Lecture 19 Overview Ch. 4-5
List of topics
1.
2.
3.
4.
Heat engines
Phase transformations of pure substances, Clausius-Clapeyron Eq.
van der Waals gases
Thermodynamic potentials, chemical reactions
P
e  1
 QC
 QH
2
1
Problem 1 (heat engine)
Working substance for the cycle shown in the
Figure is 1 mole of ideal gas. Find the efficiency
of this heat engine (in terms of TH and TC).
TH
3
V1
V2
TH
Q 12  CP dT CP TH  TC   0
1–2
TC
TC
TC
V
2–3
Q 23  CV dT CV TC  TH   0
TH
 V1 
 TC


Q 31 W31   PdV R TC ln    R TC ln 
 V2 
 TH
V2
V1
3–1
 QH  Q12
 QC  Q23  Q31
T
f
RTH  TC   R TC ln  C
2
 TH
e  1
f

1   RTH  TC 
 2
CV 
f
R
2

  0

f

C P  CV  R    1 R
2 

f

TH  TC1   TC ln  TC
  1 2
 TH
f

1  TH  TC 
 2



Problem (vdW)
V, P1
U vdW
V, P2
An insulating membrane (does not conduct
heat) divides an insulated tank into two
equal volumes. Each volume contains one
mole of the same van der Waals gas (the
constants a and b are known). The
pressure in volume I is P1, in volume II –
P2 . The piston has been removed. Find
the pressure, explain your reasoning.
N 2a f
N 2a
 U ideal 
 NkBT 
V
2
V
2 N  a
f
N 2a f
N 2a f
Nk BT1 
 Nk BT2 
 2 N k BT f 
2
V
2
V
2
2V
2
2

2N  a 1
P

 P1  P2 
2
2V  2 Nb 2V  2
2 Nk BT f
1
T f  T1  T2 
2
Isothermal Process for a vdW gas
U   Q   W
U vdW T
1
1 

N a

V V 
f 
 i
2

 V f  Nb 
 Nk BT N 2 a 
1 
2  1




W    PdV    
 2  dV   Nk BT ln 
  N a V  V 
V

Nb
V
V

Nb

f 
 i

Vi
Vi 
 i
Vf
Vf
 V2  Nb 

 V1  Nb 
 Q12  NkBTH ln 
Latent heat, dS for phase transformations
10 kg of water at 200C is converted to ice at - 100C by being put in contact with a reservoir
at - 100C. This process takes place at constant pressure and the heat capacities at
constant pressure of water and ice are 4180 and 2090 J/kg·K respectively. The heat of
fusion of ice is 3.34·105 J/kg.
(a) Calculate the heat absorbed by the cold reservoir.
(b) Calculate the change in entropy of the closed system “reservoir + water/ice”.
The conversion consists of three processes: (a) water at 200C  water at 00C; (b)
water at 00C  ice at 00C; (c) ice at 00C  ice at -100C:
(a) the heat absorbed by the cold reservoir


Qres  10 kg  4180  20  3.34 105  2090 10 J/kg  4.385 106 J
(b) the change in entropy of the sub-system “water/ice”:
water
Mc
dT
 273 
P
cooling of water
S a  
 McP ln 
  2955 J/K
T
293


293K
MLvap
10 kg  3.34 105 J/kg
Sb  

 12,234 J/K
forming ice
T0
273K
263K
ice
Mc
 263 
ice
P dT
cooling of ice
Sc  
 McP ln 
  780 J/K
T
273


273K
Qres 4.385 106 J
S res 

 16,673 J/K
The increase of entropy of the reservoir:
Tres
263 K
The total change of entropy of the whole system: S  S a  Sb  Sc  S res  704 J/K
273K
Vapor equation
1.
For Hydrogen (H2) near its triple point (Ttr=14K), the latent heat of vaporization
Lvap=1.01 kJ/mol. The liquid density is 71 kg·m-3, the solid density is 81 kg·m-3,
and the melting temperature is given by Tm =13.99+P/3.3, where Tm and P
measured in K and MPa respectively. Compute the latent heat of sublimation.
2.
The vapor pressure equation for H2: P  P0 exp  

 Lvap 
 where P0 = 90 MPa .
 RT 
Compute the slope of the vapor pressure curve (dP/dT) for the solid H2 near
the triple point, assuming that the H2 vapor can be treated as an ideal gas.
1. Near the triple point:
P
liquid
solid
Lmelt
Ptr
gas
Ttr
 2 10 3 kg/mol 2 10 3 kg/mol
14  
3
71 kg/m
81 kg/m 3
Ttr VL  VS 



dT / dP
1/3.3 106
Lvap  Ttr S G  S L 
T
Lmelt  Ttr S L  S S 


  162 J
Lsub  Ttr SG  S S 
Lsub  Ttr SG  S S   Lvap  Lmelt  1010  162 J/mol  1172 J/mol
Vapor equation (cont.)
2. At the solid-gas phase boundary:
Lsub
L
dP

 sub
dT Ttr VG  VS  TtrVG
Assuming that the H2 vapor can be treated as an ideal gas
VG 
PtrVG  RTtr
RTtr
RTtr
8.3 J/K  mol 14 K


Ptr
P0 exp  Lvap / RTtr  9 107 Pa  exp - 1010 J/ 8.3 J/K  mol 14 K 
 7.69 10-3 m3 / mol
dP Lsub
1172 J/mol
4



1
.
09

10
Pa/K
3
3
dT TtrVG 14 K  7.69 10 m / mol
Phase transformations
The pressure-temperature phase diagram of carbon is shown below. For simplicity, assume
that the molar volumes of graphite and diamond are independent of temperature and
pressure at 5.3×10-6 and 3.4×10-6 m3, respectively. 1 kbar = 108 N/m2 .
(a) Determine the latent heat per mole of transformation at T = 1000 K.
(b) Sketch the graph of Gibbs free energy of carbon at the constant temperature T = 2000
K as a function of pressure between P = 50 and 70 kbar. Mark the transition pressure.
Explain the changes, if any, of the slope of G.
(c) Sketch the graph of entropy per mole of the material at the constant pressure P = 90
kbar as a function of temperature between 4000 and 5500 K. State any assumptions
you make and explain your graph.
(a)
dP
L

dT TV
L  TV
dP
dT
 90 108 Pa 
  48.9kJ
 1000 K  5.3 10 m  3.4 10 m 
3500
K



(b)
G.
6
6
3
3

dG T , N   SdT  VdP  dN T , N  VdP
the slope of G changes at
the
phase
transition
because of the change in
volume
63 kbar
P
Phase transformations, cont.
(c)
S
S 
4000
L
Tmelt
5500 T
C.-C. Equation
The vapor pressure of solid ammonia is given by the relation: ln P  23.03  3754 / T
where units –mm of Hg, T – absolute temperature.
The vapor pressure of liquid ammonia is given by the relation: ln P  19.5  3063 / T
(a)
(b)
(c)
(a)
What is the temperature of triple point?
Compute the latent heat of vaporization at the triple point. (Assume that the
vapor can be treated as an ideal gas, and the density of vapor is negligibly
small compared to that of the liquid)
The latent heat of sublimation at the triple point is 31.4 kJ/mol. What is the
latent heat of melting at the triple point?
Ttr – from the equation
23.03  3754 / Ttr  19.5  3063 / Ttr
Ttr = 195 K
ln P  19.5  3063 / T  dP / dT  3063P / T 2
dP
L
L


The Clausius-Clapeyron eq.
gives L  3063PVG / T  3063R  25.4 kJ/mol
dT TV TVG
(b) From the equation for liq. ammonia
(c)
Denote Sg, Sl, and Ss as the entropy for vapor, liquid and solid at triple point.
The latent heat of vaporization -
The latent heat of sublimation The latent heat of melting -
Ttr S g  Sl 
Ttr S g  S s 
Ttr S L  S S 
 Ttr SG  S S   Ttr SG  S L   31.4 kJ/mol - 25.4 kJ/mol  6 kJ/mol
C-C eq., phase transformation)
A cylinder closed with a piston is filled with the saturated water vapor at T = 1000C.
The vapor is heated up by 10C, and, at the same time, the piston is moved to prevent
condensation and to keep the vapor saturated (the system is “moving” along the
coexistence curve). Find the relative change in the vapor volume, V/V. Assume that
the vapor is an ideal gas, the latent heat of vaporization Lvap=40.7 kJ/mol, and the
vapor density is negligible in comparison with the water density.
The Clausius-Clapeyron equation for the coexistence curve “liquid-gas”:
For an ideal gas:
Lvap
Lvap
Lvap
P
P 
T


TVG
T T VG  VL  TVG
PiVi Pf V f
P
V

For a small temperature changes :

Ti
Tf
P
V
Lvap T
Lvap T
V
P
40.7 kJ/mol 1 K





 0.035  3.5%
2
2
V
P
TVP
RT
8.3 J/mol  K  373 K 
phase transformations
The triple point for water corresponds to Ttr=0.010C and Ptr =0.006 bar. At T~Ttr,, the
latent heat of melting is 335 kJ/kg, and the slope of the solid-vapor phase coexistence
curve is dP/dT=50 Pa/K . Assume that water vapor behaves as an ideal gas near its triple
point. Find the latent heat of vaporization.
Along the solid-gas phase equilibrium curve:
Lsub Ttr   Lmelt Ttr   Lvap Ttr 
Lsub
L
dP

 sub
dT T vgas  vsolid  Tvgas
- the latent heat of sublimation at the triple point
 dP 
Lvap Ttr   Lsub Ttr   Lmelt Ttr   Ttr vgas    Lmelt Ttr  
 dT  sub
3
nRTtr  dP 
2 10 g / kg 8.3 J / K  mol
3
Ttr
50
Pa
/
K

335

10
J / kg 
   Lmelt Ttr   273K 
2
Ptr  dT  sub
18 g / mol 6 10 Pa
2863 103 J / kg  335 103 J / kg  2529 103 J / kg
Chem. equilibrium
Consider the following equilibrium at 500 K:
CO(g) + 2 H2 (g)  CH3OH(g)
The equilibrium concentrations are: [CO] = 0.0911 M, [H2] = 0.0822 M, [CH3OH] =
0.00892 M, what is the value of the equilibrium constant? Is G positive or negative
when the reactants are transformed into products?
K
CH3OH  
0.00892
 14.5
2
2
COH 2  0.0911 0.0822
Since the value of the equilibrium constant is greater than one, G <0.
Consider the reaction:
N2(g) + 3H2 (g)  2NH3 (g)
Suppose we mix the following initial concentrations:
Keq = 0.5 at 400 K
nNH3  1 M
nN 2  1 M nH 2  1 M
In which direction will the reaction go?
NH3 2  1 2  1 K
eq
N 2 H 2 3 1 1 3
Thus, the reaction will go to the left (some
ammonia molecules will be transformed into
hydrogen and nitrogen).
Problem (chem. equilibrium)
Consider the following reaction:
CO(g) + H2O(g)  CO2 (g) + H2 (g)
Keq = 23.8 at 600 K
If the initial amounts of CO and H2O were both 0.100 M, what will be the amounts
of each reactant and product at equilibrium?
CO
start
change
finish
H2O
CO2
H2
0.1
0.1
0
0
-x
-x
+x
+x
0.1-x
0.1 - x
x
x
We start with n0CO and n0H20 moles of the reacting gases and define as the yield x
the number of moles of CO2 and H2 that the reaction will produce at equilibrium:
The mass action law requires:
x2
x2

K
2
0
0
nH 2  x nCO 2  x 0.1  x 

This is a quadratic equation with respect to x :
Thus, in equilibrium,
nH 2  0.083 M


x  K 0.1  x 
x
0.1 K
 0.083
1 K
nCO 2  0.083 M nH 2 0  0.017 M nCO  0.017 M
Chem. equilibrium
Consider the following reaction: A (g) + 2B (g)  C (g) +3D (g), where A, B, C, D
are some molecules. In equilibrium,
nA  1M
nB  1M nC  1M nD  1M
(a) Calculate the equilibrium constant and the “standard” Gibbs energy for this
reaction at a given temperature.
(b) To the above equilibrium system we add an extra 1M of A. What are the new
equilibrium concentrations of each component?
3

C D 
K
1
2
AB 
 G 0 
  1
K  exp  
 k BT 
(a) The mass action law:
G 0  0
(b)
A
B
C
D
initial
2
1
1
1
change
-x
- 2x
+x
+ 3x
final
2-x
1 - 2x
1+x
1+3x
nA  2 M - x
1  x 1  3x 3
2  x 1  2 x 2
nB  1M - 2x nC  1M  x nD  1M  3x
1