong mt kmol chnh l l
ni nng ca mt kmol kh, k hiu l UO, ta c:
02
A KTi
U N hay 0 ( R N )k2
ATi
U R (9.3)
c. Ni nng ca khi lng kh l tng bt k.
Trong khi lng m kh l tng bt k c cha n (kmol) kh, mi kmol kh
c khi lng , do vy s kmol c trong m (kg) cht kh l:
m
n .
Nng lng tng cng ca cc phn t c trong khi lng m c gi
l ni nng ca khi lng kh bt k, k hiul U th biu thc ca U l:
0.
2T
m iU nU R
(9.4)
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d. bin thin ni nng ca kh l tng.
Xt khi lng kh bt k hai trng thi nhit l T 1 v T2 , ni nng
tng ng vi hai trng thi l U1 v U2: 11 TR2im
U
;22
TR2
imU
bin thin ni nng gia hai trng thi l:
)2 1
( 2 2
T T Tm i m i
U R R
(9.5)
9.3. CC NH LUT PHN B PHN T KH
Do tnh cht hn n v khng ng u ca cc phn t kh nn cc phn
t kh khng hon ton ging nhau. Bng phng php thng k v phng php
tnh ton trn thc nghim, ngi ta ch c th xc nh c thng s trng thi
c tnh xc sut m thi. Di y l cc phn b phn t kh theo vn tc, p
sut v cao
9.3.1. nh lut phn b ht theo vn tc ca Maxwell.
Th nghim ca Stern chng minh rng, cc phn t kh c vn tc ly cc
gi tr t 0 n nhng gi tr ln nht. Thc nghim cng ch ra rng, s chuyn
ng ca cc phn t kh l v cng hn lon (chuyn ng Brown) v do vy
vn tc ca phaan t kh khc nhau trong cng mt khi kh s rt khc nhau.
Nu gi dn l s phn t kh c vn tc t v n v + dv, n l s phn t kh ca
khi kh th xc sut phn tkh c vn tc t v n v + dv l: dn/n. Mc-xoen
chng minh c rng:
( )dn
F v dvn
(9.5)
Trong hm F(v) c gi l hm phn b phn t kh theo vn tc, theo
Mc-xoen hm ny c xc nh:
23/2
22( ) 42
v
F v e v
vi:kT
m (9.6)
23/2
2242
vdn
e v dvn
(9.7)
Biu thc (9.7) chnh l biu thc nh lut phn b phn t kh theo vn
tc ca Mc-xoen. nh lut ny cho bit xc sut mt phn t kh c vn tc
nm trong khong (v, v +dv).9.3.2.nh lut phn b ht theo th nng ca Bn-z-man.
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Bn-z-man tm ra s ph thuc ca mt phn t kh vo v tr ca n
trong mt trng lc m n ang tn ti v c th l trng trng ca Tri t.
Xt mt khi kh l tng ti mt cao h no trong mt hnh tr
tng tng t thng ng, c y l S v ng sinh l dh. Ngi ta chng
minh c rng chnh lch v p sut gia hai y tr l: .dp g dh (9.8)
Du (-) biu th khi tng cao th p sut gim
Thayp
RT
vo ta c: .
pdp g dh
RT
Hay:.dp g dh
p RT
. Ly tch phn hai v ca biu thc ta c:
( )
(0) 0
.( ) (0)
p h h gh
RT
p
dp g dhp h p e
p RT
(9.9)
Cng thc (9.9) c gi l cng thc kh p, n cho ta bit s bin i p
sut kh quyn theo cao trong trng trng u ca Tri t. Theo phng
trnh c bn ca thuyt ng hc phn t ta thy n 0 t l thun vi p, do t
(9.9) ta suy ra c biu thc tng ng l:
n( ) (0)gh
RTh n e (9.10)
Mt khc ta thy:. . .
Am N g hgh
RT RT
, trong m l khi lng ca mi phn
t kh.
Suy ra:W
tgh mgh
RT kT kT
, vi Wt= mgh l th nng tng tc ca phn t
kh ti cao h.
Vy ta c:W
n( ) (0)t
kTh n e
(9.11)
Cng thc (9.11) chnh l biu thc ca nh lut phn b phn t kh theo
cao ca Bn-z-man. nh lut cho php ta xc nh s ph thucca mt
phn t vo cao.
CU HI N TP CHNG 9
1. Trnh by cc ni dung c bn ca thuyt ng hc phn t cc cht kh2. Thit lp phng trnh c bn ca thuyt ng hc phn t cc cht kh. Nu cc h qu 3. T phng trnh c bn ca thuyt ng hc phn t cc cht kh v biu thc ca
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ng nng tnh tin trung bnh. Hy tm li phng trnh trng thi ca kh l tng.
4. Ni nng l g? Ti sao ni ng nng ca kh l tng chnh l tng ng nngchuyn ng ca cc phn t?
5. Khi nim bc t do. Pht biu nh lut phn b u nng lng theo bc t do6. Thit lp biu thc ni nng ca kh l tng. Ti sao ni trong qu trnh ng nhit
ni nng ca h khng thay i?
CHNG X. NGUYN LY TH NH T NHIT NG LC HOC
10.1. NNG LNG, NHIT VA CNG
10.1.1. Nng lng.Nng lng l i lng c trng cho mc vn ng ca vt cht,
mi trng thi khc nhau ca vt cht th dng vn ng ca h cng khc nhau.
Khi trng thi ca h thay i th nng lng ca h cng thay i theo, do vy
nng lng l hm ca trng thi.
Trong nhit ng hc ta ch kho st nng lng bn trong h, nng
lng chnh bng ni nng ca h.
10.1.2. Nhit v cng.
Nhit v cng cng l thc o mc nng lng truyn t h ny cho h
khc, hoc t dng nng lng ny sang dng nng lng khc. Ta c th biu
th hai i lng cng v nhit thng qua m hnh trao i nng lng sau: hnh
v m hnh truyn nng lng ca h A cho h B
Hnh 10-1
Cng v nhit lng l hai i lng khc nhau nhng n tng ng
nhau, v d mun lm cho mt khi lng kh nng ln (ni nng tng), ta c thtin hnh theo hai cch: cch th nht l truyn nhit , cch th hai l dng lc
nn kh.
HA HB
Phng tin: Lc
Phng tin: Nhit (Nng lng o bng Nhitlng)
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10.2. NGUYN LY TH NH T
10.2.1. Pht biu
Nguyn l th nht l mt trng hp ring ca nh lut bo ton v bin
i nng lng vn dng vo cc qu trnh v m (qu trnh nhit ng hc).
bin thin nng lng ton phn W ca h trong mt qu trnh
bin i v m c gi tr bng tng cng v nhit ca h trao i trong qu
trnh :
W = A + Q (10.1)
trn ta bit nng lng ca mt h nhit ng chnh l ni nng ca h:
W = U, do W = U
H thc (10.1) khi tr thnh:
U = A + Q (10.2)
Ngha l:Trong qu trnh bin i trng thi, bin thin ni nng
ca h c gi tr bng tng cng v nhit ca h trao i trong qu trnh .
y l pht biu ca nguyn l th nht nhit ng hc
Trong mt s trng hp, tin tnh ton, ngi ta dng cc k hiu v
pht biu nh sau:Nu A v Q l cng v nhit m h nhn c th A = - A v Q = - Q l
cng v nhit m h sinh ra. Do (10.2) cng c th c vit:
Q = U + A (10.3)
(10.3) cho ta mt cch pht biu khc ca nguyn l th nht: Nhit
truyn cho h trong mt qu trnh c gi tr bng bin thin ni nng ca
h v cng do h sinh ra trong qu trnh .
Cc i lng Q, U, Ac th dng hay m.
-Nu A > 0 v Q > 0 th U > 0, ngha l khi h thc s nhn cng v
nhit t bn ngoi th ni nng ca h tng.
-Nu A < 0 v Q < 0 th U < 0, ngha l khi h thc s sinh cng v ta
nhit ra bn ngoi th ni nng ca h gim.
10.2.2. H qu ca nguyn l th nht.
Ta xt mt vi trng hp ring ca nguyn l:a. Qu trnh bin i trng thi h theo mt qu trnh khp kn (chu trnh):
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U1 = U2, do vy: 0A Q A Q
Nh vy: h nhn cng truyn nhit, hoc h nhn nhit thc hin cng.
iu ny cho bit:mun sinh cng th h phi nhn nhit lng t ngoi vo.
H qu ny ca nguyn l cho ta mt ngha: Khng c my thc hincng m khng cn tiu th nng lng (khng c ng c vnh cu loi 1)
b. Xt h c lp gm hai vt khng trao i cng v nhit vi bn ngoi:
Q12 = A12 = 0, do vy U12 = 0
Nu h khng sinh cng th Q1 + Q2 = 0 ta c Q1 = - Q2
Trong mt h c lp ch gm hai vt, nu vt ny to nhit th vt kia phi
thu nhit, Nhitlng to ra bng nhit lng thu vo
10.3. NG DNG NGUYN L TH NHT
10.3.1. Qu trnh cn bng.
Trng thi cn bng:L trng thi trong mi thng s ca h c xc
nh v tn ti khng i.
Qu trnh cn bng: L mt qu trnh bin i, gm mt chui lin tip
cc trng thi cn bng. Thc t khng c qu trnh cn bng, v trong qu trnh
bin i: khi trng thi cn bng trc b ph v th trng thi cn bng sau licthit lp. Tuy nhin nu qu trnh bin i xy ra rt chm th c th xem
nh mt qu trnh cn bng.
Cng trao i ca h trong qu trnh cn bng: Xt h l mt khi lng
kh nht nh ng trong xi lanh c gii hn bi pt tng, bin i trng thi
theo qu trnh cn bng.
+ Nu kh gin n: ln cng trong mt s dch chuyn nh ca pt
tng dV.Pdl.S.Pdl.FdA . va tho mn dV > 0 (do gin n), va thomn dA < 0 (h nhn cng) th vit biu thc gi tr i s cng trong qu trn h
nn l dV.PdA .
+ Nu kh b nn: ln cng trong mt s dch chuyn nh ca pt
tng dV.Pdl.S.Pdl.FdA . va tho mn dV < 0 (do b nn), va tho mn
dA > 0 (h nhn cng) th vit biu thc gi tr i s cng trong qu trnh nn
l dV.PdA .
Vy xt mt cch tng qut, biu thc gi tr i s cng trong mt qutrnh cn bng l:
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2 2
12
1 1
A A PdV (10.4)
Nhit lng trao i ca h trong qu trnh cn bng:Gi Q l nhit
lung m h trao ivi bn ngoi gia hai trng thi, th biu thc tnh nhit
lng trao i ca h l:
+ .Q C m Vi C l nhit dung ca h c khi lng m.
+ dT.m.C)TT(m.CdQ *12
* Vi C*l nhit dung ring ca h.
+ dT.Cm
)TT.(Cm
dQ12
Vi C l nhit dung mol phn t ca h.
Nu xt trong mt qu trnh phc tp t trng thi 1 n trng thi 2 th:
2 2
1 1
mQ Q C dT
10.3.2. ng dng nguyn l th nht vo cc qu trnh bin i.
a. Qu trnh ng tch:
+ th biu din qu trnh
+ Th tch khng i V1 = V2 , p sut, nhit thay i.
+ Phng trnh trng thi:2
2
1
1
T
P
T
P
+ Cng trong qu trnh: t trng thi 1n trng thi 2:
A12= 0PdV2
1
do dV=0
+ bin thin ni nng (kh l tng):
T2
iRmU
+ Nhit lng trao i: p dng nguyn l th nht U = Q + A = Q (v A = 0)
Vi: T.Cm
QV
, do ta c biu thc nhit dung mol phn t trong qu trnh
ng tch l:
2iRC
V
P
V
T2P2
P1 T1
P
V
NM
T2T1
V2
P
V1
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b. Qu trnh ng p:
+ p sut khng i P1=P2 , th tch v nhit thay i.
+ Phng trnh trng thi:2
2
1
1
TV
TV
+ Cng trong qu trnh: t trng thi 1 n trng thi 2
12
2
1
( )2 1
A PV PVPdV
Ta c: 11 RTm
PV
v 22 RTm
PV
T.Rm
TT(Rm
A )1212
+ bin thin ni nng (kh l tng): T2
iRmU
+ Nhit lng trao i: p dng nguyn l th nht AUQQAU
Suy ra: TR
2
iRmT.R.
mT
2
iRmAUQ
Mt khc: TCm
QP
ng nht biu thc, ta rt ra nhit dung
mol phn t trong qu trnh ng p:
R2
iRC
P
c. Qu trnh ng nhit:
+ th biu din qu trnh ( hnh v )
+ Nhit khng i T1=T2 , p sut v th tch thay i.
+ Phng trnh trng thi: PVVPVP 2211
+ Cng trong qu trnh: t trng thi 1n trng thi 2:
2
1
1212 12
. .
V
V
m dV m dV
A P dV RT RTV V
P1
V
N
M
V2
P2
V1
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212
1
.lnVm
A RTV
+ bin thin ni nng (kh l tng): 0T2
iRmU
v T = const
+ Nhit lng trao i: p dng nguyn l th nht
AQ0QAU
+ Nhit lng trao i:1
212
V
Vln.RT
mAQ
d. Qu trnh on nhit:
+ nh ngha: l qu trnh h bin i nhng khng trao i nhit vi bn ngoi.
+ Nhit lng trao i Q=0.
+ Xt trong qu trnh nh: dU = dA + dQ = dA
Trong : dT.C.m
dT2
iR.
mdU V
v dV.PdA
ta c: . . -PdVVm mRT dV
C dTV
Do :2 2
1 1
0
T V
V VT V
dT R dV dT R dV T C V T C V
Ly tch phn v bin i: 1 1ln ln ln ln 0V V
R RT T V V
C C
*
1 1ln ln ln lnV V
R RT V T V const
C C
( ) ( )**
ln . .V V
R R
C C
T V const T V const
1 1P V P
V V V
C C CR
C C C
Vi l hs Poat xng. Thay vo ta c phng trnh theo V, T:
( 1) ***TV const
hay ( 1) ( 1) ( 1)1 1 2 2. . ......... .T V T V T V
Nu thay mRPV
T
v bin i ta c phng trnh theo P, V:
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( 1) ( 1) ***. .T V PV V const mR
1 1 2 2. . . ................ .PV const P V P V PV
+ Cng trong qu trnh on nhit:
Theo biu thc tnh cng: 2
1
V
V
dV.PA
Qu trnh on nhit p sut thay i do vy ta phi biu din P nh l
mt hm s ca th tch V trc khi thc hin php tch phn. Trong qu trnh
on nhit th rt P ta c:V
VPP 11
Thay vo biu thc tnh cng ta c:
)1(
V.PV.PA
V.VPV.VP)1(
1A
)1(
VlVP
V
dVVPA
1122
1
111
1
222
V
V
V
V
1
1111
2
1
2
1
CU HI N TP V BI TP CHNG 10
1.Trnh by cc khi nim, cng thc tnh v: nng lng ca chuyn ng nhit
2.Nu nguyn l, h qu, ngha ca nguyn l th nht nhit ng hc?
3.Th no l trng thi cn bng, qu trnh cn bng? Cng v nhit trong mt
qu trnh cn bng trng qut?
4.Vn dng nguyn l nguyn l th nht nhit ng hc kho st cc qu trnh
cn bng
5.Mt khi kh dn n on nhit sao cho p sut ca n gim t P1= 2at n
P2= 1at, V2=2lt, sau h nng ng tch n nhit ban u th p sut l P3
= 1,22at.
a. V th v xc nh t s Cp/Cv.b. Nhit m khi kh nhn vo trong cc qu trnh trn?
P S: a) = 1,4 ; b) Q = 64,75J
6.Mt khi kh ban u c th tch V1= 0,39 m3
v p sut P1=1,55.105
N/m2
,c dn ng nhit sao cho th tch tng 10 ln. Sau kh c t nng ng
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tch trng thi cui p sut ca khi kh bng p sut ban u. Bit trong ton
b qu trnh ny, nhit lng phi truyn cho khi kh l 1,5.106J.
a. V qu trnh trn th OPV.b. Xc nh t s Potxng .c. Tnh bin thin ni nng, cng do khi kh sinh ra trong qu trnh trn.
P S: b) = 1,4 ; c) U = 13,61.105J; A = 1,39.105J
7.Ngi ta mun nn 10 lt khng kh n th tch 2 lt. Hi nn nn ng nhit
hay nn on nhit? HD: Qu trnh no t tn cng hn th c li hn
8.Nn ng tch 3lkhng kh p sut 1at. Tm nhit ta ra bit rng th tch
cui cng bng 1/10 th tch ban u. Q = 676J
9.Mt bnh kn th tch 2l, ng 12g kh nit nhit 10oC. Sau khi h nng,
p sut trung bnh ln ti 104mmHg. Tm nhit lng m khi kh nhn c,bit bnh gin n km. P S: Q = 4,1 kJ
10. H nng 16 gam kh xy trong mt bnh kh gin n km nhit 370C, t
p sut 105N/m2ln ti 3.105N/m2. Tm:
a. Nhit ca khi kh sau khi h nng;b. Nhit lng cung cp cho khi kh.
P S: a) T2= 930K ; b) Q = 6,4 kJ
11. Sau khi nhn c nhit lng Q= 150cal, nhit ca m = 40,3g kh Oxi
tng t t1= 16oC ti t2=40
oC. Hi qu trnh h nng c tin hnh trong iu
kin no?
12.2 kmol kh ccbonic c h nng ng p cho n khi nhit tng thm
50oC. Tm
a. bin thin ni nng ca khi khb. Cng do kh gin n sinh rac. Nhit lng truyn cho kh
P S: U = 2500 kJ ; A = 830 kJ ; c) Q = 3330 kJ
13.7 gam kh ccbonic c h nng cho ti khi nhit tng thm 10oC trong
iu kin gin n t do. Tm cng do kh sinh ra v bin thin ni nng ca
n. P S: A = 13,2 J ; U = 39,7 J
14.2m3kh gin n ng nhit t p sut p=5at n p sut 4at. Tnh cng do kh
sinh ra v nhit lng cung cp cho kh trong qu trnh gin n.
P S: 2,2.105J
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CHNG XI. NGUYN LY TH HAI NHIT NG LC HOC
11.1.NH NG HN CH CU A NGUYN LY TH NH T
Tt ccc qu trnh xy ra trong t nhin u phi tun theo nguyn l th
nht, nhng ngc li, mt qu trnh tho mn nguyn l th nht c th vn
khng xy ra trong thc t. V d: S truyn nhit t vt lnh sang vt nng tuntheo nguyn l th nht; vin n gm vo tng c nng chuyn ho thnh nhit
nng. Ta hy tng tng qu trnh ngc li, mc d vn tun theo nguyn l
th nht, nhng li khng th xy ra, l do l nguyn l th nht cha cp ti
chiu din bin ca qu trnh. Nh vy: mt hn ch ca nguyn l th nht
NH l cha cp ti chiu din bin ca qu trnh trong mt h c lp.
Nhit v cng u l thc o ca mc nng lng, theo nguyn l th
nht hai i lng ny tng ng nhau, c th chuyn ho ln nhau. Nhng
trong thc t, cng c th chuyn ho hon ton thnh nhit, ngc li nhit
khng th chuyn ho hon ton thnh cng. Nh vy: mt hn ch na ca
nguyn l th nht l cha nu ln c im khc bit gia cng v nhit.
Nguyn l hai nhi t ng lc hoc s b sung v khc phc nhng hn ch trn.
11.2. NGUYN LY TH HAI NHIT NG LC HOC
11.2.1. Qu trnh thun nghch v qu trnh bt thun nghch
nh ngha 1:Mt qu trnh bin i ca mt h t trng thi A sang trng thi B c
gi l mt qu trnh thun nghch nu h c th t thc hin theo chiu ngc
li; v trong qu trnh ngc li h qua cc trng thi trung gian m qu
trnh thun qua.
i vi qu trnh thun nghch, nu lt i h nhn cng A th lt v
h tr ng cng A cho mi trng. Nh vy A=0, U=0,Q=0.
Vy:i vi qu trnh thun nghch, sau khi thc hin qu trnh thun vqu trnh nghch mi trng khng b thay i.
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C th ly v d v qu trnh thun nghch nh cc con lc dao ng iu
ho, chuyn ng ca piston trong xilider (xilanh).v.v...
nh ngha 2:
Mt qu trnh bin i ca mt h t trng thi A sang trng thi B c
gi l mt qu trnh bt thun nghch nu h khng th thc hin c theochiu ngc li; hoc nu c th h khng qua hoc khng i ht cc trng thi
trung gian m qu trnh thun qua.
Qu trnh thun nghch l qu trnh l tng, thc t khng xy ra. Mi
qu trnh xy ra trong thc t u l qu trnh bt thun nghch.
11.2.2.My nhit.
Khi nim:L my bin cng thnh nhit, hoc bin nhit thnh cng.
Trong my nhit c cht vn chuyn nhit gi l tc nhn lm nhim v
trao i nhit gia cc vt c nhit khc nhau gi l cc ngun nhit. Cc
ngun nhit c coi nh ngun c nhit khng thay i, s trao i nhit
gia cc vt khng lm thay i ti nhit ca n, ngun c nhit cao c
gi l ngun nng, ngun c nhit thp c gi l ngun lnh. Qu trnh hat
ng ca my nhit, tc nhn trong my nhit c vn hnh mt cch tun hon
(hay theo mt chu trnh).
Phn loi:
+ ng c nhit: Chuyn ho nhit nng thnh c nng.
Hnh 11-1 trnh by lu ca mt ng c nhit.
Hnh 11-1
Chu trnh lm vic ca ng c nhit c th hin trn (hnh 11-2):
ng cong pha trn biu din qu trnh thun, tc nhn nhn nhit t ngun
nng Q1 v nh nhit cho ngun lnh Q2 , ng thi dn n sinh cng (Adn
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lm lnh l t lnh dng trong gia nh. Ngun lnh T2l bung lnh (ngn cha
lm lnh). Cng ca t lnh nhn c cng l do m t nn kh. Ngoi ra cc
my iu ha cng thuc my lm lnh.
Mc ch ca cc my lm lnh l chuyn nng lng di dng nhit t
ngun lnh n ngun nng di tc dng ca cng ngoi lc ln tc nhn. nh gi hiu sut camy, ngi ta a ra i lng gil h s lm
lnh :
2 2
1 2
Q Q
A Q Q
(11.3)
H s lm lnh cng cao th my lm lnh cng tt i vi t lnh m
khng cn nhn cng t bn ngoi. V A = 0 do c gi tr ln v cng.
Trong thc t khng th ch to c my lm lnh vnh cu.
11.2.3. Pht biu nguyn l th hai nhit ng hc
Nguyn l th hai c pht biu di nhiu cch khc nhau. Cc cch pht
biu ny v thc cht l hon ton nh nhau. Trong mc ny, chng ta s trnh
by hai cch pht biu tiu biu: pht biu ca Thompson v pht biu ca
Clausius v chng ta s ch ra rng hai cch pht biu l hon ton tng
ng.
a. Cch pht biu nguyn l II ca Thompson :
Cch pht biu nguyn l II ca Thompson c lin quan n ng c
nhit.
Cc bng chng thc nghim xc nhn rng khng th no ch to c
mt ng c nhit m ng c ny bin i hon ton lng nhit m n nhn
vo thnh cng c hc, ni khc i khng th no ch to c mt ng c
nhit c hiu sut 100%.Trong cc ng c nhit vic chuyn mt phn nhit lng d tha cho
ngun lnh l mt iu bt buc. iu va nu trn c tha nhn nh l mt
nguyn l v c Thompson pht biu di dng nguyn l II nhit ng hc
nh sau :
Mt ng c nhit khng th sinh cng nu nh n ch trao i nhit vi
mt ngun nhit duy nht
Ngi ta gi mt ng c nhit hot ng tun hon sinh ra cng bngcch ch trao i nhit vi mt ngun nhit duy nht l ng c vnh cu loi hai,
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ni khc i ng c vnh cu loi II l ng c nhit c hiu sut 100%. Nguyn
l II khng nh rng khng th ch to c ng c vnh cu loi II.
b. Cch pht biu nguyn l II ca Clausius:Cch pht biu nguyn l II ca Clausius c lin quan n my lm lnh.Chng ta c th hnh dung mt my lnh nh l mt thit b lm vic nh
mt ng c nhit nhng trong cc qu trnh u din ra theo chiu ngc li.
Mt ng c nhit ly nhit t mt ngun nng v nh mt phn nhit cho mt
ngun lnh sau khi sn ra mt cng c hc no , ngha l mt ng c nhit
to ra mt cng c hc u ra. Mt my lnh lm vic hon ton ngc li. N
ly nhit t mt ngun lnh (tc l cc vt nm trong bung lnh ca mt t
lnh) v nh nhit cho mt ngun nng hn (tc mi trng khng kh xung
quanh) nhng my lnh i hi mt cng c hc u vo.Nh vy trong my lnh, nhit c truyn t vt lnh sang vt nng
nhng khng phi l mt qu trnh din tin mt cch t nhin m l mt
qu trnh c s can thip t bn ngoi: ta phi a mt cng c hc vo h th khi
mi c qu trnh truyn nhit t vt lnh sang vt nng hn.
Cc phn tch v nguyn l lm vic ca my lnh dn n mt cch pht
biu khc ca nguyn l II nhit ng hc. l cch pht biu ca Clausius:
Khng th tn ti mt qu trnh nhit ng m kt qu duy nht l struyn nhit t mt ngun lnh cho mt ngun nng.
c. S tng ng ca hai cch pht biu nguyn l II :
Cch pht biu ca Clausius mi thot nghe th c v nh khng lin quan
g vi cch pht biu ca Thompson, nhng tht ra hai cch pht biu trn u
tng ng nhau. Chng ta s chng minh iu .
Cch chng minh ca chng ta nh sau: nu mt my vi phm cch pht
biu ca Clausius th n cng vi phm cch pht biu ca Thompson v ngc li
nu n khng vi phm cch pht biu ca Clausius th n cng s khng vi phm
cch pht biu ca Thompson.
Tht vy, gi s nu ta c th ch to
c mt my lnh m khng nhn cng t
bn ngoi (my lnh ny r rng vi phm cch
pht biu ca Clausius).
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+ Qu trnh nn on nhit (4 - 1): Trong qu trnh ny cht vn chuyn
cch nhit vi bn ngoi, khng trao i nhit, nhn cng t bn ngoi.
Hiu sut ca chu trnh Ccn thun nghch:
+ Biu thc: Gi thit cht vn chuyn l kh l tng, ta c hiu sut
ca ng c nhit:
1
2
1
21
Q
Q1
Q
Ngi ta chng minh c hiu sut ca chu trnh l:
2 2
1 1
1 1Q T
Q T (11.3)
Tng qut:1
2
1
2
TT1
QQ1 (11.4)
V my nhit chy theo chu trnh Cc-n thun nghch nn nu ta cho chy
theo chiu nghch ta s c my lm lnh Cc-n, ngi t cng chng minh
c hiu sut my lm lnh Cc-n cho bi biu thc:
2
1 2
TK
T T
(11.5)
nh l Ccn: Hiu sut ca tt c cc ng c chy theo chu trnhCcn ch ph thuc vo nhit ngun nng v ngun lnh m khng ph
thuc vo tc nhn v cch ch to my. Hiu sut ca chu trnh Ccn thun
nghch ln hn hiu sut ca ng c chy theo chu trnh Ccn bt thun
nghch cng lm vic vi hai ngun nhit trn.
11.4. Biu thc nh lng ca nguyn l th hai
T biu thc tnh hiu sut ca ng c nhit thun nghch chy chu trnh
Carnot, ta c:
2 2
1 1
'T Q
T Q
T biu thc nh ngha hiu sut v ca chu trnh Carnot, ta c:
1 2 1 2
1 1
'Q Q T T
Q T
(11.6)
l biu thc nh lng ca nguyn l th 2.Ta thit lp biu thc tng qut ca nguyn l th 2:
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T (11.6) ta c:
2 2
1 1
'T Q
T Q (11.7)
Ta c: Q2= - Q2
Suy ra: 1 21 2
0Q Q
T T (11.8)
Trng hp tng qut:cc qu trnh ng nhit ln lt tng ng vi nhit :
T1, T2, Ti ca ngun nhit bn ngoi v ng vi nhit lng Q1, Q2, Qi m
h nhn c t bn ngoi, t (11.8) ta c:
1
0n
i
i i
Q
T (11.9)
Nu trong mt qu trnh bin thin lin tc, ta c th coi h tip xc vi ln lt
v s ngun nhit c nhit T v cng gn nhau v bin thin lin tc. Mi qu
trnh tip xc l mt qu trnh vi phn trong h nhn c nhit Q, h thc
(11.9) tr thnh:
0Q
T
(11.10)
+ Du = ng vi chu trnh thun nghch,+ Du < ng vi chu trnh khng thun nghch.
(11.10) l biu thc nh lng tng qut ca nguyn l th 2.
11.4. HM ENTROPI V NGUYN L TNG ENTROPI
11.4.1. Cng thc lin h gia S, Q v T.
a. Cng thc.
Mt h c nhit T, khi trao i nhit vi bn ngoi l Q th binthin entrpi ca h l S. Theo l thuyt thng k chng minh c cng
thc lin h sau:
T
QS (11.11)
V mt nh tnh cng thc trn cho ta bit: Nhit lng Q ca h nhn
c s lm tng mc chuyn ng nhit cc phn t, trng thi hn lon ca s
phn b cc vi thi tng ln, entrpi ca h s tng ln mt lng S t l vi Q.Nu nhit ca h cao th nhit lng Q ca h nhn c s t lm thay i
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trng thi hn lon ca h, xc sut nhit ng W s tng t, do S ca h nh.
Ngc li nu nhit ca h thp th nhit lng Q ca h nhn c s lm
trng thi hn lon ca h thay inhiu, xc sut nhit ng W s tng nhiu,
do S ca h ln.
b. n v
T biu thc trn ta nhn thy th nguyn ca entrpi S l Jun/ (J/).
c. Cc tnh cht Entrpi.
+Entrpi S l mt hm s ca trng thi, n khng ph thuc vo qu trnh
ah t trng thi ny qua trng thi khc. Do vy ta c th thay th vic tnh
bin thin S ca h gia hai trng thi trong nhng qu trnh bt thun
nghch bng vic tnh bin thin S ca h gia hai trng thi trong nhng
qu trnh thun nghch gia cng hai trng thi .
+Entrpi S l mt i lng c tnh cht cng. Chng minh tnh cht ny
nh sau:
Ta c: S = k.lnW vi Wh = W1.W2....WN
thay vota c: S = k.ln W1.W2....WN = k.ln W1+ k.lnW2 +....+ k.ln WN
Suy ra: S = S1+ S2+ S3+........+ SN = Si
Cng thc cho ta xc nh entrpi S ca h phc tp gm nhiu phn hoc
gm nhiu qu trnh bin i.
d. bin thin entrpi trong mt qu trnh.
Trong mt qu trnh thun nghch v cng nh bin thin entrpi l dS
c vit l:T
dQdS
(11.12)
ly tch phn hai v ng thc: qtTN
S
ST
dQdS
2
1
vi S1 v S2 tng ng l
cc gi tr ca hm s S ti trng thi u v trng thi cui. Do vy v tri biu
thc l: SSSdS 12S
S
2
1
Vy bin thin entrpi trong qu trnh thun nghch l:
qtTN T
dQS
(11.13)
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Tnh bin thin entrpi trong mt s qu trnh:
bin thin entrpi ca h khng c lp trong qu trnh thun nghch.
Xt mt khi lng kh l tng thc hin qu trnh bin i t trng thi
1 (P1V1T1) sang trng thi 2 (P2V2T2). Ta tin hnh bin i nh sau:
Theo nguyn l th nht NH v phng trnh trng thi kh l tng ta
c: PdVdT2
iRmdAdUdQ
Suy ra: dV
V
RTmdT
2
iRmdQ
thay vo trn, ta c:V
dVR
m
T
dT
2
iRm
T
dQdS
tch phn hai v v bin i:
2 2 2
1 1 1
2 212
1 1
2
ln ln2
S T V
S T V
m iR dT m dV dS RT V
T Vm iR mS R
T V
2 2
12
1 1
ln lnVT Vm m
S C RT V
(11.14)
+ Vy nu biu din bin thin S theo V,T th ta c biu thc:
1
2
1
2V
VVlnRm
TTlnCmS
+ Nu biu din bin thin S theo V,P th ta p dng phng trnh
trng thi bin i:
11
22
1
2
VP
VP
T
T
ta c:
1
2
1
2V
1
2V
1
2
11
22
V
V
VlnR
m
V
VlnC
m
P
PlnC
mS
V
V
lnR
m
VP
VP
lnC
m
S
suy ra:1
2P
1
2V
V
VlnC
m
P
PlnC
mS
Cc trng hp c bit:
+ Qu trnh ng nhit (T1=T2), biu thc c vit thnh:
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2
1
1
2
P
PlnR
m
V
VlnR
mS
(11.15)
+ Qu trnh ng p (P1 = P2), biu thc c vit thnh:
1
2
p
1
2
P
T
TlnC
m
V
VlnC
mS
+ Qu trnh ng tch (V1=V2), biu thc trn