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Transcript
Topic 2 The first law of
thermodynamics
2012. 02.15
2.1 Introduction of thermodynamics
Therme = Heat
Dynamic = Power
The study of energy and its transformation
Four laws:
The zeroth law---temperature
The first law---internal energy enthalpy
The second---entropy S
The third law---S=0
Energy transformations in our life
History and Progress
Stage I:end of 17th –mid 19th Century
Character of “heat”
Steam engine
Heat engine theory
Large amounts of phenomenon
Stage II: mid19th-1870s
The first and second laws of thermodynamics
Stage III:1870s-beginning of 20th century
The third law of thermodynamic
Statistic thermodynamic
Boltzmann
Stage IV: 1930sQuantum statistical mechanics
Non-equilibrium thermodynamics
Basic concepts

System and the surroundings

States and properties of systems: thermodynamic
parameters

Process and paths

Thermodynamic equilibrium
Systems and surroundings
An open system can exchange matter and energy with its
surroundings
A closed system can exchange energy with its surroundings, but it
cannot exchange matter boundary
An isolated system can exchange neither energy nor matter with its
surroundings
Process and path
P1, V1, T1




Initial state(T1,P1,V1)
state (T2,P2,T2)
Initial state
final
Path 1: T1,P1,V1→T1,P2,Va
(Constant-temperature process)
→T2,P2,V2 (Constant-pressure
process)
Path2: T1,P1,V1→Tb,P2,V1
(Constant-volume process)
→T2,P2,V2 (Constant-pressure
process)
Path3:T1,P1,V1→T2,P1,Vc
(Constant-presure process)
→T2,P2,V2 (Constanttemperature process)
1
2
( )T
P2, Va, T1
3
( )V
P2, V1, Tb
( )P
( )P
P2, V2, T2
( )P
P1, Vc, T2
( )T
Final state
State and path functions




State properties: temperature, pressure, volume
density, viscosity
Extensive properties: dependent on amount
Intensive properties: independent on amount
State function is a property that is independent of how a
sample is prepared, completely differential ,single valued
Properties that relate to the preparation of the state are
called path functions
Question: T, P, V, ρ, Vm……
W, Q
Extensive or intensive?
State function or path function?
Thermodynamic equilibrium
No pure inter- or Intra- energy transformation




Thermal equilibrium
Temperature is equal everywhere
Mechanical equilibrium
No work
Phase equilibrium
No phase transition
Chemical equilibrium
No component change
2.2 The zeroth law of thermodynamics--temperature






If two systems are in thermal equilibrium with a third
system, they are also in thermal equilibrium with each
other.
If A and C are each in thermal equilibrium with B, A is
also in equilibrium with C.
Temperature as a quality of heat, by Galileo and Newton
The temperatures are equal for all systems in thermal
equilibrium.
Temperature scale
Thermometers
2.3 The first law of thermodynamics
The derivation: a long zigzag story……
A design of 1st generation
perpetual motion machine
J.R. Mayer
James Prescott Joule
Lord Kelvin
States in words:

The total energy of system plus surroundings remains
constant.

The total energy of an isolated system is conserved.
(The first generation) Perpetual motion machines do not exist
Mathematical Expression:

E(total energy)= U(internal energy) +
T (kinetic energy/energy of motion) +
system
Q/W
V (potential energy)
surrounding
In
equalibrium state: T=0, V=0, E=U
U = Q + W
closed system
dU = δQ + δW
closed system
U--- Internal energy of the system
Q--- Heat
W--- Work by the system, -; to the system, +
About U:
Internal energy
Unit :J/kJ
U  UT  U e  UV  U R  U N  U ES
U  U sys  U final  U initial
•State function
U  f (T , V )
U
U
dU  (
)V dT  (
)T dV
T
V
About Heat




The energy of a system changes as a result of a
temperature difference
Heat flow from the hotter body to the colder one
Unit: J/kJ
Q>0 endothermic
Q<0 exothermic
cooler
heater
Insulating
shield
Q=0
Thermal shield
Heat capacity
U
U
dU  (
)V dT  (
)T dV
T
V
U
The heat capacity at
CV  ( )V
T
constant volume
dU=dQ=Qv, at constant volume,
no additional work
QV=CvdT
Calorimetry

Qv=Cv.△T
About Work:


Work is done when an object is moved against an opposing force
Giving out work: W<0; Given work: W>0
Type of work
dW
comments
units
Expansion
-psurrdV
Psurr is the external pressure
dV is the change in volume
Pa
m3
Surface
expansion
γdσ
extension
Electrical
γ is the surface tension
dσ is the change in area
N.m-1
m2
fdl
f is the tension
dl is the change of length
N
m
Фdq
Ф is the electric potential
dq is the change in
charge
V
C
2.4 Expansion work/p-V work

The work arising from a change in volume
W= -Psurr.ΔV (?)
W   Wi   PSurrdV   PSurrdV
V2
V1
Think about:
If Q=0, ΔU=-pdV ?
When would it be right?
Example: Calculation of expansion work
δWsurr,1   psurr dV  0

Free expansion
Psurr=0

等外压膨胀(psurr=常数)Expansion against constant pressure
Wsurr,2   psurr (V2  V1 )

多次等外压膨胀
Wsurr,3   p '(V ' V1 )
 p "(V " V ')
 p2 (V2  V ")

准静态膨胀 reversible expansion
W   Wi   PSurrdV   PSurrdV
V2
V1
Reversible expansion


The system is always infinitesimally close to equilibrium,
and an infinitesimal change in conditions can reverse the
process to restore both system and surrounding to their
initial state.
A reversible process is obviously an idealization
Reversible isothermal process in a perfect gas
2
V1
P2
W    pdV  nRT ln  nRT ln
V2
P1
1
Features of reversible expansion
quasi-static (quasi-equilibrium)
Carried out infinitesimally slowly so that the
driving force is only infinitesimally greater than the
opposing force
Return to the original system state without having had a
change, either in the system or the surroundings


The work done by a system during a reversible process
is the maximum work we can get.
The work done on a system in a reversible process is the
minimum work we need to do to achieve that state
change.
判断下列哪个过程是可逆过程?

(1)摩擦生热;

(2)室温、标准压力下一杯水蒸发为同温同压的水;

(3) 373K,P0下,一杯水蒸发为气体;

(4)手电筒中干电池放电使灯泡发亮;

(5)N2和O2混合;

(6)恒温下将1mol水倾入一溶有不挥发溶质的大量溶液中
2.5 Enthalpy 焓



△U= Qv (Closed system, only expansion work
exchanging, constant volume)
Qv , heat transactions at constant volume
Qp, heat transactions at constant pressure
ΔU= Q+W=Qp- PsurrΔV
Psurr= P1= P2
∴ΔU= Qp – P2 V2 + P1V1= U2-U1
∴QP=(U2+ P2 V2)-(U1+ P1V1)
Enthalpy H≡U+PV,∴Qp=H2-H1=ΔH
•State function, extensive
Comparing H and U



H=U+PV
State function
Same units
Heat capacity

Heat capacity at constant pressure
Qp
H
Cp 
(
)p
T
dT

Heat capacity at constant volume
QV
U
CV 
 ( )V
T
dT

H  Qp   CpdT
U  QV   CV dT
Closed system, in equilibrium, PV work only
Relating ΔH and ΔU

W’=0 , ΔH=Qp and ΔU=Qv

For process only involve solids or liquids
Δ(PV) is negligible
ΔH ≈ ΔU




For gases:
Δ(PV)=nRΔT
Qp  QV  nRT
For example:

H2O(l,0℃)→H2O(l,50℃)
System: Water; Water+ electric line
W, Q, △U
?
U = Q + W
Home work:


A: P66
2.7(a,b)
Y:P12
1, 5, 9
Preparation for next class;
Y: 1.6-1.9
A: 2.6, 3