化學(xué)勢(shì)外文翻譯

1、<p>　　本科畢業(yè)論文外文翻譯</p><p><b>　　化學(xué)勢(shì)</b></p><p>　　院（系、部）名 稱 ： 理化學(xué)院 </p><p>　　專 業(yè) 名 稱： 物理學(xué) </p><p>　　學(xué) 生 姓 名：

2、 </p><p>　　學(xué) 生 學(xué) 號(hào)： 1112080119 </p><p>　　指 導(dǎo) 教 師： </p><p>　　2012年3月15日</p><p>　　河北科技師范學(xué)院教務(wù)處制</p><p>　　目前為止，我們的討論只局限于封閉的物理

3、系統(tǒng)，即不能與周圍環(huán)境進(jìn)行物質(zhì)交換的系統(tǒng)?，F(xiàn)在我們把注意力轉(zhuǎn)到這章的開放系統(tǒng)，這個(gè)系統(tǒng)中的物質(zhì)不是封閉的。</p><p>　　假設(shè)在系統(tǒng)中引入dn摩爾質(zhì)量的物質(zhì)，加入的物質(zhì)每千摩爾所含的內(nèi)能很有可能在一個(gè)化學(xué)反應(yīng)中被釋放到該系統(tǒng)的其余部分。增加的能量正比于dn且可以寫成.數(shù)量被稱為化學(xué)勢(shì)。</p><p>　　一個(gè)帶電極性分子受到庫(kù)侖引力，當(dāng)它被吸引到另一個(gè)分子的附近區(qū)域時(shí)，這種力量表現(xiàn)

4、為負(fù)的勢(shì)能，即“勢(shì)阱”。 當(dāng)新的粒子靠近它的鄰居，它失去勢(shì)能的同時(shí)獲得動(dòng)能。通過碰撞動(dòng)能傳遞給其他粒子，使系統(tǒng)在此過程中獲得內(nèi)能。假設(shè)一個(gè)遠(yuǎn)離其它粒子無限遠(yuǎn)處的一個(gè)靜止粒子，它的動(dòng)能和勢(shì)能全為零 。把該分子移動(dòng)到第二個(gè)分子立場(chǎng)中，理論上，這可以被慢慢做以致動(dòng)能可以被忽略不計(jì)，這個(gè)分子增加的動(dòng)能大小等于勢(shì)阱的深度，定量，</p><p>　　中，是總的能量，為動(dòng)能，為勢(shì)能；為分子之間的距離。</p>

5、<p><b>　　當(dāng)，且時(shí)</b></p><p><b>　　，，所以</b></p><p>　　圖9.1 分子間作用力產(chǎn)生勢(shì)阱的示意圖</p><p>　　能量是守恒的，勢(shì)能轉(zhuǎn)化為動(dòng)能，動(dòng)能被添加到系統(tǒng)的內(nèi)能中。</p><p>　　探求的大小是多少是合理的。在一個(gè)標(biāo)準(zhǔn)的實(shí)驗(yàn)室中

6、實(shí)驗(yàn)，硫酸加入水中，導(dǎo)致了溫度的上升。假設(shè)在室溫下有千摩爾硫酸被加入到一升水中，我們確定出硫酸在水中的化學(xué)勢(shì)。</p><p>　　相比于硫酸分子與水分子間的相互作用，我們可以假設(shè)硫酸分子間的相互作用是可以忽略的（所有的酸分子和水分子式呈現(xiàn)電極化的）水的比熱容量是。</p><p>　　因此水所獲得的熱量隨增加的酸為</p><p>　　由于水的質(zhì)量是酸的質(zhì)量的倍，

7、我們可忽略后者的比熱容。化學(xué)勢(shì)為</p><p><b>　　每千摩爾</b></p><p>　　因?yàn)闊崃繌乃徂D(zhuǎn)移到水轉(zhuǎn)移，符號(hào)為負(fù)。</p><p>　　我們好奇每部分的化學(xué)勢(shì)，即勢(shì)阱深度的本質(zhì)。由于一千摩爾擁有分子，這個(gè)值為焦或電子伏特（一電子伏特等于焦?fàn)枺┐蟛糠只瘜W(xué)勢(shì)為這個(gè)量級(jí)。</p><p>　　為了解釋加入

8、到系統(tǒng)中的量的影響我們需要添加一項(xiàng)到熱力學(xué)基本方程中</p><p><b>　?。?.1）</b></p><p>　　在這里是加入的物質(zhì)（每千摩爾中）的增量，為化學(xué)勢(shì)每千摩爾焦耳。</p><p>　　如果在一個(gè)開放系統(tǒng)中且</p><p><b>　　（9.2）</b></p>

9、<p>　　得到 （9.3）</p><p>　　也就是說，化學(xué)勢(shì)被定義為恒定熵和體積下的每千摩爾的內(nèi)能增量。</p><p>　　如果有多個(gè)種類的粒子被加入到系統(tǒng)中，則（9.1）式可寫成</p><p><b>　?。?.4）</b></p&g

10、t;<p><b>　?。?.5）</b></p><p>　　表示除了保持不變的外的所有的集合，</p><p>　　另一種呈現(xiàn)與間的關(guān)系的為（9.4）式。這可以通過歐拉定理來求得齊次函數(shù)。歐拉定理表明，如果</p><p><b>　　（9.6）</b></p><p>　　則

11、 （9.7）</p><p>　　這個(gè)定理很容易地被差分方程所證明，設(shè)置（9.6）式的為1，現(xiàn)在假設(shè)所有種類物質(zhì)的量，即所謂的成分在系統(tǒng)中增加了一倍或減半，推廣來說，改變因子而不改變?nèi)魏蔚幕緺顟B(tài)的變量。然后廣延量將隨著產(chǎn)生改變，并且所有獨(dú)立的廣延量都將隨因子的變化而變化。因此是適用于歐拉定理的齊次函數(shù)：</p><p>&

12、lt;b>　?。?.8）</b></p><p>　　從的微分中我們知道，</p><p>　　，， （9.9）</p><p>　　把（9.9）式中的關(guān)系代入（9.8）式，我們得到</p><p><b>　?。?.10）</b></p><p>　　我們

13、已知吉布斯函數(shù)被定義為，立即可得</p><p><b>　?。?.11）</b></p><p>　　如果只有一個(gè)組成部分存在，則或，在此種情形下就是每千摩爾單位的這種物質(zhì)的吉布斯函數(shù)。</p><p>　　最后我們采用差分方程（9.10），得到</p><p><b>　?。?.12）</b>&

14、lt;/p><p>　　方程（9.4）和（9.12）等同，我們得到</p><p><b>　?。?.13）</b></p><p>　　吉布斯-迪昂方程的一個(gè)關(guān)系式。</p><p>　　差分方程（9.11）給出</p><p><b>　?。?.14）</b></p&g

15、t;<p>　　注意到。因此，如果我們讓某個(gè)過程在恒定的溫度和壓強(qiáng)下發(fā)生，（9.13）式成立的前兩項(xiàng)為零，所以和隨之為零。</p><p>　　隨后，（9.14）式得到重要結(jié)論</p><p><b>　?。?.15）</b></p><p>　　9.1 The chemical potential </p><

16、;p>　　Until now we have confined our discussion to closed physical systems，which cannot exchange matter with their surroundings. We turn our attention in this chapter to open systems, in which the quantity of matter is

17、 not fixed. </p><p>　　Supposed that dn kilomoles of matter are introduced into a system. Each kilomole of added matter has its own internal energy that is released to the rest of the system, possibly in a ch

18、emical reaction . the added energy is proportional to dn and may be written as . The quantity is called the chemical potential.</p><p>　　The chemical potential is associated with intermolecular forces. An e

19、lectrically polarized molecule experiences a coulomb attraction when it is brought into the vicinity of another such molecule,*this force is expressed as a negative potential energy, a sort of “potential well.” As the ne

20、w particle approaches its neighbor, it gains kinetic energy while losing potential energy. The kinetic energy is imparted to other particles through collisions, so the system gains internal energy in the process</p>

21、;<p>　　Consider a motionless motionless molecule infinitely distant from other molecules. Its kinetic energy and potential energy are both zero. The molecule is moved into the force field of a second molecule. Thi

22、s can , in principle, be done slowly so that the kinetic energy is negligibly small. Left by itself, however, the molecule picks up kinetic energy in magnitude to the depth of potential well(figure9.1)quantitatively,

23、 </p><p>　　Where E is the totle energy, K is the kinetic energr, and is the potential energy; is the distance between the molecule and its neighbor, at ,and so </p><p>　　everywhere. At ,</p

24、><p><b>　　, so </b></p><p>　　Figure 9.1 Schematic diagram of a potential well due to intermolecular forces. </p><p>　　Energy is conserved, but a conversion from potential e

25、nergy to kinetic energy takes place. The kinetic energy is added to the internal energy of the system.</p><p>　　It is reasonable to ask what the magnitude of is. In a standard laboratory experiment, sulfuri

26、c acid is added to water , producing an increase in temperature. Imagine that kilomoles of acid at room temperature are added to a liter of water , also at room temperature. the temperature is observed to rise . we wish

27、to determine the chemical potential of acid in water. </p><p>　　We shall assume that the interaction among the acid molecules is small compared with their interaction with the water molecules (both acid and

28、water molecules are electrically polarized.) the specific heat capacity of water is . thus the heat gained by the water through the addition of the acid is </p><p>　　Since the mass of the water is more than

29、times the mass of the acid, we can ignore the heat capacity of the latter. The chemical potential then ,is </p><p>　　The sign is negative because heat is transferred from the acid to the water.</p>&l

30、t;p>　　We are interested in the chemical energy per particle, which is essentially the depth of the potential well. Since a kilomole has molecules, this value is J or (one electron volt is equal to joules.) most chemi

31、cal potentials are of this order of magnitude.</p><p>　　To account for the effect of adding mass to a system, we need to add a term to our fundamental equation of thermodynamics: </p><p><b&g

32、t;　　(9.1)</b></p><p>　　Here dn is the increment of mass added (in kilomoles) and is the chemical potential in joules per kilomole. If, in an open systkem, and </p><p><b>　　(9.2)</

33、b></p><p>　　Then </p><p><b>　　(9.3)</b></p><p>　　That is, the chemical potential is defined as the internal energy per kilomo

34、le added under conditions of constant entropy and volume.</p><p>　　If there is more than one type of particle added to the system (say mtypes), then Equation(9.1)becomes </p><p>　　,

35、 (9.4)</p><p><b>　　With </b></p><p><b>　　(9.5) </b></p><p>　　The subscript means that all other except are held constant.</p>&l

36、t;p>　　Another way of seeing the relationship between and is integrate Equation (9.4).This can be done by using Euler’s theorem for homogenous functions. Euler’s theorem states that if </p><p>　　,

37、 (9.6)</p><p><b>　　Then </b></p><p><b>　　(9.7)</b></p><p>　　The theorem can be easily proved by differentiating equation (9.6) wit

38、h respect to and then setting equal to unity. Now . Suppose the amounts of all the types of substance, called constituents, in the system were doubled or halved or, more generally, changed by the factor without changing

39、any of the fundamental state variables. then the extensive variable would be changed by and all the independent, extensive state variables would also be changed by the factor .thus is a homogeneous function an</p>

40、<p><b>　　(9.8)</b></p><p>　　From the differential of we know that </p><p>　　, , (9.9)</p><p>　　Substituting the relations of Equation (9.9) in t

41、he Equation(9.8),we have </p><p><b>　　(9.10)</b></p><p>　　Recall that the Gibbs function is defined as . It is therefore immediately clear that </p><p><b>　　(9.11)

42、</b></p><p>　　If only one constituent is present, then or ; so in this case is simply the Gibbs function per kilomole of the substance.</p><p>　　Finally ,if we take the differential of E

43、quation(9.10), we obtain</p><p><b>　　(9.12)</b></p><p>　　Equating Equations(9.4) and (9.12), we have </p><p><b>　　(9.13)</b></p><p>　　A relation

44、 known as the Gibbs-Duhem equation . Taking the differential of equation (9.11) gives</p><p><b>　　(9.14)</b></p><p>　　Note that .thus if we have a process that takes place at constan

45、t temperature and pressure, the first two terms of Equation (9.13) vanish so the sum also vanishes.</p><p>　　Equation(9.14) then yields the important result </p><p><b>　　(9.15)</b>&l