10.2 Consider two systems, A and B. each composed of the same single particle type. The two systems arc contained in a chamber surrounded by rigid adiabatic walls and they are separated from each...

1 answer below »
10.2 Consider two systems, A and B. each composed of the same single particle type. The two systems arc contained in a chamber surrounded by rigid adiabatic walls and they are separated from each other within the chamber by a rigid diathermal wall which is also permeable to the particles (see Fig. D.12). Show, using an argument similar to the one used in section 10.2, that the condition for equilibrium against particle exchange is the equality of the chemical potentials.
10.3 Consider the system of question 10.2. Suppose that the two systems, composed of the same single type of particle, are both in the same phase, e.g. a gas on each side of the separating wall. Show that the pressures arc equal. Would the pressures be equal if different
phases existed on either side of the wall? (Hint: For a system consisting of a single type of particle, 14 = G(T, P, = 0(T, P). If the phases are the same on either side of the wall, the function (/) must also be the same on either side.)



ˉ --==⋯ ˉ ˉ ˉ ■ 0====■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ w 188 0pc″ ′,sre,· t、 nn‘】扌l·cc” c″氵icd′ po氵 cJ∶ liα ′ eXamplc cousider u As u is extensiVe we must Wr∶ tc "012) wbere thc ncW lt nc1ion φ ′。oflhc cnt「 opy and vdvme pCr pa:ttle Di晚 rentiatinε 、Ⅳith respcct to lV at C0nstant s and /gives∫ [again∶ (1013) so〃 is not simply ⒍/N bu1involvCs extfa tenns dcpendCnt()n thc dσ/V=ds′S=dll/=dⅣ∥Nj=α “ 014) for a"i substitutinε 1hesc rela1ionsfof dU ds and dNiin cqua1i1丬 l “03)and canceuing the common1erm α, ∪=TS— P/+∑卩N: "Os| “ 丨囝 ⑾Φ ∶δ∴叫Ⅲ刂:扌f1:拈芥‖帑辟:∴::湍烈 i品 丨引腊:J丨∵ ⒑2孑 品:ξl糅t书志:P):rⅠ品〗f° R EXCHANCE PHAsE EQVⅡ IBRIUIⅥ 7乃c召〃″″ j乃 `nr,″ (k″∶″∫rⅡ ),iΙ″∶″C` P四 ```c lt,c、 ch″″。t, 189 P}刍 `oh° ● "Ja`ˇ— / 人 jVA V. 戗 P,r, RigⅡ●●d● b● "● 秽括J%:℃ ;c艹:蓍勰卢 ⑾提品憔 %凡∷$栌Ⅹ:紧;J:百 ⅣA+Ⅳ:=Ⅳ “he totaI number of particlcs) (1017) %+‰ =/(the total v。lume) (IO18) 〃Λ+tJ:=u “hct。 taIinternalcnergy) (t019l 懈 鞘 斌 擀 抻 麒 侧 黼 宀 械 s=s(I/A‰ ⅣA〃:‰ ,Ⅳ :) =s人(u^./A,Ⅳ A)+sRlV:,‰、Ⅳd ˉ a max〗 mum 拥鲰 :I:眢l忠rw铞腽盅fm:lt ˇ N s Ν φⅣ 〓Ⅳ吆●亠u〓U 耐 ∷ Ⅳ N+ U N 〓Ⅲ ~ d Ν+φ〓μ 190 opefl s'st纟 Wts曰Ⅱd I此纟 c乃e,,,;c夕 ′j,o】 eⅡ f沏 J from thei【 equⅡibrium vaIues, ds=dsA+dsB=o Applying equation(10D10tllc tWo phascs A and B, ds^=⊥¢u^+PAdlAˉ ll【dⅣV dsI,=壳(dU:+P:dv。 ūBdⅣ:) 耀鲲攒盟拣嚣以$扩渊 (甘 i一 壳 )dI/^+(号:— ;专)d/人 ˉ (号 :ˉ 茕 )dⅣ A=0 (1023) This must be tfue for any dLl^,dˇ Aand dⅣ ^and so their C灏 0en玲 in equadon(I023)must all be zero Thus∶ rA=T:, which is the Cond"ion for thC[mal Cqu"ibrium; PA=P:, which is the C0ndition for mcchanical equuibrium;and `Ι ^=u:、 which is‘ he c0ndition fQr cqunibl△ um agains1par“ Cle exchangc We conc1udc fnalIy∶ 舳社i批鞭韶黼谢鞲橛擀钭 eqlla1 甜骨:皆絷发揪紫丨l∶讠Ψ惑 t:l:l丨 勐 · 笫 ‘ 嬲 蓝 1tira扌 雨i蕊:、品品盂飞e llave a std倥 ve氵 dose to tM rlt砧 cqtlll儿num“ 扯c,山饰 Ⅱ g叩9血 tll煎 thtr o asm钔 lpσ“Ⅱuc C× 00| ●Ⅱ ̄_~ ^i ^ `j 。r⒈ ^0● 棘 :刂踽姓跳囔湓i滞嬲屮琊J芈捌 轴王恧锶嚅Ⅰ渴 q〖 黜氵棣王r扌甜%「目 (1020l (lO21) (1022) on tl△ e ngllt of equatlon(1021)and(102勾 Wc mtlst Iemember tllat 舳杏莒蚤哕链猡1∵ir掷:馓坍s找 j嘿锣: δs=-lt^(~挚 )— u:(+立兮|)>o (1020l (1025) 'Pp″ ‘西汤Jls犭 IJlf c乃enlzcn氵 pore,lJ杨 I 191 lll^ˉ u:)5ⅣA>o 锯‰ 鞫 t皙i氵 甜找芒:潇岁坩⒊糊 邪甲 袋 ξ∶甘11f::I:fa;;号 :∶ ∏ 翟 s⒊ h蹴 帑 s摁 丨暂 肾 {景: equilibouⅢ °f two phases a1a fIxed T and P is that thc Cibbs ftulctions per unit mass are equal N° w this means that the Cibbs l∶巍蚤氵甫罐睽撼军擀椴械;黜 1⒍3TⅡREE APpLICΛ TIoNs OF THE CoNCEPr OF THE CHEMICAL poTENΠ A△ 奋胳艹夥榉镞萝在l茁硭i蹄:j∶l黼奄 }
Answered Same DayDec 31, 2021

Answer To: 10.2 Consider two systems, A and B. each composed of the same single particle type. The two systems...

Robert answered on Dec 31 2021
119 Votes
Problem 10.2
Consider two systems A and B each composed of the same single particle typr. The two systems are contained in a
chamber surrounded by rigid adiabatic walls and they are searated from each o
ther withing the chamber by a rigid
diathermal wall which is also permeable to the particles. Show using an argument similar to the one used in section
10.2, that the condition for equilibrium against partcile exchange is the euality of the chemical potential.
Solution
Rigid adiabatic wall: A wall which doesnot allow the heat exchange throug it.
Diathermic wall: A wall which allows the heat exchange through it.
Since the above question is based on single particle system, hence we can take an example of ice melting in water
then there will be only one type of particles that is of water only.
Let us suppose there are two type of phases A and B as shown in the diagram of the same substance that is water as
we have considered before.
Since the two phases are separated by the diathermic wall then they will be in thermal equilibrium with each other
hence we can take
TA = TB = T ( In thermal equilibrium )
Let the two phases have separate volume VA and VB and are seprated by a diathermic wall which is also permeable
then there can be heat and particle exchange across the walls. Let there be NA particles in pahse A having internal
energy UA and pressure PA and similarly for phase B let NB , UB and PB are the number of particles , internal energy
and pressure respectively. Hence
N = NA + NB ( total no of Particles)
V = VA + VB ( total Volume)
UA + UB = U ( total internal energy)
Since there will be no exchange of heat with the surroundings then internal energy will remain constant.
Now the total entropy for the combined system will be
S = SA + SB
Where S is a function of U , N and V of the system.
Now from the second law of thermodynamics we know that enetropy should be maximum
So SA + SB = maximum.
As all the above mentioned quantities will be change by a very infinitesimal amount hence we...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here