Transcript thermodynamics - La Salle High School

THERMODYNAMICS
thermo – heat
dynamics – changes
Laws of Thermodynamics:
0th, 1st, 2nd, 3rd
Note: Thermodynamics ……..

is not concerned about rate of changes
(kinetics) but the states before and after
the change

does not deal with time
Classical Thermodynamics
Marcoscopic observables
T, P, V, …
Statistical Thermodynamics
Microscopic details dipole
moment, molecular size, shape
Joule’s experiment
Thermometer
w
adiabatic wall
T  mgh
h
(adiabatic process)
U (energy change) = W (work) = mgh
Page 59
w
T  time interval of
heating
U = q (heat)
Conclusion: work and heat has the same effect
to system (internal energy change)
*
FIRST LAW: U = q + W
U: internal energy is a state function
[ = Kinetic Energy (K.E.)
+ Potential Energy (P.E.) ]
q: energy transfer by temp gradient
W: force  distance
E-potential  charge
surface tension  distance
pressure  volume
First Law:
The internal energy of an isolated system is constant
Convention: Positive: heat flows into system
work done onto system
Negative: heat flows out of system
work done by system
Pressure-volume Work
d  work   Fext dl
  M piston  gdh
weight

 Adh
A
  Pext  Adh
  Pext dV
work   Pext V2  V1 
 Mgh
P1 = P2
M
V1
M
V2
If weight unknown, but only properties of system
are measured, how can we evaluate work?
P
M
V1
M
V2
P
Assume the process is slow and steady,
Pint = Pext
Free Expansion:
Free expansion occurs when the external pressure
is zero, i.e. there is no opposing force
Reversible change: a change that can be
reversed by an infinitesimal modification of a
variable.
Quasi equilibrium process:
Pint = Pext + dP (takes a long time to
complete)
infinitesimal at any time
W   Pext dV
  Pint dV
quasi equilibrium process
Page 64-66
P 1
2
Example:
P1 = 200kPa = P2
V1 = 0.04m3 V2 = 0.1m3
V
*
2

W12    PdV   P V2  V1
1

 200kPa 01
.  0.04m
3
 12kJ
what we have consider was isobaric expansion
(constant pressure) other types of reversible
expansion of a gas: isothermal, adiabatic
Page 65-66
Isothermal expansion
remove sand slowly
at the same time
maintain temperature
by heating slowly
2
Wrev    PdV
21
nRT
 
dV
V
1
2
 nRT 
1
P
*
T
dV
V
 V2 
V1
 nRT ln 
 V1  area under curve
V2
V
Ex. V1 = 0.04m3 P1 = 200kPa V2 = 0.1m3
Wrev
 V2 
 nRT1 ln 
 V1 
 V2 

  PV
1 1 ln
 V1 
 0.1 
 200kPa  0.04m ln

 0.04 
3
 7.33 kJ
PV
P2  1 1  80kPa
V2
Adiabatic Reversible Expansion
For this process
PV = constant for ideal gas
(proved later)
 
Cp
(slightly larger
than 1)
Cv

PV
1 1 dV
W    PdV   

V
1
1

1
1
PV
V

V
1 1
2
1

1 
2
2

P2V2  PV
1 1

1 

V2
V1
P1
T1 P2
T2
Ex. V1 = 0.04m3, P1 = 200kPa, V2 =
0.1m3,
 = 1.3
 0.04 
P2  200

 010
. 
W
1.3
 60.77 kPa
 60.77  01
.  200  0.04
1  13
.
d: pressure drop without
volume change
 6.41kJ
200kPa
const
P
isothermal
W    PdV  0
|Wa|>|Wb|>|Wc|>|Wd|
a
b
d
adibatic
V
State Function Vs Path Function
State function: depends only on position in
the x,y plane
e.g.: height (elevation)
300
1
B
2
200
100m
Y
X
A
Path Function: depends on which path is taken
to reach destination
from 1  2, difference of 300m (state function)
but path A will require more effort.
Internal energy is a state function,
heat and work are path functions
P
200kPa 1
2
481.13K
3
0.04
0.1
192.45K
V/m3
5 moles of monoatomic gas
3
CV  R
2
U  nCV T  3 2 R 289  5  18kJ
W1 2  12 kJ
Q1 2  U  12 kJ  30kJ
W1 3 2  7.33kJ  0  7.33kJ
Q1 3 2  7.33kJ  18kJ  25.33kJ
Joule’s second experiment
Energy UU(V,T) ???
thermometer
V1
V2
adiabatic wall
At time zero, open valve
thermometer
No temperature
change
adiabatic wall
After time zero, V1 V1+V2
T=0
 Q=0
W=0 no Pext  U=0
U=U(T) Energy is only a function of
temperature for ideal gas *
 U 

 0
 V  T
 U 
 U 
dU  
 dT  
 dV  CV dT
 T V
 V  T
0 (for ideal gas)
CV is constant volume heat
capacity
*Enthalpy
Define
H = U + PV
state function
intensive variables
locating the state
Enthalpy is also a state function
H = U + PV + VP
At constant pressure
H = U+PV
= U - W
=Q
H = QP constant pressure heating
H  H T , P 
H is expressed as a functional of T and P
 H 
 H 
dH  
 dT  
 dP  C P  0  C P
 T  P
 P  T
C P  CV  R
CP
 
CV
for ideal gases
Thermochemistry
Heat transferred at constant volume qV = U
Heat transferred at constant pessure qP = H
Exothermic
H = -
Endothermic H = +
Standard states, standard conditions
do not measure energies and enthalpies
absolutely but only the differences, U or H
The choice of standard state is purely a
matter of convenience
What is the standard state ?
The standard states of a substance at a specified
temperature is its pure form at 1 atmosphere
25oC, 1 atmosphere:
the most stable forms of elements assign
“zero enthalpy”
H
o
298
= 0 used for chemical reactions
Standard enthalpy of formation
Standard enthalpy change for the formation
of the compound from its elements in their
reference states.
Reference state of an element is its most stable
state at the specified temperature & 1 atmosphere
C (s) + 2H2 (g)  CH4 (g)
289K, 1 atm
o
Hf
= -75 kJ
From the definition, Hfo for elements  0
Hess’s Law
The standard enthalpy of an overall reaction is the sum of the
standard enthalpies of the individual reactions into which a
reaction may be divided.
Standard reaction enthalpy is the change in enthalpy when the
reactants in their standard states change to products in their
standard states.
Hess’ Law
H2
H3
H4
R 
 X 
Y 
 P
H1
H1 = H2 + H3 + H4
function
state
Hess’s law is a simple application of the first law of thermodynamics
e.g. C (s) + 2H2 (g)  CH4 (g)
298K, 1 atm
H1o = ?
C (s,graphite) + O2 (g)  CO2 (g)
Ho = -393.7 kJ
H2 (g) + ½O2(g)  H2O (l)
Ho = -285.8 kJ
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
Ho = -890.4 kJ
H1o = -393.7 + 2(-285.8) - (-890.4)
= -75 kJ/mole
Heat of Reaction (Enthalpy of Reaction)
Enthalpy change in a reaction, which may be
o
obtained from Hf of products and reactants
Reactants  Products
H r 
o

  H
n p H f , prod 
o
products

i
 nR H f ,reaction
o
reaactan ts
o
f
i
products
 reac tan ts
I stoichiometric coefficient, + products,
- reactants
E.g. CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)
Hfo/ kJ
CH3Cl
-83.7
HCl
-92.0
Cl2
0
CH4
-75.3
Hro = (-83.7-92.0) - (-75.3+0) = -100.4 kJ
 n
Reactants
R
H f , R o
elements
Products
 n H
elements
p
o
f ,p
Page 83
2A + B  3C + D
0 = 3C + D - 2A - B
Generally,
0 =  J J J
J denotes substances,
J are the stoichiometric numbers
*
H r   j H
o
o
f ,j
Bond energy (enthalpy)
Assumption – the strength of the bond is
independent of the molecular environment
in which the atom pair may occur.
o
C (s,graphite) + 2H2 (g)  CH4(g)H = -75.4 kJ
H2 (g)  2H (g)
Ho = 435.3 kJ
o
C (s,graphite)  C (g)
H = 715.8 kJ
 C (s,graphite) + 2H2 (g)  C (g) + 4H (g)
Ho = 2(435.3)+715.8 = 1586.5 kJ
C (g) + 4H (g)  CH4 (g) H = -75.4-1586.5
= -1661.9 kJ
CH Bond enthalpy = 1661.9/4 kJ = 415.5 kJ
Temperature dependence of Hr
T
H r  H r   C p dT
T
o
To
C P 


C p i ,productsn p 
products
reactants
P
HrT
CP,P
Hro
R
CP,R
298
C p i ,reaction nR 
T
 C
i
reactants
 products
pi
What is the enthalpy change for vaporization
(enthalpy of vaporization) of water at 0oC?
H2O (l)  H2O (g)
Ho = -241.93 - (-286.1) = 44.01 kJmol-1
o
H2 = H + CP(T2-T1) assume CP,i
constant wrt T
H (273) = Ho(298) + CP(H2O,g) - CP (H2O,l)(273-298)
= 44.10 - (33.59-75.33)(-25)
= 43.0 kJ/mole
Reversible vs Irreversible
Non-spontaneous changes vs Spontaneous
changes
Reversibility vs Spontaneity
First law does not predict direction of changes,
cannot tell which process is spontaneous.
Only U = Q + W
Second Law of Thermodynamics
•Origin of the driving force of physical and chemical change
•The driving force: Entropy
•Application of Entropy:
•
Heat Engines & Refrigerators
•
Spontaneous Chemical Reactions
•Free Energy
Second Law of Thermodynamics
No process is possible in which the sole result is the absorption
of heat from a reservoir and its complete conversion into work
Hot Reservoir
q
w
Engine
Direction of Spontaneous Change
More Chaotic !!!
Spontaneous change is usually accompanied by a
dispersal of energy into a disorder form, and its
consequence is equivalent to heating
Entropy (S) is a measurement of the randomness
of the system, and is a state function!
SQ
S 1/T
Page 122
Entropy S
For a reversible process, the change of entropy is defined as
dqrev
dS 
T
(thermodynamic definition of the entropy)
*
Another expression of the Second Law:
The entropy of an isolated systems increases in the course of a
spontaneous change:
Stot > 0
where Stot is the total entropy of the isolated system
Entropy S
The entropy of an isolated systems increases in the course of a
reversible change:
Stot = 0
where Stot is the total entropy of the isolated system
Carnot’s Theoretical Heat Engine
Heat flows from a high temperature reservoir
to a low temperature body. The heat can be
utilized to generate work.
e.g. steam engine.
The efficiencies of heat engines
Hot Reservoir
q
w
Engine
S = - |q|/Th < 0
not possible! contrary to the second law
The Nernst Theorem
The entropy change accompanying any physical or
chemical transformation approaches zero as the
temperature approaches zero.
S  0 as T0
Third Law of Thermodynamics
If the entropy of each element in its most
state is taken as zero at the absolute zero
of temperature, every substance has a
positive entropy. But at 0K, the entropy of
substance may equals to 0, and does
become zero in perfect crystalline solids.
Implication: all perfect materials have the same
entropy (S=0) at absolute zero temperature
Crystalline form: complete ordered,
minimum entropy
Statistical Interpretation of S
S=0
at 0K for perfect crystals
S = k ln 
Boltzmann
postulate of
entropy
number of arrangements
Boltzmann constant
Entropy Change of Mixing
one distinguish
arrangement
4!
4!
S A  k ln
 0, S B  k ln
0
4!
4!
 A B
N
 N B !
number of
arrangement
increased
87  65


 70
NA!NB !
4  3 2 1
A
S A B  k ln 70
In general mixing NA, NB
S mix

N A  N B !
 k ln
N A! N B !
Entropy change of mixing
Stirling’s approximation: ln N!  N ln N + 0(N)
for large N
S mix  k  N A  N B  ln  N A  N B   N A ln N A  N B ln N B 
 k  N A  N B ln  N A  N B   X A ln N A  X B ln N B 
 k  N A  N B  X A ln X A  X B ln X B   0
Extensions of 2nd Law
TdS dq
Clausius Inequality
For adiabatic process,
TdS  0 or dS  0
Entropy will always attain maximum in
adiabatic processes.
A similar function for other processes?
Define Helmholtz free energy
A = U - TS
Thermodynamic
State Function
dA = dU - TdS - SdT
Substitute into Clausius Inequality
0  dq  TdS
0  dq  dA  dU  SdT
0  PdV  dA  SdT
for isothermal, isochoric (constant volume)
process,
0  dA
change in Helmholtz free energy = maximum
isothermal
work
Example of isothermal, isochoric process:
combustion in a bomb calorimeter
O2 +
fuel
Temp. bath
heat given
out
O2, CO2,
H2O
Higher P
Gibbs Free Energy
Define Gibbs free energy
G = H - TS
Thermodynamic
state function
= U + PV -TS
dG = dU +PdV + VdP -TdS -SdT
constant pressure, constant temperature
0  dG
G will tend to a minimum value
dG  0 equilibrium
dG  0 spontaneous change
More applications since most processes are
isothermal, isobaric
chemical reactions at constant T, P
Reactants  Products
endothermic H is positive
exothermic
H is negative
Change of Gibbs free energy with temperature
(constant pressure)
dG   SdT VdP  0
G2  G1    SdT
 G 

  S
 T  P
H vap PH vap
dP H vap



dT TVvap
TVvap
RT 2
H vap
dP

 d  ln P  
2 dT
P
RT
H vap
sat
 ln P  
sat  const .
RT
P2 H vap T2  T1 
ln

P1
RT1T2
In P
1/T
Example: What is the change in the boiling
point of water at 100oC per torr
change in atmospheric pressure?
Hvap = 9725 cal mol-1
Vliq = 0.019 l mol-1
Vvap = 30.199 l mol-1
sat
dP
dT sat
9725 cal mol 0.04129 l atmcal 


T V  V 
37315
. K 30180
.
l mol 
H vap
1
1
1
v
l
 0.03566 atm K 1
 27.10 torr K 1
dT
 0.0369 K torr 1
dP