u, q, w(we+wf), cv( ) cp, h=u+pv
DESCRIPTION
Example 1: U = Q + W H2O(l,0℃)→H2O(l,50℃) System: Water W, Q, △U ? For the first case, Q=0, ()V, QV=△U=0, is it right? Why?TRANSCRIPT
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Brief reviewU = Q + W
U, Q, W(We+Wf), Cv( ) Cp, H=U+PV
Initial state A Final state B
State function
Path function
Reversible process
Irreversible process
( )dV
V VQ UC TT
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Example 1:
H2O( l, 0℃)→H2O( l, 50℃)System: Water
W, Q, U △ ? U = Q + W
For the first case, Q=0, ()V, QV= U=0, is it right? Why?△
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Example 2:
Some air in a bicycle pump is compressed so that its volume decreases and its internal energy increases.
If 25 J of work are done by the person compressing the air, and if 20 J of thermal energy leave the gas through the walls of the pump, what is the increase in the internal energy of the air?
U = Q + W
If psur=constant the whole pump as a system, Qp=△H=0?What happens if we release the pump? How about the process happen reversibly?
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Discussion 1: About the variation of internal energy dU, △U
At constant volume
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For ideal gas:
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Good approximation for real gases under most conditions
For liquids and solids
So , for all substances without phase transformation
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dH, H△
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For all substances
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Discussion 2: About Cv, Cp
2,mpC a bT cT
2,m / '/pC a b T c T or
Dependence of heat capacity on temperature for real substances
State properties Extensive function
Cv,m, Cp,m Intensive function
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Example
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Differential scanning calorimetry DSCDifferential thermal analysis DTA
Qualitative and quantitative analysis depending on heat capacity
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2.6 Relating Cp and Cv( ) ( )p pV VH UC CT T
( )( ) ( ) p VU PV UT T
( ) ( ) ( )p p VU V UpT T T
( ) ( ) ( ) ( )p pV TU U U VT T V T
( ) ( ) ( )p p pV TU V VC C pV T T
[ ( ) ]( ) pTU Vp V T
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For ideal gases( ) 0, TUV
( ) /pV nR pT
p VC C nR
By statistical mechanic CV,m CP,m
Monoatomic 3/2R 5/2R diatomic (or linear molecule ) 5/2R 7/2R polyatomic molecule(or nonlinear molecule) 6/2R=3R 4R
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real gas
For any substance other than an ideal gas
For liquids and solids ≈ 0
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( ) ( ) ( ) ( )p pV TU U U VT T V T
( , ), ( , )U U T V V V T p
d ( ) d ( ) dV TU UU T VT V
d ( ) d ( ) dp TV VV T pT p
d ( ) d ( ) [( ) d ( ) d ]pV T TU U V VU T T pT V T p
d ( ) ( ) d [( ) ( ) ( ) ]dT T V T pU V U U VU p TV p T V T
=( ) d [( ) ( ) ( ) ]dT V T pU U U Vp Tp T V T
( ) ( ) ( ) ( )p V T pU U U VT T V T
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2.7 Gay-Lussac-Joule experiment Q=0 W=0 ΔU=0 Constant energy process
dU=0, dT=0, dV≠0
dVVUdT
TUdU TV )()(
0)(
TVU
Ideal gases
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Properties of ideal gases0)(
TVU
ΔH=ΔU + ΔPV =ΔU + nR( T2-T1)dP
PUdT
TUdU TP )()(
0)(
TPU
U is the function of T only, U(T)
VV TUC )(
( )P PHCT
U, H, Cv, Cp of ideal gases are only the function of T
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2.8 Adiabatic processes of ideal gasesdU Q W = 0W Q ( )
If W>0, U>0, △ △ T>0, T↑
If W<0, U△ <0, △ T<0, T↓
Free expansion: W=0 △U=0, H=0△
If wf=0 dU+pdV=0 dU=CVdT, p=nRT/V
C vdT = - dVVnRT
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, ,2 1 1
1 , 2 2
( 1)P m V m
V m
C CT V Vln ln lnT C V V
V
P
mV
mP
CC
CC
,
,
1
2
TT
= 1
2
1 )(
VV
C v,m ln(1
2
TT )=Rln(
2
1
VV
)
T1 V1 γ-1= T2 V2γ-1
1pV K 12TV K 1
3p T K
Adiabatic Process Equation
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Comparing with other processesnpV K n= 0 (pressure constant),
1 (isothermal),
γ (adiabatic) ….
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A-B Isotherm
1
2
2
12
1
lnlnPPnRT
VVnRTpdVW
2
1
dV
VW p V
2
1
= dV
V
K VV
( )pV K 1 1
2 1= 1 1( )
(1 )K
V V
1 1 2 2pV p V K
2 2 1 1= 1p V pVW
2 1( )1
nR T T
A-C Adiabatic
P
V
C(P2,V2”) B(P2,V2)
A(P1,V1)
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Example(a) The pump is operated quickly so the
compression of the air in the cylinder before the valve opens can be considered adiabatic.
At the start of a pump stroke, the pump cylinder contains 4.25 × 10-4 m3 of air at a
pressure of 1.01 × 105 Pa and a temperature of 23 °C. The pressure of air in the dinghy is 1.70 × 105 Pa. When the valve is about to open, the volume of air in the pump is ?.
γ for air = 1.4
(b) Calculate the temperature of the air in the
pump when the valve is about to open.
V2 = 2.94 x 10-4 m3
T2 = 344 K
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Homework A: P88 3.2, 3.3 3.6 3.7 Y: P21 16 P25 23
Preparation for next class:
The working principle of an refrigerator
A : P 55-69 2.7-2.9
Y:33-47 1.10-1.12