Formule Biofizica
Transcript of Formule Biofizica
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Formule Biofzica
CURS I
∆U = U 2 − U 1 = Q − L
(1.1)
Q
L
∆U
Fig. 1.1 – Ilustrarea Principiului I al termodinamicii
Schematic, relaţia este repreentat! "n #igura 1.1.
$elaţia (1) se mai poate scrie %i su& #orma'
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L = Q − ∆U
(1.)
U 2 = U 1 , ∆U = 0 si L = Q(1.)
∆U = 0 , U 2
= U 1
(1.*)
U 2 = U 1 si Q = L (1.+)
L = 0 ; ∆U = QV
Q p
= ∆U + p ∆V = (U 2 −U 1 ) + p (V 2
−V 1
)
(1.)
Q P = (U 2 + pV 2 ) − (U 1 + pV 1 )
(1.-)
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Q P = H 2 − H 1 = ∆ H
(1.)
unde H = U + pV .
η =
Lu
=
Q1 − Q2
= 1 −
Q2
= 1 −
T 2
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Qc
Q1
Q1
T 1
(1.1/)
L = Q %i Q > L
(1.11)
Qdegradat 0 S·T
Intrun sistem iolat'
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∆S ≥ /
U − T ⋅ S = F
sau'
∆U − T ⋅ ∆S = ∆ F
∆ F = −T ⋅ ∆S < 0
deoarece ∆S 2 /.
CURS II
S = k · ln P
(2.1)
S = − K ∑
ni
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log
ni
i N
N
(.)
S = − K ∑ pi log pi
i
(.)
(2.4)
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∆ S
=
∆e S
+
∆i S
∆t
∆t
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∆t
(.-)
σ e 0 - σ i
(.3)
Φ = T ⋅σ = −
∆i F
∆t
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(.1/)
∆4 0 56 ∆S 0 4P – 4$
(.11)
56∆S 0 4$ – 4P
η 0 Lutil 7 Qconsumat
8in punct de 9edere &iochimic, randamentul unui sistem meta&olic este '
η 0 4stocat 7 4consumat
Fotosistemul I – captarea energiei radiante, hν , pe dou! c!i' 1. transport neciclic deelectroni'
2 Clorofila A + 2 h ν (700 nm) 2 Clorofila A + NAD! + A" + !2#
. transport ciclic de electroni'
2 Clorofila A + 2 h ν (700 nm) 2 Clorofila A + A" + !2#
Fotosistemul II – #otolia apei'
!2# + 2 h ν ($%0 nm) + 2 Clorofila & 2 Clorofila & + ' #2 + 2 !+
II. Faza de sinteză (:iclul :al9in)
$ C#2 + 12 NAD! + 1% A" + 12 !2# C$!12#$
$ C#2 + 24 !2# + $0 h ν C$!12#$ + $ #2 + 1% !2#
:;<1=; > ;= ;:= > ;<= ∆4 0 ;-; ?cal7mol
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@tapele principale ale respiraţiei celulare sunt'
1. Glicoliza aerobă'
C$!12#$ + 2 NAD + 2 AD 2 ir*a +2 NAD! + 2 A"
. Sinteza AcetilCoA'
ir*a + NAD A-ilCoA + NAD! + C#2
. Ciclul acizilor tricarboxilici (ciclul Krebs)'
A-ilCoA / NAD! + AD!2 + A" + 2C#2
*. Lanţul respirator (transportul de electroni)'
NAD! + !+ + ' #2 NAD+ + / A"
AD!2 + ' #2 AD + / !2# + 2 A"
5. Pompa de protoni
:;<1=; > ;= ;:= > ;<= > ; A5P
A5P > <= A8P > P ∆4 0 . ?cal7mol
BA8< > <> > C = BA8> > <= ∆4 0 +.* ?cal7mol
CURS III
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f
=
na
a
N
Prin de#iniţie, probabilitatea ( pa) de apariţie a &ilei al&e se 9a eDprima prin'
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pa
= lim
na
N →∞ N
Ip f (1 p)
(.)
I = log 2
1
= −log 2 p
(.*)
p
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I = − log 2
1
= log 2 2 = 1bit
2
Un multiplu al &itului este &Eteul sau octetul'
1 Byte % bit log2 2% log2 23$
unde N = n1 > n > ... > nh, sau'
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I
=
n1
⋅ I
+
n2
⋅ I
+ +
nh
⋅ I
m
N
1
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I m = p1 I 1
+ p2 I 2 ++ ph I h
(.-)
:um I 0 log p, reult!'
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I m = − p1 ⋅log p1 − p2 ⋅log p2 −− ph ⋅log ph
(.)
sau'
h
I m
= −∑ pi ⋅ log pi
(.3)
i=1
h
1
1
1
1
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I m = −∑(
⋅log
) = −(h ⋅
) log
= log h
(.1/)
h
h
1
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h
h
h
1
1
1
1
I m = −∑(
⋅log
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) = −(h ⋅
) log
= log h
(.1/)
h
h
1
h
h
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I ma = log h
(.11)
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h
n
n
S = − k ∑
i
⋅ log
i
(.1)
N
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N
i=1
:nd N → ∞, atunci
ni
→ pi ast#el'
N
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h
S = − k ∑ pi
⋅ log pi
(.1)
I m = H
(.1*)
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R H ma 5 H
(.1+)
R =
R
=
H ma4 − H
= 1−
H
H ma
H ma
H ma
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(.1;)
CURS IV
µ = ! " #
(4.1)
$ % =
υ
=
m&
(*.)
V
µ ⋅V
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$
= 100 ⋅
m&
6
(*.)
P
V
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$
=
echi'a#en(i
= υ ⋅ ) =
m& ⋅ )
(*.*)
N
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µ = µ 0 + RT ln
c
(*.;)
c0
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n
( µ i + ) i F Ψ)
H = ∑υ i
(*.)
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K =
[$+ ]⋅ [ *
− ]
(*.)
[$*]
&
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R = ρ
#
S
Λ
∞
= Λ
+
+ Λ
−
(*.1)
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unde'
Λ+ = k ⋅u+
iar
Λ− = k ⋅u−
u =
'
(*.1*)
,
a = # 6 c
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I =
1
∑$ i ⋅ ) i2
(*.1)
2
i
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unde :i este concentra ia molar! a ionuluiț i , iar z este num!rul de sarcini
(9alen a ionului).ț
8e eDemplu, pentru o solu ie de /,/1ț
B
deBa:l' I =
0,01⋅12 + 0,01⋅12
= 0,01, iar
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2
pentru o solu ie de /,/1 B de <S=*'ț I =
0,01⋅ 22
+ 0,01⋅12
+ 0,01⋅12
= 0,0/ , adic! de )
2
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ori mai mare
K =
[ H + ]⋅[ *−
]
(*./)
[ H*]
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pH = pK
− log
[ H*]
= pK
+ log
[ *−
]
[ *− ]
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sau'
pH = pK a + log
accept- &e p-t-ni
&-n- &e p-t-ni
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!
Q 0 q p + pn 0 q α q + α qn 0 q G 1 > α (n1)H
(*.+)
sau'
i =
Q
= 1 + α (n −1)
(*.;)
!
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i 1 0 α (n1)
(*.)
sau'
α =
i −1
n −1
α =
i −1
=
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1,3 −1
= 0,3
(*.-)
n −1
2 −1
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CURS V
∆G=∆Wch +∆Wel = 0,
astfel:∆W
= −∆W , sau RT ⋅ ln
a2
= − z ⋅ F ⋅ ∆ E
ch
el
a1
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MM+z + z·e-
E = E 0 + RT ⋅ ln a, undea = f ⋅[ M + z ] zF
½ H 2<=> H+ + e-
Potenialul electrodului de hidrogen se exprim! prin:
E H
= E 0
+
RT
⋅ ln
[ H+ ]
zF
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pH 2
unde pH2 este presiunea hidrogenului. Dac! pH2 = 1 atm, atunci:
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α =
[α ]25 D ⋅ l ⋅ c
I = I 0 ⋅10−a ⋅b⋅c
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T =
I
E = −lgT = −lg
I
= −lg10−a ⋅b⋅c = a ⋅ b ⋅ c
I 0
I 0
dW =σ ·dS
dW =σ ·L·dl =F·dl
cosθ = γ sv − γ sv
σ
σ · cosθ =γ sv -γ sl
mg= σ · l= σ · 2πr
p = pin – pext
Energia necesar! meninerii bulei de aer este:
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dW = p ⋅ dV = p ⋅ 4π R2 ⋅ d R
undeinem seama c! volumulV =4π R3 /3, iar prin derivare:dV = 4π⋅ R2⋅d R
Energia de meninere este egalat! de energia de suprafa!:dW =σ ·dS =σ · 8π R ·d R
undeS = 4π R2, iar prin derivare: dS = 8π R · d R
Egalând cele dou! expresii obinem:
p · 4π R 2 ·d R = σ ·8π R · d R
deci:
p =
2σ
R
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1
1
p = σ
+
R1
R2
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p = 4σ R
p0 + ρ ⋅ gh = p0 + 2σ r
r = R cosθ
σ = ρ ⋅ ghR
2 cosθ
CURS VI
dm = − DS ⋅ dc dt d x
D = KT
6π η r
undeK este constanta lui Boltzmann (1,38·10-23 J/K),T este temperatura absolut!,η
este vâscozitatea mediului (N·s/m2), iarr este raza particulei.
v= s · pi · V
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π =
n
RT
V
π ⋅V = n ⋅ R ⋅T
W ch1 = µ 0 + RT ⋅ ln c1 c0