复旦大学物理化学AII 15-1 The phase rule and its deduction 2015
第四章,液体混合物与溶液
得
若溶质不挥发
§4.6
稀溶液的依数性
稀溶液的依数性: 稀溶液具有的某些性质仅与一定量溶液中 溶质的质点数有关而与溶质的本性无关.
1.溶剂蒸气压下降
对二组分稀溶液:
p p* x p* ( 1 x ) p* p* x
A A A A B A
A B
p A p* p A p* x B A A
2.单选题: (1)1molA与 nmol B组成的溶液,体积为0.65 dm3 ,当xB = 0.8 时,A的偏摩尔体积VA=0.090dm3· -1,那么B的偏摩尔VB 为: mol (A) 0.140 dm3· -1 ; mol (B) 0.072 dm3· -1 ; mol (C) 0.028 dm3· -1 ; mol (D) 0.010 dm3· -1 。 mol
X X X dX dT dp dnB B nB T , p , n T p ,nB ,nC p T ,nB ,nC C
X nB T , p , n C
XB
def
偏摩尔量
偏摩尔量XB是在T, p 以及除B外所有其他组分的物质的量 都保持不变的条件下, 任意广度性质X 随nB的变化率.
X X nB dX B dT dp T p ,nB p T , n B B
物理化学(下)答案(完整版)
第七章 电化学
7.1 用铂电极电解CuCl 2溶液。通过的电流为20A ,经过15min 后,问:(1)在阴极上能析出多少质量的Cu?(2)在的27℃,100kPa 下阳极上能析出多少体积的的Cl 2(g )?
解:电极反应为:阴极:Cu 2+ + 2e - → Cu 阳极: 2Cl - -2e - → Cl 2(g ) 则:z= 2 根据:Q = nzF =It
()22015
Cu 9.32610mol 296500
It n zF -⨯=
==⨯⨯ 因此:m (Cu )=n (Cu )× M (Cu )= 9.326×10-2×63.546 =5.927g 又因为:n (Cu )= n (Cl 2) pV (Cl 2)= n (Cl 2)RT
因此:3
223
Cl 0.093268.314300Cl 2.326dm 10010
n RT V
p ⨯⨯===⨯()() 7.2 用Pb (s )电极电解PbNO 3溶液。已知溶液浓度为1g 水中含有PbNO 3 1.66×10-2g 。通电一定时间后,测得与电解池串联的银库仑计中有0.1658g 的银沉积。阳极区的溶液质量为62.50g ,其中含有PbNO 31.151g ,计算Pb 2+的迁移数。
解法1:解该类问题主要依据电极区的物料守恒(溶液是电中性的)。显然阳极区溶液中Pb 2+的总量的改变如下: n 电解后(12Pb 2+)= n 电解前(12Pb 2+)+ n 电解(12Pb 2+)- n 迁移(1
2Pb 2+)
则:n 迁移(12Pb 2+)= n 电解前(12Pb 2+)+ n 电解(12Pb 2+)- n 电解后(1
物理化学英文6相平衡 Phase Equilibrium
dp/dT = DSfus/ DVfus
Clausius-Clapeyron Equation
dp DS DH for phase equilibrium dT DV TDV DV RT / p if one phase is gaseous
the only pressure (for a given temperature) at which liquid and vapor coexist.
p vapor liquid
Phase Equilibrium
m
solid
G T
p
S
liquid
Sg Sl Ss
gas
Tf
Tb Temperature
/
p0 )
m
A (l)
m
0 A
RT ln( p*A
/
p0 )
RT
ln xA
m
A
(l)
m
* A
RT
ln
xA
where mA* is the chemical potential of the pure liquid
Ideal Heat of Mixing
39届国际物理竞赛试题IPhO(英文)理论第一题
39届国际物理竞赛试题IPhO(英⽂)理论第⼀题
WATER-POWERED RICE-POUNDING MORTAR
A. Introduction
Rice is the main staple food of most people in Vietnam. To make white rice from paddy rice, one needs separate of the husk (a process called "hulling") and separate the bran layer ("milling"). The hilly parts of northern Vietnam are abundant with water streams, and people living there use water-powered rice-pounding mortar for bran layer separation. Figure 1 shows one of such mortars., Figure 2 shows how it works.
B. Design and operation
1. Design .
The rice-pounding mortar shown in Figure 1 has the following parts: The mortar , basically a wooden container for rice.
The lever , which is a tree trunk with one larger end and one smaller end. It can rotate around a horizontal axis. A pestle is attached perpendicularly to the lever at the smaller end. The length of the pestle is such that it touches the rice in the mortar when the lever lies horizontally. The larger end of the lever is carved hollow to form a bucket. The shape of the bucket is crucial for the mortar's operation.
硫化物电解质li6ps5cl锂离子传输机制
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浙江大学物理化学(甲)第五章(相平衡)
2. 几点说明: (1)系统中若存在化学反应:
对于每一个独立的化学 反应,应该满足: BB 0 在化学平衡中将会讨论
B
或者理解为每一个化学 平衡都有一个平衡 常数来联系物质的浓度 关系
如果有R个相互独立的平衡关系式,则自由度要减去R 注意:系统中的化学反应并不全是独立的
1 2 3
相律(phase rule)是多相平衡系统研究的热力学基础,它 讨论相平衡系统中相数、独立组分数与描述该平衡系统的 独立变量数之间的关系。它是由Gibbs提出的,是物理化 学中最具有普遍性的规律之一。
相(phase)是指系统内部物理性质和化学性质完全均匀的 一部分。相与相之间在指定的条件下有明显的界面,在界 面上,从宏观性质的角度来看,性质的改变是飞跃式的。
f S F 2 如果令
考虑到(1) (2)两种 情况
f F S R R 2
C S R R C称为独立组分数 number of independen component t
def
则相律可以表示为: C F 2 f
G1 ( , T , p) G2 ( , T , p) G1 dG1 G2 dG2
dG1 dG2
由热力学基本方程式: dG SdT Vdp 代入上式并整理: V2 V1dp S2 S1dT
9701_s15_ms_41 a-level化学
1 1 1 1 1
(c)
(d)
1 1
1
(e) (i) (ii)
F = Le moles of O2(g) = 130 / 24 000 = 5.417 x 10–3 mol moles of electrons needed = 4 x 5.417 x 10–3 or 2.17 x 10–2 mol no. of coulombs passed = 1.2 x 30 x 60 or 2160 C no. of electrons passed = 2160 / 1.6 x 10–19 or 1.35 x 1022 no. of electrons per mole = 1.35 x 1022 / 2.17 x 10–2 = 6.2 x 1023 (mol–1)
Eocell (= 0.80 – 0.77 =) (+)0.03V and Ag+ / Ag or Ag / silver or right Ecell would be less positive / more negative because the [Ag+(aq)] (in the Ag electrode) is less than 1.0 mol dm–3
Syllabus 9701
Paper 41 1 1 1 1
6
(a) (i) (ii) (iii) (iv)
应用化学专业本科课程设置 - 江南大学图书
新增专业设置情况
调研(三)
应用化学专业
江南大学图书馆
2006年12月
本次调研以教育部直属211高校中的44所大学(下附详细名单)的网站内容为基础,对应用化学专业的课程设置、教学计划等情况进行了调查与总结,由于各学校网站设立的内容各不相同,有的详尽,有的粗略,所以专业课程设置的调研主要围绕几个设立比较详细的院校进行的。
44所教育部直属211高校名单
上述44所高校中共有以下26所(包括江南大学)开设了应用化学专业,此次调研根据应用化学专业所在各个高校的院系、专业方向、主要课程进行了以下总结(见下表),并列出相关大学教学计划。由于各学校院系提供的信息详略有别,所以下表中有的学校列出了详尽的介绍;有的学校没有提供某项目的情况介绍,则下表中用“未列”描述了此种状况。
应用化学专业——高校情况调研表
在设立了应用化学专业的26所高校中,以下4所高校列出了详细的本科教学计划及课程设置的相关信息,下表为这4所院校开设的相关课程。
实例1:
北京大学化学与分子工程学院——应用化学系
本科生主要课程:
说明:总学分:140
全校公共必修课30学分,专业必修60学分,合计90学分,占总学分64.3%;
通选课16学分,专业选修28学分(从以上所开选修课中选),选修课合计44学分,占总学分31.4%;毕业论文6学分,占总学分4.3%;
应用化学专业选课指导方案:应用化学基础,应用化学专题,生命化学基础,综合化学实验,物理化学Ⅱ.
实例2:
复旦大学——化学系
2006(春季)课程表
2006(春季)课程表
化学系化学、应用化学专业2004级填表日期2005 年11 月
PH01-QP-物理
*jAn21PH0101*
IB/M/Jan21/E7
PH01
For Examiner’s Use Question
Mark 1 2 3 4 5 6 7 8 9 10
11 12–25
TOTAL
Tuesday 5 January 2021 07:00 GMT
Time allowed: 2 hours
Materials
For this paper you must have:
• a Data and Formulae Booklet as a loose insert • a ruler with millimetre measurements
• a scientific calculator, which you are expected to use where appropriate • a protractor.
Instructions
• Use black ink or black ball-point pen. • Fill in the boxes at the top of this page. • Answer all questions.
• You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages. • All working must be shown.
• If you need extra space for your answer(s), use the lined pages at the end of this book. Write the question number against your answer(s).
湖北省咸宁市中考英语试题(word版,含答案及听力材料)
湖北省咸宁市2021年初中毕业生学业考试
英语试卷
考生注意:
本试卷分试题卷〔共8页〕和答题卷〔共2页〕;全卷共78个小题〔含2个选做题〕;总分值120分,考试时间:120分钟。
考生答题前,请将自己所在学校、姓名、准考证号填写在试题卷和答题卷指定的位置,并认真阅读答题卷上的考前须知。
考生答题时,请按照题号顺序在答题卷上各题的答题区域内作答,写在试题卷上无效。
第一局部听力
〔共40分〕
做题时,先将答案标在试卷上。录音内容结束后
,将试卷上的答案转涂到答题卷上。
一、情景交际听下面五个提
示句,从每题A、B、C三个选项中,选出能回应提示句的
最正确选项。听每个提示句前,你将有时间阅读各个
小题,每题5秒钟:听完后,各
个小题将给出5秒钟的作答时间。每个提示句读两
遍。〔每题分,共分〕
1.A.Yes,I’dloveto.B.Youarewelcome
. C.Nevermind.
2.A.Eatdumping
s. B.Eatmooncakcs.hamburgers.
3.A.l’mfine.B.Iagreewithyou
. C.l’mgoodatit.
4.A.AIIrig
ht. B.Notatall. C.Goodidea.
5.A.Youarekidding. B.Ofcoursenot. C.Thankyou.
二、听对话选图听下面五段对话,每段对话后有一个
问题,从下面所给的A、B、C、D、
E、F六个图片选项中,选出相应问题的最正确选项〔其中有1个为干扰选项〕。[听完每段对话
后.你将有5秒钟的作答时间。每段对话读两
ja035716jsi20030419_025246[1]
Supporting Information for
Catalytic Construction and Reconstruction of Guanidines:Ti-Mediated Guanylation of
Amines and Transamination of Guanidines
Tiow-Gan Ong,Glenn P.A.Yap,Darrin S.Richeson*
Experimental details4pages
Crystallographic details for compounds
9pages
[(Me2N)C(i PrN)2]2Ti=N(C6F5)(1b)
sample number=dr2079
8pages
(o-C6H4OMe)NC(N i Pr)2(7)
sample number=dr2112
16pages
[(Me2N)C(i PrN)2]2Ti[(C6F5)NC(N i Pr)2](2)
sample number=dr2081
General Considerations.All manipulations were carried out in either a nitrogen filled drybox or under nitrogen using Schlenk-line techniques.Toluene,hexane and ether were purified by passage through a column of activated alumina using an apparatus purchased from Anhydrous Engineering.Pentane and THF were distilled under nitrogen from Na/K alloy.Deuterated benzene and toluene were dried by vacuum transfer from Na/K alloy.Diisopropylcarbodiimide,pentafluoroaniline,2,6-dimethylaniline,p-anisidine,o-anisidine,p-aminobenzylcyanide p-chloroaniline,1,4-diaminobenzene BuLi(1.6M in hexane)were purchased from Alrich Chemical Company and used without further purification.1H and13C NMR spectra were run on a Varian Gemini-200,a Bruker300MHz or a Bruker500MHz spectrometer using the residual protons of the deuterated solvent for pounds1a and N,N',N"-triisopropylguanidine were as described in the text.The compound[(Me2N)C(i PrN)2]2TiCl2was prepared as described in Mullins,S.M.; Duncan,A.P.;Bergman,R.G.;Arnold,J.Inorg.Chem.2001,40,6952.
Physical chemistry物理化学专业英语
• Isenthalpic process (等焓~ ) • Isentropic process (等熵~ ) • Entropy (熵 ) • Adiabatic process (绝热~ ) • Cyclic process (循环~ ) • Quasistatic process (准稳态~ ) • Reversible process (可逆~ ) • Irreversible procesbs (不可逆~ ) • Spontaneous process (自发~ ) • Nonspotaneous process (不自发~ )
•Heat of fusion (熔化热) •Heat of vaporization (汔化热) •Heat of dilution (稀释热) •Heat capacity C=Q/(T2-T1) (热容) •Heat capacity at constant pressure (Cp) (等压热容)
• System (系统) • Equilibrium system (平衡~) • Nonequilibrium system (非平衡~) • Homogeneous system (均相~) • Heterogeneous system (非均相~) • Open system (敞开~) • Closed system (封闭~) • Isolated system (弧立~) • Adiabatic system (绝热~) • One-compound system (单组分~)
陕西师范大学物理化学专业硕士研究生 《量子化学与群论》课程作业解析答案
由球谐函数的加法定理
得证
5.6光谱支项
氢原子光谱项 的能级为:
第六章双原子分子
6.1(a)Li2的基态组态:
Li2+的基态组态: 所以
(b)C2的基态组态:
C2+的基态组态: 所以
(c)O2的基态组态:
O2+的基态组态: 所以
(d)F2的基态组态:
F2+的基态组态: 所以
陕西师范大学物理化学专业硕士研究生
《量子化学与群论》课程作业解析答案
第一、二章量子力学基础
1.1一维谐振子的基态波函数为:
故基态时
1.2证明:v=1时谐振子波函数为:
⑴
V=2时谐振子归一化波函数为:
⑵
由⑴,⑵式知, 时的归一化因子是正确的。
1.3
是x的偶函数,x又是奇函数
是奇函数
1.4(1)①,②,③,⑥是线性算符。
⑴
根据不同的m的波函数与 所组成的矩阵元为零和宇称性可将⑴式简化为: ⑵
即为 ⑶
第五章多电子原子
5.1
5.2交换坐标和函数名称知:
属于完全对称的:
属于反对称的:
属于非对称的:
5.3
元素
光谱支项
元素
光谱支项
元素
光谱支项
元素
光谱支项
H
复旦大学大学物理力学课件Ch1-国际单位制
Avogadro计划: 澳大利亚科学家手持一个1千克的Si单 晶球,它是世界上人造的最圆的物体。
典型的质量
已知宇宙
银河系
1053 kg
2.2×1041 kg
太阳
地球 人 灰尘 烟草花叶病毒 质子 电子
2.0×1030 kg
6.0×1024 kg 6.0×101 kg 6.7×10-10 kg 2.3×10-13 kg 1.7×10-27 kg 9.1×10-31 kg
“千克”的重新定义
一个长远的解决办法是给千克以重新定义,使得不 同的实验室都可以重现千克标准。
“Avogadro计划”
德、俄、澳大利亚等8国的科学家联合研究一个新的 标准质量。历时5年,耗资200万欧元,制成一个“完美 硅球”,它含有纯度为 99.99% 的硅 -28 原子,直径为 9.375cm。 科学家们用制造核武器的离心机来提取硅 -28,使其 纯度达到 99.99%,然后将它运到德国的国家计量研究 所制作硅晶体。 随后,他们在澳大利亚英联邦科学和工业研究组织 的实验室中将其打造成为精密的圆球。
量纲分析 表示每个物理量怎样由基本量组成的式子,称为量 纲(或量纲式)。 用途: ①用于单位换算
①用于检验公式的正误 量纲分析法是处理物理学问题的重要方法之一。
1 2 s at 2
Name meter
kilogram second ampere
50个大学校训翻译
50个大学校训翻译
1、北京大学(创建于1898年):爱国进步民主科学
Peking University(founded in1898): Patriotism,Advancement,Democracy and Science
2、清华大学(始建于1911年)自强不息德载物:Tsinghua University(founded as early as1911):Self-discipline and Social Com mitment
3、中国人民大学(Renmin University of China):实事求是
Seek Truth From Facts.
4、复旦大学(Fudan University):博学而笃志,切问而近思
Rich in Know-ledge and Tenacious of Purpose;Inquiring with Earnestness and Reflecting with Self-practice.
5、浙江大学(Zhejiang University):求
实创新:
Seek Truth and Be Creative.
6、同济大学(Tongji University):严谨求实,团结创新:
Discipline、Practicality、Unity and Creativity.
7、重庆大学(Chongqing University):耐劳苦尚俭朴勤学业爱国家:Endurance、Thrifty、Diligence、Patriotism.
8、南京大学(Nanjing University):诚朴雄伟,励学敦行:
ipho第53届题目
ipho第53届题目
The problem that Ipho第53届题目 presents is a challenging one that requires a deep understanding of physics and mathematical concepts. The problem likely involves a complex scenario or experiment that requires students to use their knowledge of physics to analyze and solve. This type of problem is common in physics competitions and is designed to test students' ability to think critically and apply their knowledge to real-world situations.
To begin with, the problem may require students to apply their understanding of fundamental physics principles such as Newton's laws of motion, conservation of energy, and the properties of waves and particles. This could involve analyzing the motion of objects in different scenarios, calculating forces and accelerations, or understanding the behavior of light and sound. The problem may also involve mathematical equations and calculations, requiring students to use algebra, calculus, and other
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Gas, no matter how many kinds of them are mixed together, there
exists only one gas phase Liquid, depending on its miscibility, one and coexistence of two or three phases are possible Solid, one solid generally constitutes one phase by itself. No matter how well-mixed are two different solid powders, there still exist two phases (Except for solid solution, which is a single phase)
9
Thermodynamics of phase equilibria
The phase rule and its deduction
Phase, number of components, degrees of freedom
degree of freedom(自由度)
Phase rule and its deduction Phase diagrams for one-component systems Second-order phase transitions Two-component phase diagrams Three-component phase diagrams
B B
B
(4) Conditions for chemical equilibrium: associated reactions have reached chemical equilibrium
B B 0 B
Physical Chemistry II
2015/6/15
2015/6/15
Physical Chemistry II
3
Thermodynamics of phase equilibria
The phase rule and its deduction
§1 Phase rule and its deduction
Phase, number of components, degrees of freedom Deduction of the phase rule Phase diagram
chemical composition and physical state. Two immiscible liquids (or liquid/solid mixtures with different compositions) separated by a distinct boundary or interfaces. The change of macroscopic properties on the interface is very huge. The number of phases in thermodynamic equilibrium with each other is denoted as
2015/6/15
Physical Chemistry II
5
Thermodynamics of phase equilibria
The phase rule and its deduction
Some basic conceptual issues need to take care when discussing ―phase‖ :
Thermodynamics of phase equilibria
The phase rule and its deduction
Physical Chemistry
The phase rule and its deduction
2015/6/15
Physical Chemistry II
1
Thermodynamics of phase equilibria
a diagram which describes how the state change of a multiphase system in response to the changes of intensive properties like temperature, pressure and constitution
2015/6/15
Physical Chemistry II
7
Thermodynamics of phase equilibria
The phase rule and its deduction
General requirements for equilibrium in heterogeneous systems In a closed heterogeneous system, heat/ material exchange and work delivery could occur between different phases. For the thermodynamic equilibrium between multiple phases, it actually contains the following four equilibrium conditions: (1) Conditions for thermal equilibrium: consider the system consisting of ,,, phases, when reaching equilibrium, each phase has the same temperature
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Thermodynamics of phase equilibria
The phase rule and its deduction
(3) Conditions for phase equilibrium: Each substance B in all phases has the same chemical potential, equilibrium phase transition is reached
T = T = … =T
(2) Conditions for baric equilibrium: when reaching equilibrium, each phase is at the same pressure
p = p = … =p
2015/6/15
Physical Chemistry II
The phase rule and its deduction
Phase equilibrium is one of the most important parts of
thermodynamics in chemistry and chemical engineering. Many processes, for example, dissolution, distillation, extraction, recrystallization, purification and polymorph analysis, needs the knowledge of phase equilibria in heterogeneous systems.
2015/6/15
Physical Chemistry II
4
Thermodynamics of phase equilibria
The phase rule and its deduction
Phase, number of components, degrees of freedom
Phase A phase is a form of matter that is homogeneous in
Extraction
2015/6/15
Recrystallization
Physical Chemistry II
2
Thermodynamics of phase equilibria
The phase rule and its deduction
Chapter 13 Thermodynamics of phase equilibria
2015/6/15
Physical Chemistry II
6
Thermodynamics of phase equilibria
The phase rule and its deduction
III) In general, mixtures of gases only form a single phase. Since
all different types of gases can be molecularly mixed in any ratio, no interface exists between them.
Liquid, depending on its miscibBaidu Nhomakorabeality, there can be one or two phases. Two totally miscible liquids form a single phase, while partly miscible ones form two phases. For example: the benzene-water system consists of two phases, whereas the ethanol-water system consists of only one phase.
I) Phase is homogeneous, but not necessarily continuous. A single phase could be separated into many particles or liquid drops, or dispersed in other phases. Therefore, a clear interface must exist between two different phases, while the existence of an interface does not necessarily mean that there are different phases. II) Being homogeneous but with different properties means that they are not the same phase. For example: iron and sulfur powers can be grinded into very fine and homogeneous state seen with the naked eye, but microscopic observation still reveal two phases. Since no matter how finely grinded, molecularly mixed state will never be reached, i.e., molecular level dispersion not possible, there still exist two phases.
Normally, unless they form solid solution, there exist many distinct solid phases. For example: alloy consists of one phase, as alloy is a solid solution formed by molecularly mixed two types of metals.