高教熱統(tǒng)答案第四章

第四章4.1UT,V,n1,…nki（1）UnUVU；（2）i

Uviii ii

i證

,1,

nki

fiii

iUi

UVii iiiii

U

UV

n(U

U)niiiii

i

i iiuUi

i

nk是n1

nknj

0n j,1

nk11nj

nj j習(xí)題4.3gi(TP)i組元的化學(xué)勢，xii組元的摩爾分?jǐn)?shù)。當(dāng)物質(zhì)的量分n1、n2的兩種純液體在等溫等壓下合成理想溶液時(shí)，試證明混合前后體積不變V

GRT(n1lnx1n2lnx2熵變SR(n1lnx1n2lnx2焓變H0解

in1g1(T,p)n1RTlnx1n2g2(T,p)n2RTlni所 GGG0n1RTlnx1n2RTln VG；V(G)0

SG；S(G)nRln

nRln

GH

HGTSn1RTlnx1n2RTlnx2n1RTlnx1n2RTlnx2（5）UHpV習(xí)題4.4

μigi(T,P)RTlnxi g'g(T,P)RTln(1 11x

1

pxp0(1p0是該溫度下溶劑的飽和蒸汽壓，pxx時(shí)的飽和蒸汽壓。該 g'

g

x

x

v

1 1

p

p

(1x)pTp xv'v

；v’ vv’-v≈v’,pv’=RT代入

x

1積分(2)

習(xí)題4.8n1molTp'VnRT;p'Vn 解：（1）pn1n2RTn1n2 V1根據(jù)SnCVlnTnRlnVnRlnnSS1S1n1CVlnTn1Rln(V1V2)n1Rlnn1n1S0n1CVlnTn1RlnV1n1Rlnn1n11nRlnV11S2n2CVlnTn2Rln(V1V2)n2Rlnn2n2S0n2CVlnTn2RlnV2n2Rlnn2n2

RlnV1V21VSnRlnV1V21V

RlnV1V2V SS2S1(n1n2)CVlnTn1RlnV1n2RlnV2n1Rlnn1n2Rlnn2(n1n2S(nn)RlnV1V2nRlnV1nRlnnn nn1 2 1 212124.9NH3N2H212N

3

NH3

Kp

p

N23H22NH3證NH3n0mol,n0εmol，未分解(1-ε)n0mol,1n

2N23nεmolH2，共有摩爾數(shù)21n 3n000 ; ;000

0 0

132Kp(xN2

)2H2H

)2

)1p2

4

N23H22NH3NH32n0mol,2n0εmol2(1-ε)n0mol,N2

molH2

2(1ε)n0

2(1ε)n0

2(1ε)n0;2(1ε)n0 )1 )3 )2

2

16(1ε2)24.10n0v1mol的氣體A1n0v2mol的氣體A2TV0,當(dāng)發(fā)生化學(xué)變化，ν3A3ν4A4ν1A1ν2A20；證

εVeV0

n0v1molA1n0v1εmol，未反應(yīng)(1-εn0v1mol,n0v2molA2εn0v2mol，未反應(yīng)(1-ε)n0v2mol,εn0v3molA3εn0v4molA4，有由式（4.10.1）比式（4.10.2）

解(4.10.3)

εVeV0

習(xí)題4.11T→0dσ0證

FSdT

FT

F;A S

T

;而實(shí)際上σAS

T→0limSTT

0 習(xí)題4.12T→0證

dpS；T→0；dp

S

dT V T

TlimST

0；原式得證p4.14pT0,L，固相的定壓熱容量Cp,C’.試求液體的絕對熵表達(dá)式。p解T溫度，pBBCA T0①A→B,SAB

p②B點(diǎn)相變過程

LTSB相變LT0TCp'T③B→C,SBC