Chapter 12 The Kinetic Theory of Gases
Classical thermodynamics has nothing to say about atoms or molecules. Its laws are concerned only with such macroscopic variables as pressure, volume, and temperature. However, we know that gas is made up of atoms or molecules (groups of atoms bound together). The pressure exerted by a gas must surely be related to steady drumbeat of its molecules on the walls of its container. The ability of a gas to take on the volume of its container must surely be due to the freedom of motion of its molecules. And the temperature and internal energy of a gas must surely be related to the kinetic energy of these molecules. Perhaps we can learn something about gases by approaching the subject from this direction. We call this molecular approach the kinetic theory of gases.
12.1 Ideal Gases
1、Our goal in this chapter is to explain the macroscopic properties of a gas, such as its pressure and its temperature, in terms of the behavior of the molecules that make it up.
2、The experiments show that, at low enough densities, all real gases tend to obey the relation )
pV , in
nRT
gas
(law
ideal
which p is the absolute pressure, A N N n /= is the number of
moles of gas present, and
R , the gas constant , has the same value for all gases, namely, K mol J R ?=/31.8. The temperature T must be expressed in kelvins. Above equation is called the ideal gas law . Provided the gas density is reasonably low, it holds for any type of gas, or a mixture of different types, with n being the total number of moles present.
3、 Work done by an ideal gas at constant temperature :
(1). Suppose that a sample of n moles of an ideal gas, confined to a piston-cylinder arrangement, is allowed to expand from an initial volume i V to a final volume f V .
(2). Suppose further that the temperature T of the gas is held constant throughout the process. Such a process is called an isothermal expansion (and the reverse is called an isothermal compression ).
4、Let us calculate the work done by an ideal gas during an isothermal expansion, we have
??==f i f i V V V V dV V nRT pdV W i f
V V nRT ln =. Recall that the symbol ln specifies a natural
logarithm, that is, a logarithm to base e .
12.2 Pressure, Temperature, RMS Speed, and Translational Kinetic Energy
1、 Let n moles of an ideal gas
being confined in a cubical box of
volume V , as in the figure.
2、 A typical gas molecule, of mass m and velocity v, is about to collide with the shaded wall. We assume that any collision of a molecule with a wall is elastic , so the change in the particle’s momentum is along the x axis and its magnitude is x x x x mv mv mv p 2)()(-=--=?. Hence the momentum x p ? delivered to the wall by the molecule during the collision is x mv 2+.
3、 The average rate at which momentum is delivered to the shaded wall by this single molecule is L
mv v L mv t p x x x x 2/22==??. 4、 F rom Newton’s second law, the rate at which momentum is delivered to the wall is the force acting on that wall. To find the total force, we must add up the contributions of all molecules that strike the wall, allowing for the possibility that they all have different speeds. Dividing the total force x F by the area of the wall then gives the pressure
p on that wall. Thus
))((///2222132222212xN x x xN x x x v v v L
m L L mv L mv L mv L F p +++=+++== , where N is the number of molecules in the box.
5、 Since A nN N = in which A N is Avogadro’s number ,
above equation can be re-expressed as
223x x A v V nM v L nmN p ==, where A mN M = is the molar mass of the gas , and V is the
volume of the box.
6、 For any molecules, 2222z y x v v v v ++=. Because there are
many molecules and because they are all moving in random direction, the average values of the squares of their velocity components are equal, so that
3/22v v x =. Thus V v nM p 32=. 7、 The square root of 2v is a kind of average speed, called the root-mean-square speed of the molecules and symbolized by rms v .
8、 Using the relation
nRT pV = for an ideal gas, it leads to M RT nM pV v v rms 332===.
See the RMS speeds of
some molecules in Table.
9、 The average
translational kinetic energy of a single molecule of an ideal gas is
kT T N R M RT m mv mv K A rms 23)(23)3)(21(212122=====
, where the constant
K J N R k A /1038.1/23-?== is the Boltzmann constant . So we come to the conclusion that at a given temperature T , all ideal gas molecules, no matter what their mass, have the same average translational kinetic energy . When we measure the temperature of a gas, we are also measuring the average translational kinetic energy of its molecules.
12.3 Mean Free Path
1、 Table gives us the RMS speed of some molecules, A question often arises: If molecules move so fast, why does it take as long as a minute or so before you can smell perfume if someone open a bottle across a room? This is because although the molecules move very fast between collisions, a given molecule will wander only very slowly away from its release point.
2、 The right figure shows the path of a
typical molecule as it moves through the gas,
changing both speed and direction abruptly
as it collide elastically with other molecules.
Between collisions, our typical molecule
moves in a straight line at constant speed.
Although the figure shows all the other molecules as stationary,
they too are moving in much the same way.
3、 One useful parameter to describe this random motion is the mean free path λ. As its name implies, λ is the average distance traversed by a molecule between collisions.
4、 The expression for the mean free path can be deduced from following steps:
(1). First we assume that our molecule is traveling with a constant speed u and that all other molecules are at rest. We assume further that the molecules are spheres of diameter d. A collision will then take place if the centers of the molecules come within a distance d of each other. A help way to look at this situation is to consider our single molecule to have radius of d and all the other molecules to be points.
(2). As our single molecule zigzags through the gas, it sweeps out a short cylinder of cross-sectional area 2d π between successive collision. If we watch this molecule for a time interval t ?, it moves a distance t u ?. So the volume of the cylinder is t u d ?2π. The number of collisions that occur is then equal to the number of (point) molecules that lie within this cylinder. It is t u d V N ?2)/(π.
(3). The mean free path is the length of the path divided by the number
of collisions
)/()/(22V N u d v V N t u d t v collisions of numbers path of length ππλ=??==.
(4). The u in the denominator is the mean speed of our single molecule relative to the other molecules, which are moving. A detailed calculation, taking into account the actual speed distribution of the molecules, gives that
v u 2=. So )/(21
2V N d πλ=.
5、The mean free path of air molecules at sea level is about m μ1.0. At an altitude of km 100, the mean free path is about 16cm. At 300 km, the mean free path is about 20 km.
12.4 The Distribution of Molecular Speeds
1、 In 1852, Scottish physics James Clerk Maxwell first solved the problem of finding the speed distribution of gas molecules. His result, known as Maxwell’s speed distribution law , is
RT v M e v RT M v P 2/22/32)2(4)(-=ππ. Here v is the molecular speed, T
is the gas temperature, M is the molar mass of the gas, and R is the gas constant. The quantity )(v P is a probability
distribution function , defined as follow: The product dv v P )( is
the fraction of molecules whose speeds lie in the range v to dv v +.
2、 As figure (a) shows, this
fraction is equal to the area of a
strip whose height is
)(v P and whose width is dv . The total
area under the distribution curve
corresponds to the fraction of
the molecules whose speed lie
between zero and infinity. All
molecules fall into this category,
so the value of this total area is
unit.
3、 There are two other
speeds. The most probable speed P v is the speed at which )(v P is a maximum. The average speed v is a simple average of the molecular speeds. We will find that
M RT v π8= and M RT v P 2=，so we have relations
rms P v v v <<.
12.5 Degree of Freedom and Molar Specific Heats
1、 The right figure shows
kinetic theory models of
helium (a monatomic gas),
oxygen (diatomic), and methane (polyatomic). On the basis of their structure, it seems reasonable to assume that monatomic molecules, which are essentially point-like and have only a very small rotational inertia about any axis, can store energy only in their translational motion. Diatomic and polyatomic molecules, however, should be able to store substantial additional amount of energy by rotating or oscillating.
2、To take these possibilities into account quantitatively, we use the theorem of the equipartition of energy, introduced by James Clerk Maxwell: Every kind of molecules has a certain number f of degree of freedom, which are independent ways in which the molecule can store energy. Each such degree of freedom has associated with it, on average, an energy of 2/
kT per molecule (or 2/
RT per mole).
3、For translational motion, there are three degrees of freedom. For rotational motion, a monatomic molecule has no degree of freedom. A diatomic molecule has two rotational degrees of freedom. A molecule with more than two atoms has six degrees of freedom, three rotational and three translational.
4、Internal energy is the energy associated with random motion of atoms and molecules. So a sample of n moles of a gas contains
nN atoms. The internal energy of the sample is then A
nRT f kT f nN E A 2)2)((int ==. Thus, the internal energy int E of an
ideal gas is a function of the gas temperature only; it does not depend on any other variable .
5、 Molar specific heat at constant volume : (1) According to the definition of molar specific heat, we have T nC Q v ?=. (2) Substituting it into the first law of thermodynamics, we find T nC T nC W Q E V v ?=-?=-=?0int . (3) So we will have the relation R f T n RT f n T n E C v 2
)2(int =??=??=. 6、 Molar specific heat at constant pressure : (1) the heat is T nC Q p ?=. (2) According to the first law of thermodynamics, we
have T nC T nR T nC V p T nC W Q E V p p ?=?-?=?-?=-=?int .
(3) So we have relation
R C C V p +=.
12.6 The Adiabatic Expansion of an Ideal Gas
1、 Using the first law of thermodynamics, we have pdV Q dE -=int pdV -=. It means pdV dT nC V -=.
2、 From the ideal gas law (
)
nRT pV = we have nRdT Vdp pdV =+ 3、 Combining these two equations, we have 0)(=+V dV C C p dp V p . Replacing the ratio of the molar specific heats with γ and integrating yield
t cons a pV tan =γ.
4、 Using the ideal gas law, we also have t cons a TV tan 1=-γ, and t cons a T p tan 1=--γγ.
5、 Free expansion: Since a free expansion of a gas is an adiabatic process that involves no work done on or by the gas, and no change in the internal energy of the gas. A free expansion is thus quite different from the type of adiabatic process described above, in which work is done and the internal energy changes. Since the int ernal energy isn’t change for free expansion, we must have
f i RT f n RT f n 22=. It means f i T T =, and
so f f i i V p V p =.
新编语言学教程Chapter 12答案 Applied Linguistics 1. Define the following terms briefly. (1) applied linguistics: the study of language and linguistics in relation to practical issues, e.g. speech therapy, language teaching, testing, and translation. More often than not nowadays, it is used in the narrow sense, and refers to language teaching in particular. (2) grammar-translation method: a method of foreign or second language teaching which makes use of translation and grammar study as the main teaching and learning activities. (3) audiolingual method: the teaching of a second language through imitation, repetition, and reinforcement. It emphasizes the teaching of speaking and listening before reading and writing and the use of mother tongue in the classroom is not allowed. (4) communicative language teaching: an approach to foreign or second language teaching which emphasizes that the goal of language learning is to achieve communicative competence.
第十二章 奶油和涂布乳制品 国际乳品协会（IDF）介绍了一种包括奶油和涂沫制品的标准，即IDF标准166： 1993，“涂布脂肪指南”，这些指南的意图是用来提供一个主要框架，在该框 架下不同国家可以根据自身需要设定更加明确详细的一组或单一的标准。
定义 涂布脂肪：“涂布脂肪”是一种以乳浊液形式存在的食物，主要是油包水型，并且主要由水相、食用脂肪和油组成。 食用脂肪和油主要由脂肪酸的甘油三酸脂组成，它们来源于蔬菜、动物、乳或海产品。 下列表(12.1和12.2)选自于该标准。 表12.1 乳脂肪和人造奶油制品的主要组成 乳脂产品 混合脂肪产品 人造奶油产品 总脂肪中含乳 总脂肪中乳脂含量 总脂肪中乳脂含量 脂100% 最高为80%，最低为15% 最高为3% 注意，按照一些国家的或其它的有关法规：脂肪含量和乳脂与其它种类脂肪的比例会有一些更严格的限制性范围。 主要的原料应该是水和/或乳制品，食用脂肪和/或油或者是它们的混合物，关于脂肪含量及涂布脂肪的标准可依据脂肪的来源不同分为三类，最高脂肪含量应达到95%。 食物的名称应符合国家法规的规定，但产品需遵守表12.2所列的一般要求，该表将所有产品全部覆盖在以下三大类中： 表12.2 乳脂肪和人造奶油制品的名称 脂肪含量% 乳脂产品 混合脂肪产品 人造奶油产品 80-95 奶油 混合物 人造奶油 762-<80 涂布乳品 涂布混合物 涂布脂肪 60-62 3/4脂肪或 3/4脂肪或 3/4稀释脂肪 稀释脂肪 稀释脂肪 稀释脂肪 奶油 混合物 人造奶油 <41-<60 减量脂肪 减量脂肪 减量脂肪涂布 涂布乳品 涂布混合物 涂布 39-41 1/2或低脂奶油 1/2或低脂混合物 1/2或低脂人造 奶油或米纳林 <39 涂布低脂肪乳品 涂布低脂肪混合物 涂布低脂肪 下列联合国粮农组织/世界卫生组织(FAO/WHO)的标准普遍适用于在国际贸易中并指明所允许使用的产品的名称： A1－奶油和乳清奶油标准 （A16－涂布低脂乳品标准——草稿） 规范标准32-1981是人造奶油标准 规范标准13-1981是低脂奶油标准
Chapter Quiz 12 1._____ theories of leadership focus on personal qualities and characteristics. A.Trait B.Path-goal C.LPC D.Contingency 2._____ is the extent to which a person is likely to have job relationships that are characterized by mutual trust, respect for employees' ideas, and regard for their feelings. A.conscientiousness B.emotional stability C.courage D.consideration 3.According to Fiedler's contingency theory, if there is NOT a match of leadership style to the group situation, what should be done? A.Replace the manager. B.Change the situation to fit the leader. C.Both A and B. D.None of the above. 4.Situational leadership theory (SLT) differs from other leadership theories most clearly because it _____. A.identifies specific leadership styles B.focuses on the followers C.makes leadership contingent on the situation https://www.doczj.com/doc/3c2804576.html,es the leadership dimensions of task and relationship behaviors 5.A _____ leadership style, identified by House in path-goal theory, leads to greater satisfaction when tasks are ambiguous or stressful than when they are highly structured and well laid out. A.directive B.supportive C.participative D.achievement-oriented 6.Which of the following is NOT true of charismatic leaders? A.They have a vision. B.They have behavior that is unconventional. C.They are willing to take high personal risk. D.They are focused on their personal needs. 7.A charismatic leader's _____ is key to follower acceptance. A.energy B.vision C.credentials D.history with the organization 8.The overall evidence indicates that transformational leadership is more strongly correlated than transactional leadership with _____. A.lower turnover rates B.higher productivity C.higher employee satisfaction D.all of the above 9._____ leaders know who they are, know what they believe in and value, and act on those values and beliefs openly and candidly. A.Transformational B.Transactional C.Charismatic D.Authentic
第十二章 卷积码的概率译码 习题 1．已知（3，1，2）码的G (D)=[1+D 2，1+D+D 2，1+D+D 2]， （1）对长为L=4的信息序列画出篱笆图； （2）求与信道序列M =（101100）相应的码字； （3）用硬判决VB 译码器译接收序列R =（111，111，000，100，000，111）。 2．第1题中的码字通过二进制输入、八电平均匀量化输出的DMC 后，得到的接收序列R = （764，565，032，530，311，477），利用最小软距离译码准则，应用软判决VB 译码器，译该接收序列。 3．找出当C 1=1，C 2=10时图12-5DMC 的整数度量表。应用这个整数度量表，用VB 译码器 译接收序列：R =（111112，121112，010202，120201，020101，121111） 。 4．若用表12-1（b ）的整数度量表，求用VB 译码器译第3题中的接收序列，并与第3题和 第2题的结果进行比较。 5．考虑一个二进制输入、八进制输出的DMC ，有转移概率如下： (|)i i P r c 6．一个生成矩阵 G (D)=[1+D 2+D 3，1+D+D 2+D 3] 的（2，1，3）码， （1）画出L=4的篱笆图； （2）设一个码组在题5中所述的DMC 传输，它的接收序列：R =（1211，1201，0301，0103， 1202，0311，0302） 。试用VB 译码器求出此码序列。 7．考虑习题6的（2，1，3）码，画出L=4的码树图。设计一个转移概率p=0.045的BSC ； （1）求出该信道的FA 比特度量表和整数度量表； （2）利用FA 算法译接收序列：R =（11，00，11，00，01，10，11）。 8．计算第5题中信道的FA 比特度量表和整数度量表。 9．利用ST 算法译第7题中的接收序列。 （1）利用第7题的FA 整数度量表； （2）利用第8题的FA 整数度量表。 10．对第5题中的DMC 信道计算R comp 1
Chapter 12 job hunting 课文语言点 1，make mistakes 意为“犯错，出错”= make a mistake make mistakes in ... 意为“在某方面犯错误” Did you make mistakes again ？你又犯错误了吗? I made a mistake in spelling 我犯了一个拼写错误 by mistake 错误地I took your book by mistake 我错拿了你的书 mistake 还可作动词,意为“弄错，误解”mistake ...for 意为“把...错认为...” she is often mistaken for a teacher 她经常被误认为是个老师 2,hunt vi 意为“搜寻，寻找”常与for 构成短语hunt for ，意为“寻找，搜寻” He has hunted everywhere for his key 。他处找他的钥匙 She began to hunt for a job after she left school 她毕业后就开始找工作 3，application 名词，意为“申请，请求” He sent in his application to the company 他向公司提交了申请书。 make an application for = apply for “申请” Who will make an application for the job ？= who will apply for the job ？ 谁会申请这份工作？ 4，interview v/ n 面试，采访do an interview “采访” The boss interviewed the person who applied for the job 老板面试了这个求职的人 interviewee n 被面试者雇员interviewer n 面试者，雇主 5，require vt 意为“需要，要求”后跟名词、代词作宾语。 The work requires care and patience 这工作需要细心和耐心。 require sb to do sth 要求某人做某事 He required us to be quiet 他要求我们安静 require doing sth 意为“需要（被）... 表示被动意思” Plants require watering regularly 植物需要定期浇水 6, as well as 意为“还有,而且” 可以用来连接两个相同的成分，如名词、形容词、动词、介词等。通常不位于句首。as well as 连接的虽然是两个并列成分，但强调的重点在前面，不在后面，意为“不但。。。而且”，翻译时要先译后面的，再译前面的，如 Living things need air and light as well as water 生物不仅需要水，还需要空气和阳光 She continued her own work as well as helped me 她除了帮助我，还继续自己的工作 The child is lively as well as healthy 这孩子既健康又活泼 拓展： 1，as well as 用于同等比较，表示“和...做得一样好” She tries to learn painting as well as her deskmate He can do everything as well as his brother 2, as well as 还可表示“除。。。这外”= besides Hiking is good exercise as well as fun 徒步旅行除了有趣以外，还是很好的锻炼 区别：as well as 和not only ...but also意为“不但...而且” as well as 强调as well as之前的部分,属插入语,谓语要与它前面的主语保持人称和数一致not only ...but also 强调.but also之后的部分,谓语要与靠近它的主语保持人称和数的一致. He as well as I likes the job 不仅我,而且他也喜欢这份工作.