1. U, Q, W(We+Wf), Cv( ) Cp, H=U+PV
Brief review
U = 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

2. 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?

3. 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
What happens if we release the pump? 
How about the process happen reversibly?
If psur=constant the whole pump as a system, Qp=△H=0?

4. About the variation of internal energy dU, △U
Discussion 1:
About the variation of internal energy dU, △U
At constant volume

5. For ideal gas:

8. Good approximation for real gases under most conditions
For liquids and solids
So , for all substances without phase transformation

9. dH, △H

12. For all substances

13. Discussion 2: About Cv, Cp
State properties Extensive function
Cv,m, Cp,m Intensive function
Dependence of heat capacity on temperature for real substances or

14. Example

16. Differential scanning calorimetry DSC Differential thermal analysis DTA
Qualitative and quantitative analysis depending on heat capacity

17. 2.6 Relating Cp and Cv

18. For ideal gases By statistical mechanic CV,m CP,m Monoatomic 3/2R 5/2R
diatomic
(or linear molecule ) /2R /2R
polyatomic molecule
(or nonlinear molecule) 6/2R=3R R

19. For any substance other than an ideal gas
real gas
For liquids and solids
≈ 0

21. 2.7 Gay-Lussac-Joule experiment
Q=0 W=0
ΔU=0
Constant energy process
dU=0，dT=0，dV≠0
Ideal gases

22. Properties of ideal gases
U is the function of T only, U(T)
ΔH=ΔU + ΔPV =ΔU + nR（T2-T1）
U, H, Cv, Cp of ideal gases are only the function of T

23. 2.8 Adiabatic processes of ideal gases
Free expansion: W=0
△U=0, △H=0
If W>0, △U>0, △ T>0, T↑
If W<0, △U<0, △ T<0, T↓
If wf=0 dU+pdV=0 dU=CVdT，p=nRT/V C v dT
= - dV V
nRT

24. V T C ln( )=Rln( ) T1 V1 γ-1= T2 V2γ-1 Adiabatic Process Equation 2
v,m
ln( 1 2 T
)=Rln( V )
T1 V1 γ-1= T2 V2γ-1
Adiabatic Process Equation

25. Comparing with other processes
n= 0 (pressure constant),
1 (isothermal),
γ (adiabatic) ….

26. A-B Isotherm
A-C Adiabatic P V
C(P2,V2”)
B(P2,V2)
A(P1,V1)

27. 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

28. Homework
A: P ,
Y: P P
Preparation for next class:
The working principle of an refrigerator
A : P Y: