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

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?

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?

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

For ideal gas:

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

dH, △H

For all substances

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

Example

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

2.6 Relating Cp and Cv

For ideal gases 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

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

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

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

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

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

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

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

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

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