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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
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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?
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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 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?
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About the variation of internal energy dU, △U
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
State properties Extensive function Cv,m, Cp,m Intensive function Dependence of heat capacity on temperature for real substances or
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Example
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Differential scanning calorimetry DSC Differential thermal analysis DTA
Qualitative and quantitative analysis depending on heat capacity
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2.6 Relating Cp and Cv
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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
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For any substance other than an ideal gas
real gas For liquids and solids ≈ 0
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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 Ideal gases
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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
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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
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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
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Comparing with other processes
n= 0 (pressure constant), 1 (isothermal), γ (adiabatic) ….
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A-B Isotherm 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: P , Y: P P Preparation for next class: The working principle of an refrigerator A : P Y:
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