Transcript The laws of thermodynamics

What is Thermodynamics
1. Understanding why things happens
2. Concerning heat, work, related temperature, pressure,
volume and equilibrium
3. Equations relate macroscopic properties
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The laws of thermodynamics
Number
Basis
Property
Zeroth Law
Thermal
Equilibrium
Temperature
First Law
Relation
between work,
energy and heat
Internal Energy
U
Second Law
Spontaneous
process
Entropy
Third Law
Absolute Zero
Degree of
Temperature
Entropy S0
as T0 Kelvin
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
Study of heat engines

Being studied by all students in physical science
and engineering
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Concept of State
V V (P,T )
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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
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For n mole gas
PV=nRT
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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
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Thermodynamics
Process
Work and Energy
Heat
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1. Open system- material and energy exchange
2. Closed system- energy exchange only
3. Isolated system- no material and energy exchange
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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.
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Why is Thermodynamics useful?
1.Qualitative explanation of materials behaviour
2.Quantitatively understanding of materials status.
3. Physical significance of thermodynamic functions.
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Applications of Thermodynamics
1.Extraction, refining
2.Corrosion
3.Phase transformation-phase diagram calculation
4.Materials processing
5.Design of new materials.
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The First Law of Thermodynamics
 Conservation of Energy Principle
 Same principle in mechanics, physics and chemistry
U U U  q  w
i
f
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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 )
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1
V ( final)
w
Pext dV

V (initial)
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Expansion against a constant external pressure
v
final
w  Pext  dV  P (v
v
)  P V
ext final initial
ext
v
initial
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Reversible process
W12
Q12
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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
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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 )
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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.
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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
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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
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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
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 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?
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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.
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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
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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.
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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
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For n moles of materials over small ranges in
temperature Cvconstant
U  nCv (T  Ti )  nCv T
f
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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
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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 


 T  v
Cv  
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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?
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