Physical Chemistry (I)
Midterm / November 9, 2013
1. (10%) There are many ways of description to elucidate the physical meanings of
the second law of thermodynamics. In the following statements, write only in one
sentence for what’s wrong of the statement in description, interpretation, or
application of the 2nd law; if you think that the statement is correct, make a short
comment for it. (2% each)
(a) The reversible isothermal expansion of an ideal gas is spontaneous.
(b) In Kelvin-Planck statement, it states that heat obtained from the hot reservoir
cannot transfer completely into work, therefore we can conclude that work
cannot transfer completely into heat as well.
(c) Clausius stated that no process is possible in which the only transfer of energy
is as heat transferred from a hotter to a colder system.
(d) Combine Clausius inequality (TdS  dq) with the 1st law of thermodynamics
(dU = dq + dw, only PV-work is involved), we conclude that dU  TdS PdV,
where the equality holds only for a reversible process.
(e) The kitchen is so hot so you can open the door of your refrigerator to cool
down the kitchen.
2. (10%) One mole of an ideal gas with Cv,m =
3
R is expanded adiabatically against
2
a constant external pressure of Pex = 1 bar until T = Tf and P = Pf. The initial
temperature and pressure are Ti and Pi, respectively; answer the following
questions.
(a) Express the internal energy change (U) in terms of temperature or pressure.
(2%)
(b) What is the maximum amount of work that can be done by this system? (2%)
(c) If Pi = 25 bar, Pf = 1 bar, and Ti = 300 K, determine the value of Tf. (3%)
(d) Compare the adiabatic process with the reversible isothermal expansion
process and explain why the indicator diagram (PV plot) shows that
Padaibatic  Pisothermal for all P < Pi (expansion), whereas Padaibatic  Pisothermal for
all P > Pi (compression). (3%)
1
3. (15%) In the first half of the 19th century, Joule tried to measure the temperature
change when a gas is expanded into a vacuum. However, the experimental setup
was not sensitive enough so that Joule found no temperature change within the
limit of his error. Soon afterward, Joule and Thomson devised a much more
sensitive method to measure the temperature change upon expansion. Answer the
following questions accordingly.
(a) According to the result of Joule’s experiment, draw a conclusion based on the
1  U 
 T 
following expression: 
 

 . (2%)
CV  V T
 V U
 U 
(b) The internal pressure of the system is defined as 
 , show that
 V T
 U 
 P 

 T 
  P . (3%)
 V T
 T V
(c) In Joule-Thompson’s experiment, a constant applied pressure P1 causes a
quantity of gas to flow slowly from one chamber to another through a porous
plug of silk or cotton. If a volume, V1, of gas is pushed through the porous
plug, and the pressure on the other side of the chamber is maintained at P2 until
the volume becomes V2, calculate the net work done by the isolated system.
(2%)
 H 
 H 
(d) Following (c) and starting with dH  
 dT  
 dP , show that
 T  P
 P T
 T 
1  H 
JT      
 . (2%)
CP  P T
 P  H
(e) Explain under what conditions the gas can be liquefied under isenthalpic
expansion. (2%)
  2V 
 CP 
(f) Show that 
  T  2  . (4%)
 P T
 T  P
4. (10%) At 298 K, the thermal expansion coefficient and the isothermal
compressibility of liquid water are  = 2.04  10 K and  = 45.9  10 bar.
 U 
 P 
Now based on the equation 
 T 
  P to answer the following
 V T
 T V
questions.
 U 
(a) Calculate 
 for water at 320 K and 1.0 bar. (2%)
 V T
2
 U 
(b) If an external pressure equal to 
 were applied to liquid water at 320 K,
 V T
how much would its volume change in percentage? (3%)
 U 
(c) Calculate 
 for N2 at 320 K and 1.0 bar, assuming that van der Waals
 V T
equation can be applied with parameters a = 1.37 dm6 bar mol and b =
0.0387 dm3 mol note that at the relatively low pressure condition (P = 1 bar),
it might be safe to use the ideal gas law to determine Vm. (3%)
(d) Compare the results obtained from (a) and (c) and make a comment for it. (2%)
5. (20%) The first law states that the energy of the universe is a constant while the
second law states that the entropy of the universe tends always towards a
maximum. Answer the following questions only based on the conditions it is
provided.
(a) If only PV-work is considered, prove that the inequality of work (dwrev < dwirr)
is always valid (e.g., expansion or compression). (2%)
(b) Prove that dqrev > dqirr based on the conclusion made from (b). (2%)
dq
(c) Prove the Clausius inequality based on the conclusion of (c), i.e., dS 
,
Tsurr
where the equality holds only for reversible processes. (2%)
(d) Following (d), prove that the entropy change of any irreversible process is
always greater than zero in an isolated system. (2%)
(e) Assuming that the system is no longer isolated, i.e., the surroundings are a
reservoir of energy and the heat entering the system is equal to the heat leaving
the surroundings. Show that the Clausius inequality can be deduced to a more
general expression for the 2nd law, i.e., dStot  dS  dSsurr  0 . (2%)
(f) Show that the Gibbs energy change is always less than or equal to zero at the
conditions of constant T and P, i.e., (dG )T , P  0 when only PV-work is
involved. (2%)
(g) Following (g), if non-expansion (non-PV) work is also considered here, i.e., w
= wPV + wnon-PV, with a short derivation to explain what is the physical meaning
of A at constant T. (2%)
(h) A box contains two identical parts with a removable partition. One part of the
box contains n moles of ideal gas A while the other part contains n moles of
ideal gas B with the same P, V, and T for both A and B. Now the partition is
removed and the gases spontaneously mix. What is the maximum amount of
work that could be obtained from this spontaneous process? (4%)
3
6. (20%) Consider a reversible Carnot cycle for a sample of an ideal gas working
substance (heat capacity Cv) running as a heat engine between two temperature
reservoirs (TH and TC ); the heat of qH is transferred from the hot reservoir (T = TH)
to the engine, the engine does mechanical work w and the heat of qC is transferred
from the engine to the cold reservoir (T = TC).
(a) Draw qualitatively the corresponding indicator diagram (PV plot) with the
(1)
(2)
(3)
(4)
 b 
 c 
 d 
 a , and state the names
following sequence, a 
for each process in segments (1)(4). (3%)
(b) Give the physical meaning for the area inside the P-V plot. (2%)
(c) Draw qualitatively the corresponding diagram for the same Carnot cycle on a T
vs. S plot with the same label as in (a). (3%)
(d) Derive an expression for the area inside the T-S plot and give the physical
meaning of it. (3%)
(e) Now combine the results obtained from (b) and (d) to prove that qH = –w –qC.
(2%)
(f) Make a derivation form (c) to prove that
qH
T
 H . (2%)
 qC TC
(g) Prove that the efficiency () of a heat engine is   1 
TC
. (2%)
TH
(h) When the heat engine runs in the reverse direction, it can be used either as a
heat pump or a refrigerator (or air conditioner). Relate the coefficients of
performance of heat pump (hp) and refrigerator (r) with respect to TH and TC.
(3%)
7. (10%) Octane (C8H18) can be used to construct a fuel cell based on the oxidation of
the compound with the energy (enthalpy) released by the following reaction:
C8 H18( l ) 
25
O2( g )  8CO2(g )  9H 2 O(l )
2
H c0  5471 kJ  mol 1
0
 5296 kJ  mol1 , calculate the maximum work that is
(a) Given that Goctane
available through the combustion of the hydrocarbon at 298.15 K and 1 bar;
note that the work includes both expansion and non-expansion types. (2%)
(b) Following (a), calculate the minimum heat that must be released into the
surroundings at 298.15 K. (2%)
(c) Following (a) and (b), calculate the internal energy change of the system at the
standard state. (2%)
4
(d) Consider a heat engine with TH = 600 K and TC = 300 K. The engine is running
with the combustion of octane as a heat source. Calculate the ratio for the
maximum work available from the heat engine to that available in an ideal
electrochemical fuel cell using octane as a fuel. (4%)
8. (10%) We define a general chemical reaction as aA + bB + …  cC + dD + …,
which for convenience we express the above reaction as cC + dD + …  aA  bB
 … = 0, or equivalently
 X
i
i
 0 , where i stands for the stoichiometric
i
coefficient of species Xi (now simplified as species i).
Answer the following
questions.
(a) Define the extent of the reaction as and the initial mole number of species i
as ni0 to satisfy ni  ni0   i , derive an expression for the reaction of Gibbs
energy (Gr) in terms of  at constant T and P. (2%)
(b) Following (a), determine the criterion in terms of G vs. for the forward
reaction to be spontaneous at constant T and P. What is the criterion for the
reaction reaching equilibrium? (2%)
(c) Considering the reaction occurring in the gas phase, derive the expression for
the thermodynamic equilibrium constant (Kp) as a function of standard state
thermodynamic functions, H r and S r . (2%)
(d) Derive an expression for the temperature variation of Kp in terms of H r
based on Gibbs-Helmholtz equation. What is the assumption you need to make
for the derivation? (2%)
(e) Express the mole-fraction equilibrium constant (Kx) in terms of Kp and ,
where the mole fraction of species i (xi) and the partial pressure of species i (Pi)
are related by Pi = xi P and   c  d  ...  a  b  ...   i .
i
5
(2%)
參考解答
1. (a) The reversible process is not spontaneous process because the system and the
surrounding are in equilibrium.
(b) Work can transfer into heat completely!
(c) Heat transferred only from a colder to a hotter system is not possible.
(d) “dU = TdS – PdV” holds for both reversible and irreversible processes since U,
T, S, P, and V are all state variables.
(e) It does not work if the refrigerator is located inside the kitchen room because
the heat released by the refrigerator is always larger than the heat taken from the
refrigerator according to the 2nd law.
2. (a) For adiabatic processes  q  0, U  w  -Pex V f -Vi 
U  n  C
-e Px
V , m Tf - iT 
Vf - iV
(b) Maximum work can be done when the final temperature reaches 0 K., i.e.,
3
3
-wmax  -U max  1  R Ti -0   RTi
2
2
 1  R T f 1  R Ti
3
R T f -300   -1  
 Pf
2
Pi

(c)



3
300 T f
T f -300  

2
25 1
2
 T f -300  8- T f
3
5
 T f  308 , T f  184.8 K ~185 K
3

(d)  For P  Pi  expansion
對外做功，等溫過程有外援但絕熱過程要消耗自己的內能而使 T，
故 Padia  Piso ( PV  nRT, P  T if V is constant)
P
Pi
Piso
adiabatic
isothermal
Padia
V
6
 For P  Pi  compression
對系統做功，等溫過程環境吸收此熱量，但絕熱過程此熱能轉變為系統
之內能而使 T，故 Padia  Piso ( PV  nRT, P  T if V is constant)
P
Padia
isothermal
Piso
adiabatic
Pi
V
3. (a) Joule’s experiment shows
 T 
T  0  
  0 , which implies
 V U
 U 

 0
 V T
(b) dU  TdS - PdV
 U 
 S 
 P 

  T
 - P  T  - P
 V T
 V T
 T  V
(c) w  -  Pex dV  -P1 0-V1  -P2 V2 -0   PV
1 1 -PV
2 2
 H 
 H 
(d)  0  dH  
 dT  
 dP
 T  P
 P T
 H 
 H 

 dT  - 
 dP
 P  P
 P T
1  H 
 T 

 

CP  P T
 P  H
1  H 
 T 
 μJT  
 

CP  P T
 P  H
or  JT
(e)
 H

 T 
 P




 H
 P  H

 T


T   1
Cp


P
 H 
 H 

 
  C p   JT
 P T
 P T
In the condition of isenthalpic expansion, gas can only be liquefied in a
cooling process when JT > 0; because when gas expansion occurs, P < 0,
the condition for  < 0 (cooling process) being valid requires
 T 
JT     0 .
 P H
7
   H  
 C     H  
(f)  P    
   
 
 P T  P  T  P T  T  P T  P
由 dH = TdS + VdP
 H 
S 
V

 T 
 V  - T 
 P T
P T
T

 V
P
   H  
  2V   V 
  2V 
 C 
 V 
故 P    

-T


-T
 2 
 2
 



 T  P  T  P  T  P
 P T  T  P T  P
 T  P

2.04  10 -4
 U 
 P 
4. (a) 

T
-P

T
-P

320

-1.0  1421 bar




45.9  10 -6
 V T
 T V
 V 
 V 
(b) dV  
 dT  
 dP
 T  P
 P T
dV 1  V 
1  V 
 
 dT  
 dP   dT -  dP
V V  T  P
V  P T
 V 
-6

  - P  -45.9  10  1421-1  -0.065  -6.5%
 V T
體積將減少 6.5%
RT a
(c) P 
Vm -b Vm2
R
 P 
  
 T  V Vm - b
RT
a
 U 
 P 

-P  2
 T 
 -P 
Vm -b
Vm
 V T
 T V
2
2
1.0
 P 


-3
 a 
  1.37  
  1.94  10 b a r
-2
RT
8.314

10

320




 U 
(d) For liquid water, 
  1421 bar
 V T
 U 
-3
For N2 gas, 
  1.94  10 bar
 V T
由於液態水中分子與分子間的引力遠大於氣態 N2 ，故前者的 internal
pressure 大於後者高達 6 個數量級之多。

5. (a) dwrev  -P dV
d wi r r - Pe x d V
dWirr - dWrev  P - Pex  dV
8
 If expansion, P  Pex and dV  0
故 P - Pex  dV  0
 If compression, P  Pex and dV  0
故 P - Pex  dV  0
In any case, P - Pex  dV  0
故 dWirr - dWrev  0
亦即 dWrev  dWirr
(b) dU  dq rev  dWrev  dq irr  dWirr
 dq r e v- dq i r r dw i r r dw r e v 0
 dq r e v dq i r r
(c) dqrev  T dS  Tsurr dS
故 TsurrdS  dq irr
and Tsurr dS  dqrev
 dS 
(d) dS 

TsurrdS  dq
dq
Ts u r r
dq irr
 0 dq irr  0 for isolated system   dS isolated system   0
Tsurr
(e) dSsurr 
dq surr - dq

Tsurr
Tsurr
dStot  dS  dSsurr 
dq - dq

 0  dStot  0
Tsurr Tsurr
(f) dU  dq  dw  dq-Pex dV
dU  Pex dV  dq  TsurrdS
at constant P and T, Pex  P, Tsurr  T
 dU  P dV - T dS  0
 d U  PV - TS T, P  0
 dG T, P  0
(g) A = U - TS
d A d U - T d S - S d T

T- dP Sd V dn W
- Sd Sd T - S d T -P d nVo n - dP VW
o n - P V - T
at constant T, dA  -P dV  dWnon-PV  dWPV  dWnon-PV
- A T  -WPV - Wnon-PV
故定溫下之 - A 即為系統對外所能做的最大功(包括 PV 及 non-PV 功)
另解
9
Ts u rdS-dU
 -dw  -dw -dw
r
P V
n o n - P V
at constant T, Tsurr  T and TdS  d TS 
-d U-TS T  -dwPV -dwnon-PV
-  dAT  -dw or -  AT  -wmax
(h)
6. (a)
PVT
n
A
P
S  SA  SB  nRln
PVT
n
B
2V
2V
 nRln
 2nRln2
V
V
wm' a x GT , P  H T S 0  T  2 n R l n2
qH
a
(1)
(1) rev. isothermal expansion
(2) rev. adiabatic expansion
(3) rev. isothermal compression
b
(4)
TH
(2)
d
(4) rev. adiabatic compression
(3)
c
TC
V
qC
(b) Area  w  w1  w2  w3  w4
  nRTH ln
Vb
V
 nRTC ln d
Va
VC
 系統對外所做的功
(c)
S
T
TH
a
(1)
b
(2)
(4)
TC
S2
d
S4
S2
(2)
S4
d
TC
S
(d) Area  TH -TC    S2 -S4   TH S  TC S  qH  qC
(e) w  qH  qC  qH  w  qC
(f) Because S is a state function,
10
b
(1)
(3)
Or
c
(3)
c
(4)
a
TH
T
 系統所吸收的熱
2nRTln2
0   dS  S1  S2  S3  S4 

qH
q
0 C 0
TH
TC
qH qC
q
T

0 or H  H
TH TC
- q C TC
q
T
-w qH  qC

 1  C  1- C
qH
qH
qH
TH
(g)  
(h) ηhp 
ηr 
 qH
 qH
qH
T
T


 H  H
w
qH  qC qH  qC TH -TC T
qC
qC
 qC
TC
T



 C
w  qH  qC
qH  qC
TH  TC  T
7. (a) G = U + PV – TS = H – TS = A + PV
G o   Ao   P V

Ao 
nRT
 25 
 8-   8.314  298.15
2 
o
o
wm a x A  G  nRT  -5296- 
 -5285 kJmol
1000
(b) G o  H o  T S o
q rev  TSo  H o - G o  -5471- (-5296)  -175 kJmol -1
(c) Uo  Ao - TSo  -5285 - 175  -5460 kJmol -1
T
300
 0.5
(d)   1 - C  1 TH
600
wh e  -q H   H o    -5471  0.5  -2735.5 kJmol
- 1
wh e -2735.5

 52%
w'fc
-5296
W e c a n d e f i nfce:
G H  T S
T S

 1
H
H
H
8. (a)
dG  -SdT  VdP   i dni
i
ni  ni  i 
dni   i d 
  dn     d   G d 
i
i
i
i
i
r
i
dG  -SdT  VdP  Gr d 
 G 
 Gr  

  T,P
11
系統無法利用的熱
反應熱
- 1
(b)
 G 
S p o n t a n e o u rs :  G  
  T , P
0
 G 
E q u i l i b r iu m
:  G 
r 
  T , P
0
(c)
Gr    
-a -bA i
i c 
C d  D
B
i
P 
P


 c  C  RTln C   d   D  RTln D
P 
P


P  
P 

-a   A  RTln A  -b   B  RTln B 
P  
P 




c
  cC  d  D 
-a -b
A  -B
   i i  RTlnQ p
i
 Gr  RTlnQ p
at equilibrium, Gr  0 and Q p  K p
 RTlnK p  -Gr  - H r  T S r
 H r S r 
 K p  exp  

R 
 RT
(d)
  G T  
H
Gibbs-Helmholtz equation: 
 - 2
T
 T  P
   Gr T  
H
  - 2r

T
T

 P
 lnK p 
H r

 no assumption
 
2
 T  P RT

T2
T1
ln
H r
H r T2dT
dT 
2
T 1 RT
R T 1T 2
H r  1 1 

 - 
R  T1 T2 
dlnK p  
K p T2 
K p T1 
d
 PC   PD 
   
  RTln  P  a  P  b
 PA   PB 
   
P  P 
T2
 H r is independent of T
12
(e)
c
d
 PC   PD 
   
P
P
K p   a  b
 PA   PB 
   
P  P 
 P
 Kx  
P 
c
d
 xC P   xD P 


 
P   P 

a
b
 x A P   xB P 

 

 P   P 

13
 xc xd
  Ca Db
 x A xB
 P
 
 P 
c  d  -a-b-