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What is Thermodynamics
1. Understanding why things happens
2. Concerning heat, work, related temperature,
pressure, volume and equilibrium
3. Equations relate macroscopic properties
The laws of thermodynamics
Number
Basis
Property
Zeroth Law
Thermal
Equilibrium
Temperature
First Law
internal energy,
work and heat
Internal Energy
U
Second Law
Spontaneous
process
Entropy
Third Law
Absolute Zero
of Temperature
Entropy S0 as
T0 Kelvin

Study of heat engines

Being studied by all students in physical science
and engineering
Concept of State
V V (P,T )
From Avogadro’s hypothesis the volume per mole of all ideal
gases at 0oC and 1atm pressure is 22.414 litres.
PoVo 1 atm 22.414 litres

To
273.16 deg ree.mole
 0.082057 litre.atm / deg ree.mole
PV  RT
R  0.082057 litres.atm / deg ree.mole
 8.3144 joules / deg ree.mole
For n mole gas
PV=nRT
For water vapour, the number of moles for Kg water
is obtained by
mass
1000 g
n

 55.56 mol
1
M
18 g.mol
Thermodynamics
Process
Work and Energy
Heat
1. Open system- material and energy exchange
2. Closed system- energy exchange only
3. Isolated system- no material and energy exchange
What we learn from this module?
1.Internal energy U and entropy S
2.Combining U and S with P, T and V gives enthalpy H=U +
PV and Gibbs energy G=H-TS
3.H is related to heat adsorption or release at constant
pressure
4. G controls the position of equilibrium in closed systems at
constant temperature and pressure.
Why is Thermodynamics useful?
1.Qualitative explanation of materials behaviour
2.Quantitatively understanding of materials status.
3. Physical significance of thermodynamic functions.
Applications of Thermodynamics
1.Extraction, refining
2.Corrosion
3.Phase transformation-phase diagram calculation
4.Materials processing
5.Design of new materials.
The First Law of Thermodynamics
 Conservation of Energy Principle
 Same principle in mechanics, physics and chemistry
U U U  q  w
i
f
Work done in an Expansion (or contraction) against an External Pressure
w  Fx
Pext  F
A
F  Pext .A
w  Pext Ax
w  PextV  Pext .(V V )
2
V ( final)
w
Pext dV

V (initial)
1
Expansion against a constant external pressure
v
final
w  Pext  dV  P (v
v
)  P V
ext final initial
ext
v
initial
W12
Q12
Reversible process and Maximum Work
Reversible process for a closed system
W system-environment =W environment-system
Q environment-system =Q system-environment
v
final
ws  e   P dV
ext
v
initial
v
final
we  s   P dV
int
v
inital
ws  e  we  s  wmax
For an ideal gas
P  nRT
int V
v
final
w   nRT dV
V
v
initial
For isothermal process,
ie. T=constant
wmax  qrev  nRT ln v( final)
v(initial)
Thermodynamics tutorial
on the 24th of Oct (Tuesday)
From 2pm D14
Questions*:
1. How to calculate W for a constant pressure
process?
2. How to calculate W for an isothermal reversible
process?
3. Is a constant pressure process an reversible
process? Explain why?
* All of these questions are concerned with ideal
gas systems.
Example:
Gas compress during quasi-equilibrium processing,
with PV=constant. The system is the gas
P1=101325 N/m2, V1=0.01m3, V2=0.005m3
Find W
W=-702J
Enthalpy
U=q-w
· Q Heat is transferred due to the presence of a temperature
difference.
· Work here is considered as the work of expansion.
· U results from the oscillation of atoms or ions in solid and
movement of the particles in gas and liquid.
· Q and w depend on how the change is carried out where
the difference between them does not.
· At constant volume, w=0 and U=q
Thursday lecture
• Sackville Street Building C9
• 11am-12:00
The Enthalpy Function
H  U  PV
When P=constant
H  H
final
H
initial
 (U
final
 PV
final
)  (U
initial
 PV
)
initial
 U  PV
U  q  w
w  P.V
H  U  P.V  q  w  P.V
 q  P.V  p.V  q
 At constant pressure, the change in enthalpy is equal
to the heat
 The change of enthalpy is independent of path.
Q: Does q or W depend on path?
 For the change involving solids and liquids,
HU, but for gases, HU
Q:explain why?
Example 1:
Given p=constant=101.3 kPa
V1=1m3, V2=2m3
Q12=200kJ
Find a) U
b) an expression for Q12 in terms of
thermodynamics properties for a quasiequilibrium process.
Example 2.
Given T=T1=T2=constant for the
process
P1=200 kPa, T1=300K
V1=2m3, V2=4m3
Find a) W and b) Q
W=277KJ
Heat Capacities (Cp and Cv)
C  q T
a) Under constant volume conditions Cv- all heat
supplied increases energy of sample
b) Under constant pressure conditions Cp- Heat
supplied increases energy of sample and provides
energy for work performed.
q
Cv 
T
Relation between Cv and U
The 1st Law
U  q  w
When V=constant
w0
Therefore
U  q
U  Cv .T
Cv
 U
 
 T



v
 U
 
 T



v
For n moles of materials over small ranges in
temperature Cvconstant
U  nCv (T  Ti )  nCv T
f
Relation between Cp and H
At constant pressure
Cp 
q
T
H  C p .T
Cp 
H
T
Cp 
H  H 

T  T  p
H  C p T
Over small range of T for n moles of materials
H  nC p (T  Ti )  nC p T
f
Summary
At constant volume
q  U
At constant pressure
q  H
Molar heat capacity at constant pressure
 H 


 T  P
C p  
Molar heat capacity at constant volume
 U 

Cv  

 T 
v
Questions:
1. For a constant temperature process of an ideal gas,
prove H=U.
2. For a gas system, explain why Cp is larger than
Cv?
3. For a solid/liquid system, explain why Cp is close
to Cv?
4. What are the equations for calculating change of
enthalpy and internal energy due to temperature
change?