物理化學 (II) 化三 第二次小考
g = 9.806 m/ s2
R = 8.3145 J/K-mol
(一)
1.
2.
3.
4.
5.
6.
7.
8.
配合題: 12 %
T
β
CV ln f  V f  V f 
Ti κ
T
C P ln f  Vβ Pf  Pf 
Ti
T f C p dT
Pf
Ti T  Pi V β dP

Tf
Ti
 Δ G  Tf  Δ H 
d r  d r 
 T  Ti  T 
qcold
Tcold

w
Thot  Tcold
qhot
Thot

w
Thot  Tcold
P
S mo P  R ln o
P
wcycle Thot  Tcold

qhot
Thot
 Δ G 


 T P
(100 %) 12/16/2009 姓名：KEY
  nR  xi lnxi
1 atm = 101.325 kPa
a)
Entropy for a change from (Pi,Ti → Pf,Tf) at constant V, ∆S
b)
Entropy of gas, ∆S for a change from (Vi,Ti → Vf,Tf)at constant P
c)
Entropy of solid, ∆S for a change from (Pi,Ti → Pf,Tf) at constant V
d)
The coefficient of performance of Carnot refrigerator, ηr
e)
The efficiency of reversible heat engine, ε
f)
The maximum coefficient of performance of heat pump, ηHP
g)
The equilibrium constant of reaction in terms of molarity, KC
h)
The variation of initial slope of compression factor with P at constant
T
i)
Molar entropy of ideal gas as function of P , Sm(P)
9.
 
10
.
μ APure T, P   R T lnx A
j)
Gibbs-Helmholtz equation under constant P
11
.
lim 
k)
chemical potential of species A in gas mixture, μ A
12
.
 c oRT
K P  o
 P
l)
entropy of mixing for the gases, ∆Smixing
m)
None of above
i
 z 

p0  P T
Ans: b c a j d



mixture
 
fielk
hg
__________________________________________________________________
(二)
1.
推導題 24%
 A T  
  U . Write an expression that
  1 T  V
would allow you to relate A at two temperatures. [6.16]
From dA = − S dT – p dV, show that 
1  A 
A
S A
A TS
U
  A T  
Ans: 
 2
  
  2   2 
2
T T
T
T
 T V T  T V T
  A T  
  A T    T

  
 
  1T  V  T V   1T
At constant V,

  A T  
  T 2 
 U
 V
 T V
T, P 
2
 1 1
A T  A T 
 A 
 U 

d


T  T  T   T 2  dT and T 2 2  T1 1  U T 2  T1 
1
1
T2
T
__________________________________________________________________
RT
a
, find an expression for the
b
P
RT 2
Boyle temperature in terms of a, b, and R. [7.17]
2.
For the Bertholet equation, Vm 
Ans: z 
pVm
z
pb
pa
b
a
 1
 2 3 and 
 2 3
 
RT
RT R T
  p  T RT R T
z
b
a
a
At T = TB, 
 2 3  0 and TB 
  0 . Therefore
RTB R TB
Rb
  p T
__________________________________________________________________
z
1 
a 

The slope of the z versus pressure curve 

 b
 as p → 0
  p T,P0 RT  RT 
for a van der Waals gas. At what temperature does the slope of the curve have
its maximum value for a van der Waals gas? Show that the value of the
 b2 
 . [P7.20]
maximum slope is 
4a
3.
z
1
Ans: 


  p  T,P0 RT
a 

 for a van der Waals gas
 b
RT 

  z  
a  1  a 
1 
1  2a 



b
b
 



2 
2 
2 
   p  
RT
RT
RT
RT
RT
RT 





T, P 0


2a
2a 
  0 or Tmax=
Setting this derivative equal to zero gives  b
Rb
RTmax 


T
z
1


  p  Tmax ,P0 RTmax
The maximum slope is 
(三)、計算題 64%
1.
Ans:
2

a  b 
 b  b
 b
 


b
a




 2 a  4 a
 RTmax  2 a 
CP ,m  Hg, l 
T
 30.093  4.944 103 .
1
1
J K mol
K
Between 0ºC and 100ºC. Calculate H and S if 1 mol of Hg(l) is raised in
temperature from 0º to 100ºC at constant P. [P5.13]
The heat capacity of Hg(l) is given by
Tf
H m   CP ,m d T / K 
Ti




 30.093 T f  Ti  2.472  103 T f2  Ti 2 J mol1
 2.84  103 J mol1
Tf
S m 
Tf
CP ,m
 T / K d T / K   30.093ln T
i
Ti
 4.944  10
3
T
f

 Ti  8.90 J K 1 mol1
__________________________________________________________________
2. The heat capacity of -quartz is given by
CP ,m  -quartz, s 
J K 1 mol1
 46.94  34.31  103
T
T2
 11.30  105 2 . The coefficient of
K
K
thermal expansion is given by  = 0.3530×10–4 K–1 and Vm = 22.6 cm3 mol–1.
Calculate Sm for the transformation -quartz (25ºC, 1 atm) → -quartz
(225ºC, 1000 atm). [P5.15]
Tf
Ans: Δ S   C P
Ti
dT
 Vβ Pf  Pi 
T
498K


 46.94 ln
 34.31 10 3  498  298   5.65  10 5  4982  2982  J K 1mol 1
298K


3
1
6
3
1
4
1
 22.6 cm mol  10 m cm  0.3530  10 K  999 atm 1.01325  105 Pa atm 1
= 21.97 K-1 mol–1 −0.0808 J K-1 mol–1 = 21.89 J K-1 mol–1
__________________________________________________________________
Calculate S, Stotal, and Ssurroundings when the volume of 85.0 g of CO initially
at 298 K and 1.00 bar increases by a factor of three in (a) an adiabatic
reversible expansion, (b) an expansion against Pexternal = 0, and (c) an
isothermal reversible expansion. Take CP,m to be constant at the value 29.14 J
mol–1 K–1 and assume ideal gas behavior. [P5.17]
Ans:
a) An adiabatic reversible expansion
Ssurroundings = 0 because q = 0;andS = 0 because the process is reversible.
Stotal = S + Ssurroundings = 0. The process is not spontaneous.
3.
b) An expansion against Pexternal = 0
T and w = 0. Therefore U = q = 0.
Vf
85.0 g
S  nR ln

 8.314 J mol1K 1  ln 3
Vi 28.01 g mol1
 3.03 mol  9.13 J mol1K 1 = 27.7 J K 1
Stotal = S + Ssurroundings = 27.7 J K–1 + 0 = 27.7 J K–1.
The process is spontaneous.
c) An isothermal reversible expansion
T = 0. Therefore U = 0.
Vf
 3.03 mol  8.314 J mol1K 1  298 K  ln 3  8.25  103 J
w   q   nRT ln
Vi
qreversible 8.25  103 J
S 

= 27.7 J K 1
298 K
T
 q 8.25  103 J
S surroundings 

 27.7 J K 1
T
298 K
Stotal = S + Ssurroundings = 27.7 J K–1 – 27.7 J K–1 = 0. The system and
surroundings are at equilibrium.
__________________________________________________________________
4.
Consider the reaction FeO(s) + CO(g) ⇄ Fe(s) + CO2(g) for which KP is found
to have the following values:
T
600ºC
1000ºC
KP
0.900
0.396



a) Calculate Greaction , Sreaction and H reaction
for this reaction at 600ºC. Assume

is independent of temperature.
that H reaction
b) Calculate the mole fraction of CO2(g) present in the gas phase at 600ºC.
[P6.12]
PCO2 P 
Ans: a) FeO(s) + CO(g) ⇄ Fe(s) + CO2(g) K P 
PCO P 
ln
O
Δ H reaction
K P 1000 C 

K P 600 C 
R

Assume that H reaction
1
1





 1273.15 K 873.15 K 
is independent of temperature
K P 1000 C  
1
1




K P 600 C   1273.15 K 873.15 K 
 0.396 
 8.3145x ln

 0.900   19.0 kJ mol 1

1
1 




 1273.15 873.15 
b) because KP = PCO2/PCO = 0.900
KP = Kx because  = 0
xCO2
 0.900 and xCO2  xCO  1
xCO
O
Δ H reaction
  R ln
xCO2  0.47
xCO  0.53
__________________________________________________________________
5. A sample containing 2.25 mol of He (1 bar, 298 K) is mixed with 3.00 mol of
Ne (1 bar, 298 K) and 1.75 mol of Ar (1 bar, 298 K). Calculate Gmixing and
Smixing . [P6.18]
Ans:
Gmixing  nRT  xi ln xi
i
 2.25 2.25 3.00 3.00 1.75 1.75 
  7.00 mol   8.314 J mol1K 1  298.15 K  
ln

ln

ln

 7.00 7.00 7.00 7.00 7.00 7.00 
 18.6 103J
Smixing  nR  xi ln xi
i
 2.25 2.25 3.00 3.00 1.75 1.75 

ln
ln
   7.00 mol   8.314 J mol1K 1  
ln


 7.00 7.00 7.00 7.00 7.00 7.00 
 62.5 J K 1
__________________________________________________________________
6. Calculate KP at 475 K for the reaction NO(g) + 1/2 O2(g) → NO2(g) assuming

that H reaction
is constant over the interval 298–600 K. Do you expect KP to
increase or decrease as the temperature is increased to 550 K? Given that
∆Gfº(kJ/mol)
∆Hfº (kJ/mol)
33.2
25.65
NO2(g)
NO(g)
91.3
87.6
[P6.23]
Ans: ∆Hreactionº = ∆Hfº(NO2, g) – ∆Hfº(NO, g)
= (33.2 – 91.3)x103 J/mol = –58.1 x103 J/mol
∆Greactionº = ∆Gfº(NO2, g) – ∆Gfº(NO, g) = (25.65 – 87.6)x103 J/mol
= –61.95 x103 J/mol
∆Greactionº (Tf) / Tf
= ∆Greactionº (Ti) / Ti – ∆Hreactionº x [(1/ Tf) – (1/ Ti)] = –34.24 x103 J/mol
Kp(Tf) = Kp(475 K) = exp[–∆Greactionº (Tf)/RTf] = 1.16 x107

Because H reaction
< 0, KP decreases as T increases.
__________________________________________________________________
7. For a gas at a given temperature, the compressibility is described by the
2
P
 P
 4.00  105    , where Pº = 1

P
P 
bar. Calculate the activity coefficient for P = 100, 300 and 600 bar. For which
of these values is the activity coefficient greater than one? [P7.23]
Ans: ln (f /p) =
empirical equation z  1  9.00  103


3
5
P 1  9.00 10 P  4.00 10  P 
1
z 1
ln   
dP  
dP
P
P
0
0
P
2
ln   9.00  103 P  2.00  105 P 2
  0.497, 0.368, 0.406, 0.670, and 1.65 at 100, 200, 300, 400, and 500 bar, respectively.
At 1 bar, ln γ = −0.7 and γ = 0.497. At 300 bar, ln γ = −0.9 and γ = 0.409. At 600
bar, ln γ = 1.8 and γ = 6.05. Only at 600 bar, γ > 1.
__________________________________________________________________
8.
For values of z near one, it is a good approximation to write
 z 
z  P  1 
 P.
 P T
If z =1.00054 at 0ºC and 1 bar, and the Boyle
temperature of the gas is 220 K, estimate the values of Vm, a and b for the van
der Waals gas.[P7.24]
Ans:
From Example Problem 7.2,
a 
1 
 z 

 
b 

P
RT
RT


T


We can write three equations in three unknowns:
a  P

z 1   b 

RT  RT

a
1 bar


5.4 104   b 

2
1
1
2
8.314 10 L bar mol K  273.15 K  8.314 10 L bar mol1 K 1  273.15 K

a
a
TB 

 220 K
2
Rb 8.314  10 L bar mol1 K 1  b
RT
a 8.314  102 L bar mol1 K 1  273.15 K a
P



 1 bar
Vm  b Vm 2
Vm  b
Vm 2
Vm
Using an equation solver, the results are
a  1.15 L2 bar mol2 , b  0.0630 L mol1 , Vm  22.72 L.
Vm