CHAPTERS 2-10 COVERED
SOLUTIONS MANUAL
, 1
Soft matter physics
Solutions to exercises
Chapter 2
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 [P a] = 2.4 [atm] (2.6)
2.2 (a) Number of polymers per unit volume is n∗ = 1/[(4π/3)Rg3 ] = 2.4 × 1020 [m−3 ]. 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[P a] (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 − Π 2 = (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
,2
where w = W/(An0 kB T ). Solution of eq.(2.11) is
x 1 p
= [ 1 + w2 − 1] (2.12)
h w
For small 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 Wef f = 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 = P V + 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 vi , we have
N0 N1 Nn 1
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 ) = V f (φ1 , ..., φn , T ) (2.17)
V V V
Pn
where we have used that Ni /V is expressed as φi /vi , and that φ0 is written as 1 − i=1 φi .
(b) The osmotic pressure is given by eq.(2.21), where Ftot is now written as
Ftot = V f (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
X ∂f
Π= φi − f (φi ) + f (0) (2.19)
i=1
∂φi
The chemical potential µi is given by µi = ∂G/∂Ni , where G is given by
G = P V + V f (φi ; T ) (2.20)
P
Using V = i vi Ni and φi = Ni vi /V , we have
X ∂f ∂φk
µi = P vi + v i f + V (2.21)
∂φk ∂Ni
k
P
Using φk = Nk vk / j Nj vj , we have
∂φk vi
= (δki − φk ) (2.22)
∂Ni V
Hence
n
" #
X ∂f ∂f
µi = v i P + f + (δki − φk ) = vi P − Π + + f (0) (2.23)
∂φk ∂φi
k=1
, 3
For i = 0, this can be written as
n
" #
X ∂f
µ0 = v0 P + f + −φk
∂φk
k=1
= v0 [P − Π + f (0)] (2.24)
(c)
n n n
" #
X φ1 X φi X ∂f
µi = P vi + vi f + vi (δki − φk )
i=0
vi v
i=0 i
∂φ k
k=1
n n
" #
X X ∂f
= P φ i + f φi + φ i (δki − φk )
i=0
∂φk
k=1
n n
n X
X X ∂f
= [P φi + f φi ] + φi (δki − φk ) (2.25)
i=0 i=0 k=1
∂φk
The underlined part is equal to zero since
n
n X n n
X ∂f X ∂f X
φi (δki − φk ) = φi (δki − φk )
i=0 k=1
∂φk ∂φk i=0
k=1
n n
" #
X ∂f X
= φk − φk φi
∂φk i=0
k=1
n
X ∂f
= (φk − φk ) = 0 (2.26)
∂φk
k=1
Therefore
X φi
µi = P + f (2.27)
i
vi
(d) Assume
X X kB T X
f = f (0) + ai (T )φi + φi ln φi + bij (T )φi φj (2.28)
i i
vi i,j
Calculate Π using eq.(2.74), and compare the result with eq.(2.78) in the text. This gives
bij = Aij . Hence for i 6= 0,
∂f
µi (φi , T ) = vi P −Π+ + f (0) (2.29)
∂φi
(0)
X vi
= µi (T ) + P vi + kB T ln φi + 2vi Aij − kB T φj (2.30)
j
vj
2.5 (a) The osmotic pressure is shown in Fig.2.1.
(b) Spinodal line is shown by the dashed line in Fig.2.2.
(c) Binodal line is shown by the solid line in Fig.2.2.
2.6 (a) f (φ) is shown in Fig. 2.3. Here to show the change of the shape explicitly, chi are chosen to
be 0.2,0.6 and 1.
(b) Spinodal line is shown in Fig.2.4.
SOLUTIONS MANUAL
, 1
Soft matter physics
Solutions to exercises
Chapter 2
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 [P a] = 2.4 [atm] (2.6)
2.2 (a) Number of polymers per unit volume is n∗ = 1/[(4π/3)Rg3 ] = 2.4 × 1020 [m−3 ]. 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[P a] (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 − Π 2 = (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
,2
where w = W/(An0 kB T ). Solution of eq.(2.11) is
x 1 p
= [ 1 + w2 − 1] (2.12)
h w
For small 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 Wef f = 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 = P V + 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 vi , we have
N0 N1 Nn 1
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 ) = V f (φ1 , ..., φn , T ) (2.17)
V V V
Pn
where we have used that Ni /V is expressed as φi /vi , and that φ0 is written as 1 − i=1 φi .
(b) The osmotic pressure is given by eq.(2.21), where Ftot is now written as
Ftot = V f (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
X ∂f
Π= φi − f (φi ) + f (0) (2.19)
i=1
∂φi
The chemical potential µi is given by µi = ∂G/∂Ni , where G is given by
G = P V + V f (φi ; T ) (2.20)
P
Using V = i vi Ni and φi = Ni vi /V , we have
X ∂f ∂φk
µi = P vi + v i f + V (2.21)
∂φk ∂Ni
k
P
Using φk = Nk vk / j Nj vj , we have
∂φk vi
= (δki − φk ) (2.22)
∂Ni V
Hence
n
" #
X ∂f ∂f
µi = v i P + f + (δki − φk ) = vi P − Π + + f (0) (2.23)
∂φk ∂φi
k=1
, 3
For i = 0, this can be written as
n
" #
X ∂f
µ0 = v0 P + f + −φk
∂φk
k=1
= v0 [P − Π + f (0)] (2.24)
(c)
n n n
" #
X φ1 X φi X ∂f
µi = P vi + vi f + vi (δki − φk )
i=0
vi v
i=0 i
∂φ k
k=1
n n
" #
X X ∂f
= P φ i + f φi + φ i (δki − φk )
i=0
∂φk
k=1
n n
n X
X X ∂f
= [P φi + f φi ] + φi (δki − φk ) (2.25)
i=0 i=0 k=1
∂φk
The underlined part is equal to zero since
n
n X n n
X ∂f X ∂f X
φi (δki − φk ) = φi (δki − φk )
i=0 k=1
∂φk ∂φk i=0
k=1
n n
" #
X ∂f X
= φk − φk φi
∂φk i=0
k=1
n
X ∂f
= (φk − φk ) = 0 (2.26)
∂φk
k=1
Therefore
X φi
µi = P + f (2.27)
i
vi
(d) Assume
X X kB T X
f = f (0) + ai (T )φi + φi ln φi + bij (T )φi φj (2.28)
i i
vi i,j
Calculate Π using eq.(2.74), and compare the result with eq.(2.78) in the text. This gives
bij = Aij . Hence for i 6= 0,
∂f
µi (φi , T ) = vi P −Π+ + f (0) (2.29)
∂φi
(0)
X vi
= µi (T ) + P vi + kB T ln φi + 2vi Aij − kB T φj (2.30)
j
vj
2.5 (a) The osmotic pressure is shown in Fig.2.1.
(b) Spinodal line is shown by the dashed line in Fig.2.2.
(c) Binodal line is shown by the solid line in Fig.2.2.
2.6 (a) f (φ) is shown in Fig. 2.3. Here to show the change of the shape explicitly, chi are chosen to
be 0.2,0.6 and 1.
(b) Spinodal line is shown in Fig.2.4.
