页数：39页文件大小：2.68 MB来源： Atkins’PhysChem物理化学ppt课件
本课件讲第二定律的方法论：由 dU、dH、dA、dG 四个基本方程出发导出麦克斯韦关系式，把不可直接测量的量转化为可测偏导数，并给出吉布斯自由能随温度、压力变化的定量表达。
基本方程
Combining the First and Second Laws 5.1 The fundamental equations 5.2 Properties of the internal energy Properties of the Gibbs energy 5.3 General considerations 5.4 The variation of the Gibbs energy with temp. 5.5 The variation of the Gibbs energy with pressure
Combining the First and Second Laws We have treated the 1 stand the 2
5.1 The fundamental equation From the 1 stLaw: dU = d q + dwFor a reversible change in a close sys
5.2 Properties of the internal energy These eqns are the purely thermodynamic definition of temperature and
5.2 Properties of the internal energy
5.2 Properties of the internal energy
5.2 Properties of the internal energy
5.2 Properties of the internal energy
(a) The Maxwell relations … To further derive from these 4 parameters:
(a) The Maxwell relations
(b) The variation of internal energy with volume Theeqn represents how the internal energy changes as the volume of a
(b) The variation of internal energy with volume
(b) The variation of internal energy with volume For a given state function of a gas, calculating π T
(c) The variation of U、HandS with T
Properties of the Gibbs energy 5.3 General consideration These relations show how the Gibbs energy v
Properties of the Gibbs energy 5.3 General consideration G of the gaseous phase substance is more se
5.4 The variation of the G with temperature In order to see how the equilibrium composition of a system depends o
5.4 The variation of the G with temperature Or we can write: That is the another form of the Gibbs - Helmholtz eq
5.4 The variation of the G with temperature The Gibbs - Helmholtz eqnis most useful when it is applied to changes
5.4 The variation of the G with pressure To find the Gibbs energy at one pressure in terms of value at another
(a) Liquids and solids For a liquid or solid, the volume changes only slightly as the pressu
(b) Gases For gases, the G and volume depends strongly on the pressure. So we c
热力学第二定律：方法论 例题
Ex: at － 5 ℃, the specific freezing enthalpy of over coldwater is Δ sH = － 322.4 J g 1. The saturated vapour pressures for over coldwater and ice are p (H 2 O, l) = 0.422 kPaandp (H 2 O, s) = 0.414 kPa, respectively at － 5 ℃.
The freezing point of water is 0 ℃ at 100 kPa. Calculate Δ G when 1 kg over coldwater freezes at constant press ur 100 kPa and constant temperature － 5 ℃. G H 2 O (l), － 5 ℃, 100 kPaH 2 O (s), － 5 ℃, 100 kPaH 2 O (l), － 5 ℃, 0.422 kPaH 2 O (g), － 5 ℃, 0.422 kPaH 2 O (g), － 5 ℃, 0.414 kPaH 2 O (s), － 5 ℃, 0.414 kPa G 1 G 2 = 0 G 3 G 4 = 0 G 5 To design the reversible path as follows:
G = G 1 + G 2 + G 3 + G 4 + G 5 G 3
(b) Gases ifwesetpi = p (the standard pressure of 1 bar), the the Gibbs energ
(b) Gases At various stages it is necessary to switch from a idealized systorea
(b) Gases The fugacity is defined as a function of pressure and temperature. We
(b) Gases Although expressions in terms offuga cities are exact, they are usefu
(b) Gases
(b) Gases When p’ 0, the gas behaves perfectly, and f’ becomes equal top’. f’
(b) Gases From the Fig. it can be seen that for most gases Z < 1 up to moderate
(b) Gases
The equations and formulas In this chapter, we have learned many equations and formulas. Do you
Chapter 5 Summary for the 2 nd Law 1. 热机效率: 2. 卡诺定理及推论不可逆可逆任何物质, 任何变化3. 熵，熵变的计算：不可逆过程须设计成可逆过程计算。对封闭系统， W’ = 0, 一切可逆过程的熵变皆可由上式导出。
(1) pVT 变化熵变(理想气体) (2) 相变化熵变不可逆相变：设计可逆过程计算(3) 环境的熵变：
Exercises (a) 5.4, 5.7, 5.8 Problems: 4.11, 4.13 Homework
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