第三章热力学第二定律天津大学物理化学第五版部分习题答案!

1、习题3 3r -7；r) = L = 0.666( irj =. Q、= 300 kJ , -TV = Q】+ Q? = 200 kJ习题3 7fl； cTSsys = m #dT = mCpln -=- = 1155 J IL】1斗=警丝=叫耳TJ = -1009 J L amoS“o = ASjj 4- ASamb = 146 J - K_1C) Ssys = m=mCplnmCpln刃qT3 r 3-;X r2 T z / s5 a ,+ md +-1 £pTr2 GzlKSam61155mCpSTr t mCpAT2TiTzmCp(T. _ 写)+ mOpg - 坊)T=-1

2、07S J K_1Sb。= AS护 + SSQrnb = 77 J - K_1 (3)Ssys = 171 j： *dT + m j： *dT + m £：岸dTTTT=rnCp In+ tyiCp In r- + mCp In yrT=mCpln 尹丄i=1155 J iLac _ mCpSTi | mCpATj , mCpSTz=+ FTr 1 T2T2mCT.-TJmCp(T3-T2) mCp(T,-T3)石尊Tz一 1103 J K_1Sz = HS呼 + Sam& = 52 J K-1t2习题3 10(1)恒温，At7 = 0 , Q = -WQ = -W = -R

3、Tln = 1.729 kJP2Ssy$ =学=5.763 J K_1Samb = # = -5.763 J K-iS*so =、Ss 护 + Samb = 0C)系统s是状态函数，故Ss" = 5.763 J - K_1Q = -W = peAV=Pe x (V> - V.)(RT RT、=O'(瓦-瓦丿=1.247 kJS“ = = 4.157 “ILS® =隠护 + ASarnb = 1.606 J -I：-1 G)Ssg = 5.763 J - K_1Q = -W = -p.AV = 0-QSom6 = -Tj7 = 0S”。= Ss3 + Samb

4、= 5.763 J -K-1习题3.15=395.9 KV = “= 0.1646 m3" P2Z1 =型=204.5 K3 nR7*TTnS = nCp m In * + 池斤 In 冬 + nCp m In = nCp m In * + 位斤 In 冬=-6S.65 J K“ 丄1Pl丄2'上1PlU = nCVm(7； 一 7；) = -5.957 kJAZf = nCpm(T3 -7；) = -9.928 kJW = nCVm(T2 - I；) -p2(V； - K) = 13.93 kJ (W = 十 W2 , Wx = AUX)Q = nCprn(T3 -T2)

5、= -19.S9 kJ (Q = Q】+Q? , Q】=0 , Q, = AK)习题3 20温度、压力相同，所以K = =(% +SJR7 =+ %(chj = 150 dm3 pp=31.83 JK-iS = ASH) + AS(C/4) = % R in 為 + 於中 R In习题3 341叩T=iwrc|P-120 kPai-+ I T-IOWC|P=I2<» kPa>H/> (g) (g)T-100CT-100hCP=120 kPaP=120kPaI、H 、S11Jk1/ AS3l1j M / g1H,O (R) T=10ir( P=l20kPa+(K>

6、; r=i(Mrc P=120kPa；dapH学 / aapS早t ah2 / as21Fi)1 = IOO"CP -KH.325 kPapv«)+T-IOO°CP=120kPa»：O (g) l=100CP =101325 kPai+<8> T-io(rc P=120kP»水完全蒸发，仃:Qp = AH = AH. + n(H2O)pH学 + AH2 + SH3 =n(2foO)A Hq=122.0 kJW = p卩=-AnRT = -Sn(H2O)RT = 9.307 kJ C7 = Q + W = 112.7kJAG = AH

7、-TAS = -S.S65 kJAA = SU-TAS = -1S.17 kJV1(H20) = nn2ORT .V1(N2) = nRT ,匕=匕（凤0）+匕（迅）=（“2。S = AS. + AtapSe(H2O) + AS2 + AS3=0 + "叱卜碑+ n(H2O) R in ? + 卜(禺。)R in+ “化)R In 晶=0 + ri(H2半.。H早*斤In生+ 71(他0)斤1!1池电0十 +於(“)Rin几电。十池丄PnHOnN0 =350.7 kJ - K"1习题3 40(1)=工 i/bAzS®(B) = 256.7 J IC1 mol" Br號=刀= 247.3 kJ - mol-1B G° = TA Se = 170.7 kJ moLT mr mr m(2)B(3)CHjg>CO： (»)T-2S%?+T-25*CP =150 kPaiP«l50kPai11111ICH (g)® (g)T-25eC+T-25*CP-100 kPaP-I(MI kPaArG® = 2Ji/sAzG®(B) = 170.743 kJ - mol"1=&+怠+g用= Rin 冬(OHJ + ln(