1/30/25, 10:28 AM Masao Doi - Soft Matter Physics (Instructor's Solution Manual) (Solutions)-…
1
Softmatterphysics
Solutionstoexercises
Chapter2
2.1 (a) Assuming that the specific volume of sugar and water are 1 [cm3 /g ], the weight concentration
is estimated as
10 [g]
c =
200 + 10 [cm3 ]
= 0.048 [g/cm3 ] (2.2)
For the same specific volume, the mass fraction is
φm = 0.048 (2.3)
The molar fraction is
10
500
xm = 10 200
500
+ 18
−3
= 1.8 × 10 (2.4)
(b) The gas constant is RG = 8.3[J/molK ].
10 1
Π = [mol] ∗ 8.3[J/molK] × × 300[K] (2.5)
500 210 × 10−6 [m3 ]
= 2.4 × 105 [Pa] = 2.4 [atm] (2.6)
2.2 (a) Number of polymers per unit volume is n = ∗
1/[(4π/3)R3g ] 20 −3
= 2.4 × 10 [m ]. The weight
concentration is
c∗ = n∗ M/NA = 0.012[g/cm3 ] (2.7)
where M is the molecular weight and NA is the Avogadro number.
(b) Osmotic pressure is given by
Π = n∗ kB T = 1[Pa] (2.8)
2.3 (a) The difference in the pressure across the semi-permeable membrane must be equal to the
difference in the osmotic pressure, (because the chemical potential of the solvent, given by
eq.(2.27), must be continuous across the semi-permeable membrane.) Let Π1 , and Π2 be the
osmotic pressure in the top and the bottom chamber. Then
W
Π1 − Π2 = (2.9)
A
For dilute solutions,
h h
Π1 = n 0 k B T , Π2 = n 0 k B T (2.10)
h−x h+x
Hence
h h
− =w (2.11)
h−x h+x
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2
where w = W/(An0 kB T ). Solution of eq.(2.11) is
x 1
=
h p
[ 1 + w2 − 1] (2.12)
For small W, w
1 W
x = wh= h (2.13)
2 2An0 kB T
For large W, x approaches to h.
(b) If the density of the solution is considered, the change in the pressure across the semi-permeable
membrane is W/A +2ρgh. Hence the answer is given by the effective weight Weff = W +2ρghA
which replaces W in eq.(2.13).
2.4 (a) The Gibbs free energy is written as G(P,N0 ,N1 ,...Nn , T). Since ∂G/∂P= V is independent
of P , G can be written as
G = PV + F (N0 ,N1 ,...,Nn , T) (2.14)
The function F (N0 ,N1 ,...,Nn , T) satisfies the following scaling relation for any parameter α:
F (αN0 ,αN1 ,...αNn , T) = αF(N0 ,N1 ,...,Nn , T) (2.15)
P
Setting α = 1/V = 1/ i Ni viN
, 0we Nhave N 1
1 n
F( , ,..., , T) = F (N0 ,N1 ,...Nn , T) (2.16)
V V V V
Hence
N0 N1 Nn
F (N0 ,N1 ,...,Nn , T) = V F( , ,..., T ) = Vf (φ1 ,...,φn ,T) (2.17)
V V V
φi .
Pni=1
where
(b) The we have
osmotic used is
pressure Ni /Vby
thatgiven is expressed
eq.(2.21), as φi /viF,tot is
where andnow
thatwritten
φ0 is written
as as 1 −
Ftot= Vf (φi ) + (Vtot− V )f (0) (2.18)
The osmotic pressure is given by Π = −∂Ftot /∂V. Using φi = Ni vi /V, we finally have
n
∂f
Π = φi − f (φi ) + f (0) (2.19)
X
i=1 ∂φi
The chemical potential µi is given by µi = ∂G/∂Ni , where G is given by
G = PV + Vf (φi ; T ) (2.20)
P
Using V = i vi Ni and φi = Ni vi /V, we have ∂f ∂φk
(2.21)
X k
∂φk ∂Ni
µi = Pvi + vi f + V
P
Using φk = Nk vk / Nj vj , we have
j ∂φk vi
= (δki− φk ) (2.22)
∂Ni V
Hence
n
∂f
" (δki− φk ) #
X
k=1 ∂f∂φi
µi = vi P + f + = vi P − Π + + f (0) (2.23)
∂φk
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2.7 (a) If the solution is uniform, having volume fraction φi0 for component i, the free energy of the
system is given by Ftot,0 = Vf (φi0 ). If there is a small perturbation δφi (r) = φi (r) − φi0 in
the concentration profile,the total free energy of the system is given by
Z
Ftot= drf (φi (r)) (2.31)
Expanding the right hand side for δφi , we have
∂f 1 ∂2f
Z X δφi (r) + Z δφi (r)δφj (r) (2.32)
i Xi,j
Ftot= Ftot,0 + dr dr
∂φi 2 ∂φ i ∂φ j
φi0 φi0
second term on the right hand side is zero, and R
Since the volume of the component i is conserved, drδφi (r) is equal to zero. Therefore the
1
Ftot− Ftot,0 = Z Hijδφi (r )δφj (r) (2.33)
Xi,j
dr
2
where Hij= ∂ 2 f /∂φi ∂φj |φi0 . For the solution to be stable, the right hand side of this equation
must be positive for any δφi (r). This is equivalent to the condition that all eigenvalues of the
matrix Hij is positive. At the boundary of the stable region,one solution of the eigen value
equation det(Hij− λδij) = 0 becomes zero. Therefore, at the boundary det Hij must be equal
to zero.
(b) The free energy is given by
kB T φi
f=
X Ni
(1 − φ) ln(1 − φ) + i=1,2
ln φi − χφ2 (2.34)
vc
where
φi (2.35)
X
i=1,2
φ=
The matrix Hij is calculated as
kB T N1 φ1
+ 1 − 2χ 1
− 2χ
(Hij) = 1 1−φ 1−φ 1
1
− 2χ 1
N2 φ2
+
1−φ
vc (2.36)
1−φ
− 2χ
The equation det(Hij) = 0 gives,
1 1
+ N φ + 2 (2.37)
N1 φ
11 12 2 1 − 2χ
− 2χ − 2χ = 1−φ
1 − φ 1 − φ
After some calculation, this gives the following equation:
1 φ
χ= 1 − φ +
1
(2.38)
which can be written as 2 N1 φ 1 + N2 φ 2
1 1
χ= 1 − φ +
1
(2.39)
2 Nw φ
Chapter3
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1
Softmatterphysics
Solutionstoexercises
Chapter2
2.1 (a) Assuming that the specific volume of sugar and water are 1 [cm3 /g ], the weight concentration
is estimated as
10 [g]
c =
200 + 10 [cm3 ]
= 0.048 [g/cm3 ] (2.2)
For the same specific volume, the mass fraction is
φm = 0.048 (2.3)
The molar fraction is
10
500
xm = 10 200
500
+ 18
−3
= 1.8 × 10 (2.4)
(b) The gas constant is RG = 8.3[J/molK ].
10 1
Π = [mol] ∗ 8.3[J/molK] × × 300[K] (2.5)
500 210 × 10−6 [m3 ]
= 2.4 × 105 [Pa] = 2.4 [atm] (2.6)
2.2 (a) Number of polymers per unit volume is n = ∗
1/[(4π/3)R3g ] 20 −3
= 2.4 × 10 [m ]. The weight
concentration is
c∗ = n∗ M/NA = 0.012[g/cm3 ] (2.7)
where M is the molecular weight and NA is the Avogadro number.
(b) Osmotic pressure is given by
Π = n∗ kB T = 1[Pa] (2.8)
2.3 (a) The difference in the pressure across the semi-permeable membrane must be equal to the
difference in the osmotic pressure, (because the chemical potential of the solvent, given by
eq.(2.27), must be continuous across the semi-permeable membrane.) Let Π1 , and Π2 be the
osmotic pressure in the top and the bottom chamber. Then
W
Π1 − Π2 = (2.9)
A
For dilute solutions,
h h
Π1 = n 0 k B T , Π2 = n 0 k B T (2.10)
h−x h+x
Hence
h h
− =w (2.11)
h−x h+x
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where w = W/(An0 kB T ). Solution of eq.(2.11) is
x 1
=
h p
[ 1 + w2 − 1] (2.12)
For small W, w
1 W
x = wh= h (2.13)
2 2An0 kB T
For large W, x approaches to h.
(b) If the density of the solution is considered, the change in the pressure across the semi-permeable
membrane is W/A +2ρgh. Hence the answer is given by the effective weight Weff = W +2ρghA
which replaces W in eq.(2.13).
2.4 (a) The Gibbs free energy is written as G(P,N0 ,N1 ,...Nn , T). Since ∂G/∂P= V is independent
of P , G can be written as
G = PV + F (N0 ,N1 ,...,Nn , T) (2.14)
The function F (N0 ,N1 ,...,Nn , T) satisfies the following scaling relation for any parameter α:
F (αN0 ,αN1 ,...αNn , T) = αF(N0 ,N1 ,...,Nn , T) (2.15)
P
Setting α = 1/V = 1/ i Ni viN
, 0we Nhave N 1
1 n
F( , ,..., , T) = F (N0 ,N1 ,...Nn , T) (2.16)
V V V V
Hence
N0 N1 Nn
F (N0 ,N1 ,...,Nn , T) = V F( , ,..., T ) = Vf (φ1 ,...,φn ,T) (2.17)
V V V
φi .
Pni=1
where
(b) The we have
osmotic used is
pressure Ni /Vby
thatgiven is expressed
eq.(2.21), as φi /viF,tot is
where andnow
thatwritten
φ0 is written
as as 1 −
Ftot= Vf (φi ) + (Vtot− V )f (0) (2.18)
The osmotic pressure is given by Π = −∂Ftot /∂V. Using φi = Ni vi /V, we finally have
n
∂f
Π = φi − f (φi ) + f (0) (2.19)
X
i=1 ∂φi
The chemical potential µi is given by µi = ∂G/∂Ni , where G is given by
G = PV + Vf (φi ; T ) (2.20)
P
Using V = i vi Ni and φi = Ni vi /V, we have ∂f ∂φk
(2.21)
X k
∂φk ∂Ni
µi = Pvi + vi f + V
P
Using φk = Nk vk / Nj vj , we have
j ∂φk vi
= (δki− φk ) (2.22)
∂Ni V
Hence
n
∂f
" (δki− φk ) #
X
k=1 ∂f∂φi
µi = vi P + f + = vi P − Π + + f (0) (2.23)
∂φk
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2.7 (a) If the solution is uniform, having volume fraction φi0 for component i, the free energy of the
system is given by Ftot,0 = Vf (φi0 ). If there is a small perturbation δφi (r) = φi (r) − φi0 in
the concentration profile,the total free energy of the system is given by
Z
Ftot= drf (φi (r)) (2.31)
Expanding the right hand side for δφi , we have
∂f 1 ∂2f
Z X δφi (r) + Z δφi (r)δφj (r) (2.32)
i Xi,j
Ftot= Ftot,0 + dr dr
∂φi 2 ∂φ i ∂φ j
φi0 φi0
second term on the right hand side is zero, and R
Since the volume of the component i is conserved, drδφi (r) is equal to zero. Therefore the
1
Ftot− Ftot,0 = Z Hijδφi (r )δφj (r) (2.33)
Xi,j
dr
2
where Hij= ∂ 2 f /∂φi ∂φj |φi0 . For the solution to be stable, the right hand side of this equation
must be positive for any δφi (r). This is equivalent to the condition that all eigenvalues of the
matrix Hij is positive. At the boundary of the stable region,one solution of the eigen value
equation det(Hij− λδij) = 0 becomes zero. Therefore, at the boundary det Hij must be equal
to zero.
(b) The free energy is given by
kB T φi
f=
X Ni
(1 − φ) ln(1 − φ) + i=1,2
ln φi − χφ2 (2.34)
vc
where
φi (2.35)
X
i=1,2
φ=
The matrix Hij is calculated as
kB T N1 φ1
+ 1 − 2χ 1
− 2χ
(Hij) = 1 1−φ 1−φ 1
1
− 2χ 1
N2 φ2
+
1−φ
vc (2.36)
1−φ
− 2χ
The equation det(Hij) = 0 gives,
1 1
+ N φ + 2 (2.37)
N1 φ
11 12 2 1 − 2χ
− 2χ − 2χ = 1−φ
1 − φ 1 − φ
After some calculation, this gives the following equation:
1 φ
χ= 1 − φ +
1
(2.38)
which can be written as 2 N1 φ 1 + N2 φ 2
1 1
χ= 1 − φ +
1
(2.39)
2 Nw φ
Chapter3
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