首页化学物理化学Atkins’PhysChem物理化学 10 Chapter 6 Physical Transformations of Pure Substances2.pptx

10 Chapter 6 Physical Transformations of Pure Substances2.pptx

页数:61页文件大小:5.67 MB来源: Atkins’PhysChem物理化学ppt课件

本课件先讲界面效应:弯曲液面的附加压力与蒸气压变化、毛细上升与下降;随后转入混合物,引入偏摩尔量与化学势,并列出物质的量浓度、质量分数、摩尔分数等溶液组成表示方法。

纯物质的物理转变图示页,源文件仅提取到章节标题,图示内容需对照原图补充
相与相界页,Phases, Phase boundaries Three typical phase diagrams P
克拉佩龙方程页,Clapeyron equation: d p / d T = Δ trsS / Δ trsV Clausiu
查看本课件的文字版(共 61 页,供检索与复制)

第1页:纯物质的物理转变

第2页:相与相界

Phases, Phase boundaries Three typical phase diagrams P

第3页:克拉佩龙方程

Clapeyron equation
d p / d T = Δ trsS / Δ trsV Clausiu

第4页:亥姆霍兹自由能

Now we link surface effects to the Helmholtz and Gibbs

第5页:表面张力 例题

Ex
using the surface tension Calculating the work need

第6页:弯曲液面

Minimization of the surface are a of a liquid 6.9 Curved surfaces A curved surface Vapour pressure Ca

第7页:弯曲液面

(a) Bubbles, cavities, and droplets Bubble: vapour and

第8页:弯曲液面

(a) Bubbles, cavities, and droplets In bubble or cavity

第9页:弯曲液面

(a) Bubbles, cavities, and droplets Curvature radius 

第10页:弯曲液面

(a) Bubbles, cavities, and droplets Nano bubbles: The B

第11页:弯曲液面

(a) Bubbles, cavities, and droplets These images of nan

第12页:弯曲液面

(a) Bubbles, cavities, and droplets Kelvin equation: Δ

第13页:弯曲液面

(b) Nucleation For a droplets of water at 25 ° Cr = 1 μ

第14页:毛细现象

6.10 Capillary action Capillary action: liquids tend to

第15页:毛细现象

(a) Capillary rise The adhesive force < cohesive force

第16页:毛细现象

(b) The contact angle Between the edge of the meniscus

第17页:毛细现象

(b) The contact angle

第18页:毛细现象

粘附功(work of adhesion) (b) The contact angle

第19页:毛细现象

粘附功(work of adhesion) 在等温等压条件下,单位面积的液面与固体表面粘附时对外所作的最大功称为粘附功,它是液体能否润湿固体的一种量度。粘附功越大,液体越能润湿固体,液- 固结合得越牢。在粘附过程中,消失了单位液体表面和固体表面,产生了单位液- 固界面。粘附功就等于这个过程表面吉布斯能变化值的负值。 (b) The contact angle

第20页:毛细现象

浸湿功(work of immersion) (b) The contact angle

第21页:毛细现象

浸湿功(work of immersion) 等温、等压条件下,将具有单位表面积的固体可逆地浸入液体中所作的最大功称为浸湿功,它是液体在固体表面取代气体能力的一种量度。只有浸湿功大于或等于零,液体才能浸湿固体。在浸湿过程中,消失了单位面积的气、固表面,产生了单位面积的液、固界面,所以浸湿功等于该变化过程表面自由能变化值的负值。 (b) The contact angle

第22页:毛细现象

内聚功(work of cohesion) (b) The contact angle

第23页:毛细现象

内聚功(work of cohesion) 等温、等压条件下,两个单位液面可逆聚合为液柱所作的最大功称为内聚功,是液体本身结合牢固程度的一种量度。内聚时两个单位液面消失,所以,内聚功在数值上等于该变化过程表面自由能变化值的负值。 (b) The contact angle

第24页:毛细现象

第25页:简单混合物

We now go from the single systems to mixtures. Non - reacting substances Energetics

第26页:偏摩尔量

Thermodynamic Description of Mixtures Partial Molar Quantities Thermodynamics of Mi

第27页:偏摩尔量

(a) Partial Molar Volume 17.1 Partial Molar quantities the contribution that one component in a mixture makes to the total volume of a sam

第28页:偏摩尔量

(a) Partial Molar Volume 2 The different increase in total volume in the H 2 O / E tO H depends on the identit

第29页:偏摩尔量

(a) Partial Molar Volume 3 The partial molar volume of components of a mixture vary as the mixture goes from p

第30页:偏摩尔量

(a) Partial Molar Volume 4 The partial molar volume is the slope of a plot of total volume as the amount of J

第31页:偏摩尔量

(a) Partial Molar Volume 5 For a binary mixture, the composition can be changed by addition of dnAofAanddnBofB

第32页:偏摩尔量

(a) Partial Molar Volume 6 For example, if V has the following function V = A + B n A + C (n A 2 – 1) The part

第33页:偏摩尔量

(b) Partial Molar Gibbs Energies The concept of partial molar quantity can be applied to any extensive state functio

第34页:偏摩尔量

(b) Partial Molar Gibbs Energies Gibbs energy depends on composition, pressure and temperature (so, G may change if

第35页:偏摩尔量

Partial Molar quantities

第36页:化学势

(c) More on chemical potential Chemical potential tells us even more than just about variation inGG = U + Vp – ST

第37页:化学势

(c) More on chemical potential Chemical potential tells us even more than just about variation in G

第38页:简单混合物

(d) Gibbs - Duhem Equation Since G = n A μ A + n B μ B, and µ J depend on composition, for a binary mixture G

第39页:简单混合物

(d) Gibbs - Duhem Equation So, for the binary mixture, if one partial molar quantity increases, the other must

第40页:简单混合物

(d) Gibbs - Duhem Equation The experimental values of the partial volume of K 2 SO 4 (a q) at 298 K are given

第41页:(d) Gibbs - Duhem Eq

(d) Gibbs - Duhem Equation dVA / db = 9.108 b - ½ The 1 st step is to change the variable V A to the molalityb

第42页:化学势

7.2 The thermodynamics of mixing Dependence of Gibbs energy on mixture composition is: G = n A μ A + n B μ BAt const

第43页:化学势

(a) The Gibbs energy of mixing The chemical potentials are at their “ pure values ” at this point. Gibbs energy is

第44页:化学势

(a) The Gibbs energy of mixing After mixing, the partial pressures of the gases are pAandpB, where the total press

第45页:化学势

Calculating Gibbs Energy of Mixing Two containers of equal volume are partitioned from one another, with one containin

第46页:化学势

Calculating Gibbs Energy of Mixing

第47页:化学势

(b) Other Thermodynamic Mixing Functions Since (∂ G / ∂ T) p, n = - S and It follows that for a mixture of perfect gases, th

第48页:化学势

(b) Other Thermodynamic Mixing Functions The isothermal isobaric enthalpy of mixing of two gases is found from Δ G = Δ H - T

第49页:液体的化学势

7.3 Chemical Potentials of Liquids (a) Ideal solutions Relating to pure substances will be denoted with “ * ”. The che

第50页:液体的化学势

7.3 Chemical Potentials of Liquids (a) Ideal solutions If another substance is present (e. g., a solute in the liquid)

第51页:液体的化学势

7.3 Chemical Potentials of Liquids Raoult’ s Law 拉乌尔定律After experimenting with mixtures of similar liquids, the French

第52页:液体的化学势

7.3 Chemical Potentials of Liquids Raoult’ s Law 拉乌尔定律Some mixtures obey Raoult’ s law very closely, especially when t

第53页:液体的化学势

7.3 Chemical Potentials of Liquids Molecular Interpretation of Raoult’ s Law To consider the rate at which molecules l

第54页:液体的化学势

7.3 Chemical Potentials of Liquids Molecular Interpretation of Raoult’ s Law The rate at which molecules return is pro

第55页:液体的化学势

7.3 Chemical Potentials of Liquids Departure from Raoult’ s Law, Dilute Solutions Some solution mixtures be have very

第56页:液体的化学势

7.3 Chemical Potentials of Liquids Departure from Raoult’ s Law, Dilute Solutions Ideal solution, consider the solvent

第57页:液体的化学势

7.3 Chemical Potentials of Liquids (b) Ideal - dilute solutions: Henry’ s law The English chemist William Henry found

第58页:液体的化学势

7.3 Chemical Potentials of Liquids (b) Ideal - dilute solutions: Henry’ s law The ideal - dilute solutions p B = xBKB

第59页:化学势

7.3 Chemical Potentials of Liquids The Validity of Raoult’ s & Henry’ s Laws

第60页:液体的化学势

Using Henry’ s Law Estimate the molar solubility of oxygen in water at 25 o C and partial pressure of

第61页:简单混合物 习题

Exercises (a) 7.5, 7,9, 15 Problems: 7.8 Homework

顶部目录