物化上双语课件24

1、Chapter 2 Laws and Fundamental Equations in Thermodynamics II.Variations of Thermodynamic Functions in Various Processes 3 2.1 IntroductionI. Fundamentals of Thermodynamics II. Variations of Thermodynamic Functions in Various Processes 2.8 Variations of Thermodynamic Functions in pVT Changes 2.9 Jou

2、le-Thomson Effect 2.10 Variations of Thermodynamic Functions in Phase Transitions 2.11 The Third Law of Thermodynamics 2.12 Variations of Thermodynamic Functions in Chemical ReactionsIII. Two Categories of Application Framework of the Chapter4Variations of Thermodynamic Functions Connections with th

3、e pVT relations and the heat capacities 5pVT Changes of Ideal Gases 6(1) Isothermal Process T = Tsur = C pVT Changes of Ideal Gases 7(2) Adiabatic Process Q = 0，U=Wdef2.8 Variations of Thermodynamic Functions in pVT ChangesAdiabatic Reversible Process Adiabatic Irreversible Process 8Comparison betwe

4、en Isothermal and Adiabatic Processes 123pHpLReversible AdiabaticIrreversible AdiabaticReversible Isothermal02.8 Variations of Thermodynamic Functions in pVT Changes9(3) Isochoric and Isobaric Processes Example1mol of a diatomic ideal gas at 0.1MPa undergoes successively the following changes: (a) I

5、sochoric heating from 25 to 100, (b) adiabatic expansion against vacuum with the volume doubled, (c) isobaric cooling to 25. Calculate the total Q, W, U, H, S. Given =5R/2, +R=7R/2.2.8 Variations of Thermodynamic Functions in pVT ChangespVp1, V, 25p2, V, 100p3, 2V, T3p3, V4, 25Q=0ABCD10Example2.8 Va

6、riations of Thermodynamic Functions in pVT ChangespVp1, V, 25p2, V, 100p3, 2V, T3p3, V4, 25Q=0ABCDp1=0.1MPaT1=298KV1=24.78dm3p3=0.063MPaT3=373KV3=49.55dm3p4=0.063MPaT4=298KV4=39.6dm3p2=0.125MPaT2=373KV2=24.78dm3(a)(b)(c)11Solution2.8 Variations of Thermodynamic Functions in pVT Changes12Example2.8 V

7、ariations of Thermodynamic Functions in pVT Changes13(4) Isothermal Mixing of Ideal Gases ExampleA container is separated by a barrier into two parts at 0, one is filled with 0.2mol of O2, the other is 0.8mol of N2, both with a pressure of 0.1MPa. Removing the barrier, the two gases mix homogeneousl

8、y. Calculate Q, W, U, H, S, G, and judge the reversibility. (4) Isothermal Mixing of Ideal Gases 2.8 Variations of Thermodynamic Functions in pVT Changes14解：(4) Isothermal Mixing of Ideal Gases Solution2.8 Variations of Thermodynamic Functions in pVT Changes152. Non-ideal Gases, Liquids and Solids 2

9、.8 Variations of Thermodynamic Functions in pVT Changes16Isochoric Process Isobaric Process 2.8 Variations of Thermodynamic Functions in pVT Changes2. Non-ideal Gases, Liquids and Solids 17The enthalpy change of an ideal gas 2.8 Variations of Thermodynamic Functions in pVT Changes2. Non-ideal Gases,

10、 Liquids and Solids For real fluids, we need the enthalpy difference between the real fluids and corresponding ideal gases. Since is just the H of a real fluid when the pressure approaches zero, Departure Function18 2.1 IntroductionI. Fundamentals of Thermodynamics II. Variations of Thermodynamic Fu

11、nctions in Various Processes 2.8 Variations of Thermodynamic Functions in pVT Changes 2.9 Joule-Thomson Effect 2.10 Variations of Thermodynamic Functions in Phase Transitions 2.11 The Third Law of Thermodynamics 2.12 Variations of Thermodynamic Functions in Chemical ReactionsIII. Two Categories of A

12、pplication Framework of the Chapter19Joule Experiment The containers with gas and the water bath are taken as the system. 2.9 Joule-Thomson EffectFor ideal gas CH2O Cgas, the temperature change can not be detected The accuracy of the experiment should be improved! 20The improvements： The free expans

13、ion changes into the throttling expansion with a porous plug.2. the temperature of the gas is determined directly.1. Throttling Process Joule-Thomson Experiment 2.9 Joule-Thomson Effect21The gas in an adiabatic cylinder is separated into two parts by a porous plug. p1 p2. By pushing the piston on th

14、e left side slowly, a certain amount of gas (V1) flows into the right side with a volume increase of (V2) under the isobaric condition. Joule-Thomson Experiment It was found that the temperature changed! Joule Thomson effect2.9 Joule-Thomson Effect22The gas on the left is compressed isothermally and

15、 the work done:wL = - p1 (0- V1) = p1V1 The work done in right: wR = - p2 (V2 - 0) = -p2V2The total work done:w = wL + wR = p1V1 p2V2Joule-Thomson Experiment 2.9 Joule-Thomson Effect23U2 + p2V2 = U1 + p1V1 Joule-Thomson Experiment H2 = H1The enthalpy keeps unchanged in the throttling process2.9 Joul

16、e-Thomson Effect240 the temperature decreases after throttling2. Joule-Thomson Coefficient =0 the temperature does not change 0JT = 0 JT 0 coolinginversion Theating26For real gases2.9 Joule-Thomson Effect274. JouleThomson节流的应用：Air liquefaction28 2.1 IntroductionI. Fundamentals of Thermodynamics II.

17、Variations of Thermodynamic Functions in Various Processes 2.8 Variations of Thermodynamic Functions in pVT Changes 2.9 Joule-Thomson Effect 2.10 Variations of Thermodynamic Functions in Phase Transitions 2.11 The Third Law of Thermodynamics 2.12 Variations of Thermodynamic Functions in Chemical Rea

18、ctionsIII. Two Categories of Application Framework of the Chapter291. Reversible Phase Transitions -dn+dnWhen the state of a system is located on V-L, V-S, or L-S equilibrium lines of the phase diagram, the condition of phase equilibrium is satisfied. The corresponding phase transitions are reversib

19、le. Besides, they all satisfy the isothermal isobaric condition. ？SLV2.10 Variations of Thermodynamic Functions in Phase Transitions30Equations for Reversible Phase Transitions 1. Reversible Phase Transitions 2.10 Variations of Thermodynamic Functions in Phase Transitions311molH2O(l)1001molH2O(g)100

20、0.1MPapex=0Reversible？1. Reversible Phase Transitions Calculate directlyYesCalculate directlydesign a reversible detourNo2.10 Variations of Thermodynamic Functions in Phase Transitions32For variations of thermodynamic functions, we usually design a detour through a reversible phase transition to cal

21、culate. according to the real processIn design, using pVT relations to reach or to deviate from the phase equilibria 2. Irreversible Phase Transitions For variations of state functionsFor variations of Q and W2.10 Variations of Thermodynamic Functions in Phase Transitions33ExampleA sealed glass bulb

22、 filled with 1.802g of liquid water is put in an evacuated vessel placed in a thermostat at a temperature of 105. After the bulb is crushed, water vapor is formed in the vessel at the same temperature and with a pressure of 0.05MPa. Calculate Q, W, U, H, S, A, G, and judge the reversibility. Given t

23、hat the molar enthalpy of vaporization at normal boiling point is 40.66kJmol1, water vapor can be treated as an ideal gas, the volume of the liquid water can be neglected, =75.31JK1mol1 and =34.31JK1mol1 between 100105. 2.10 Variations of Thermodynamic Functions in Phase Transitions34MethodDesigning

24、 a detour by the following steps through a reversible vaporization at the normal boiling point 100 is a way out. Designing a detour by the following steps through a reversible vaporization at the normal boiling point 100 is a way out. 2.10 Variations of Thermodynamic Functions in Phase Transitions35

25、SolutionThe pressure of the liquid water reduces. The effect of tiny pressure change to the liquid properties can be neglected, H1=0, S1=0(2) The temperature reduces at a constant pressure. The amount of water: n=(1.802/18.02)mol=0.1mol(3) Vaporization at normal boiling point. 2.10 Variations of The

26、rmodynamic Functions in Phase Transitions36SolutionIn total (4) The temperature rises and the pressure reduces to that of the final state.2.10 Variations of Thermodynamic Functions in Phase Transitions37ExampleAt 10 and 101325Pa, 1mol of super-cooling H2O(l) solidifies isothermally to ice. Calculate

27、 Q, W, U, H, S, A, G, and judge the reversibility. Given that the solidification of 1g of water at 0 and 10 releases heat 333.4J and 312.3J, the average specific heat capacities of water and ice are 4.184JK1g1 and 2.067 JK1g1, the average densities of water and ice are 1.000gcm3 and 0.917 gcm3, respectively. 2.10 Variations of Thermodyna