化學(xué)勢(shì)外文翻譯-化學(xué)工程-資料下載頁
2025-01-19 07:50本頁面
【導(dǎo)讀】現(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。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dn?現(xiàn)為負(fù)的勢(shì)能，即“勢(shì)阱”。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。把該分子移動(dòng)到第二個(gè)分子立場(chǎng)中，理論上，這?？梢员宦鲆灾聞?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，在一個(gè)標(biāo)準(zhǔn)的實(shí)驗(yàn)室中實(shí)驗(yàn)，硫酸加入水中，導(dǎo)致了。們確定出硫酸在水中的化學(xué)勢(shì)。略的水的比熱容量是????由于水的質(zhì)量是酸的質(zhì)量的310倍，我們可忽略后者的比熱容。焦?fàn)枺┐蟛糠只瘜W(xué)勢(shì)。為化學(xué)勢(shì)每千摩爾焦耳。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。kn表示除了保持不變的jn外的所有n的集合，而不改變?nèi)魏蔚幕緺顟B(tài)的變量。產(chǎn)生改變，并且所有獨(dú)立的廣延。因此U是適用于歐拉定理的齊次函數(shù)：。就是每千摩爾單位的。（）式成立的前兩項(xiàng)為零，所以和隨之為零。
【正文】 be easily proved by differentiating equation () with respect to ? and then setting ? equal to unity. Now ).. ..,( 1 mnnVSUU ? . 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 any of the fundamental state variables. then the extensive variable U would be changed by ? and all the independent, extensive state variables would also be changed by the factor ? .thus U is a homogeneous function and Euler’s theorem can be applied to it: () From the differential of we know that TSUknV??????? ??, PVU knS ?????????? , jnVSj knU ???????????? , () Substituting the relations of Equation () in the Equation(),we have kkk nVSjmj jnSnV nUnVUVSUSU,1, ???????? ????????? ????????? ??? ?? jmj j nPVSTU ????? 1 ? () Recall that the Gibbs function is defined as PVSTUG ??? . It is therefore immediately clear that jmj jnG ??? 1? () If only one constituent is present, then nG ?? or nG?? 。 so ? in this case is simply the Gibbs function per kilomole of the substance. Finally ,if we take the differential of Equation(), we obtain jj jj jj dndnV dPP dVSdTT dSdU ?? ?? ?????? () Equating Equations() and (), we have 01 ??? ?? jmj j dnV dPSdT ? () A relation known as the GibbsDuhem equation . Taking the differential of equation () gives jj jjj j dndndG ?? ?? ?? () Note that ? ?mnnPTGG .. .., 1? .thus if we have a process that takes place at constant temperature and pressure, the first two terms of Equation () vanish so the sum also vanishes. Equation() then yields the important result ? ? jj jPT dndG ?? ?, ()
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