第二章-热力学第二定律.doc
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1、Four short words sum up what has lifted most successful individuals above the crowd: a little bit more.-author-date第二章-热力学第二定律第一章 热力学第一定律第二章 热力学第二定律练习参考答案1. 1L理想气体在3000K时压力为1519.9 kPa,经等温膨胀最后体积变到10 dm3,计算该过程的Wmax、H、U及S。解: 理想气体等温过程。U=H =0Wmax= pdV = dV =nRTln(V2/ V1)=p1V1 ln(V2/ V1)= 1519.9103110-3ln
2、(1010-3/ 110-3)=3499.7 (J) =3.5 (kJ) 等温时的公式S= pdV/ T =nR ln(V2/ V1) =Wmax/T=3.5103/ 3000 =1.17 (JK -1)2. 1mol H2在27从体积为1 dm3向真空膨胀至体积为10 dm3,求体系的熵变。若使该H2在27从1 dm3经恒温可逆膨胀至10 dm3,其熵变又是多少?由此得到怎样结论? 解: 等温过程。向真空膨胀:S= pdV/ T =nR ln(V2/ V1) (等温) =18.314ln(10/ 1) = 19.14 (JK -1)可逆膨胀: S= pdV/ T =nR ln(V2/ V1)
3、 =18.314ln(10/ 1) = 19.14 (JK -1)状态函数变化只与始、终态有关。3. 0.5 dm3 70水与0.1 dm3 30水混合,求熵变。解: 定p、变T过程。设终态体系温度为t ,体系与环境间没有热传导;并设水的密度(1 gcm-3)在此温度范围不变。查附录1可得Cp,m(H2O, l) = 75.48 JK-1mol -1。n1Cp,m(t-70)+ n2Cp,m(t-30) =00.5(t-70)+0.1(t-30) =0解得 t =63.3=336.3 KS =S1 +S2 = + = n1Cp,m ln(336.3/ 343)+ n2Cp,m ln(336.3
4、/ 303) (定P时的公式S =nCp,m ln(T1/T2)) =(0.51/1810-3)75.48ln(336.3/ 343)+(0.11/1810-3)75.48ln(336.3/ 303) = 2.36 (JK -1)4. 有200的锡250g,落在10 1kg水中,略去水的蒸发,求达到平衡时此过程的熵变。已知锡的Cp,m = 24.14 JK-1mol -1。解: 定p、变T过程。设终态体系温度为t ,体系与环境间没有热传导;并设水的密度(1 gcm-3)在此温度范围不变。查附录1可得Cp,m(H2O, l) = 75.48 JK-1mol -1。n1Cp,m1(t-200)+
5、n2Cp,m2(t-10) =0(250/118.7)24.14(t-200)+(1000/18)75.48(t-10)=0解得 t =12.3=12.3+273.2=285.5 KS =S1 +S2 = += n1Cp,m ln(285.5/ 473)+ n2Cp,m ln(285.5/ 283)=(250/118.7)24.14ln(285.5/ 473) +(1000/18)75.48ln(285.5/ 283)= 11.2 (JK -1)5. 1mol水在100和101.325 kPa向真空蒸发,变成100和101.325 kPa的水蒸气,试计算此过程的S体系、S环境和S总,并判断此过
6、程是否自发。解: 设计恒T、恒p可逆相变过程,计算S体系。已知水的蒸发热为40.67 kJmol -1。S体系 = nH蒸发/T沸点= 140.67103/373 = 109 (JK -1)p外=0, W=0,Q实际=U=H-(pV) =H-p(Vg-Vl) =H-pVg=H-nRT=140.67103 -18.314373=37.56103 (J)S环境 = -Q实际/T环境= -37.56103/373= -100.7 (JK -1)S总 =S体系 +S环境 = 109 + (-100.7)= 8.3 (JK -1) S总 0,该过程自发进行。6. 试计算10和101.325 kPa下,1
7、mol水凝结成冰这一过程的S体系、S环境和S总,并判断此过程是否为自发过程。已知水和冰的热容分别为75.3 JK -1mol -1和37.6 JK -1mol -1,0时冰的熔化热为6025 Jmol -1。解: 设计可逆过程来计算S体系。定p (101325Pa) 下:S1 = nCp,mdT/T = nCp,m ln(T2/ T1) =175.3ln(273/ 263) = 2.81 (JK -1)S2 = H /T = 1(-6025)/273 = -22.07 (JK -1)S3 = nCp,m ln(T1/ T2)=137.6ln(263/ 273) = -1.40 (JK -1)S
8、体系 = S1 +S2 +S3 = -20.66 (JK -1)H263 =H273 +Cp,mdT =(-6025)+(37.6-75.3)(263-273) = -5648 (J)S环 = -Q/T环= -(-5648)/ 263 = 21.48 (JK -1)S总 =S体系 +S环境 = (-20.66)+ 21.48= 0.82 (JK -1)S总 0,该过程自发进行。7. 有一物系如图所示,将隔板抽去,求平衡后S 。设气体的Cp均是28.03 JK -1mol -1。1mol O210, V1mol H220, V解: 纯p V T 变化。设均为理想气体,终态体系温度为t ,气体体系
9、与环境间没有热传导。n1Cp,m1(t-283)+ n2Cp,m2(t-293) =0128.03(t-283)+ 128.03(t-293)=0解得 t =15=15+273=288 KS =S1 +S2 =+ n1R ln+ n2R ln= + n1R ln+ n2R ln=1(28.03-8.314)ln(288/ 283) +18.314ln(2/1)+1(28.03-8.314)ln(288/ 293) +18.314ln(2/1)= 11.53 (JK -1)8. 在温度为25的室内有一冰箱,冰箱内的温度为0。试问欲使1kg水结成冰,至少须做功若干?此冰箱对环境放热若干?已知冰的熔
10、化热为334.7 Jg -1。(注: 卡诺热机的逆转即制冷机,可逆制冷机的制冷率)解: 水结成冰放热(冰箱得到热):Q1 = 1103334.7 = 334.7103 (J)= = 10.92至少须做功(冰箱得到功): W =334.7103/(-10.92) = -30.65103 (J)体系恢复原状,U =0,W = Q1+ Q2,冰箱对环境放热:Q2 = W - Q1 = - 30.65103 -334.7103= -365.4103 (J) 9. 有一大恒温槽,其温度为96.9,室温为26.9,经过相当时间后,有4184 J的热因恒温槽绝热不良而传给室内空气,试求:(1) 恒温槽的熵变
11、;(2) 空气的熵变;(3) 试问此过程是否可逆。解: 该散热过程速度慢,接近平衡,可视为可逆过程。(1) S恒温槽= Q /T恒温槽= (-4184)/(96.9+273)= -11.31 (JK -1)(2) S空气 = -Q /T空气= -(-4184)/(26.9+273)= 13.95 (JK -1)(3) S总 =S恒温槽+S空气= (-11.31)+ 13.95= 2.64 (JK -1)S总 0,该过程自发进行。10. 1mol甲苯在其沸点383.2K时蒸发为气,求该过程的Q、W、U、H、S、G和F。已知该温度下甲苯的汽化热为362 kJkg -1。解: 恒T、p可逆相变过程(
12、正常相变)。设蒸气为理想气体,甲苯的摩尔质量为92 gmol -1。W= p外(VgVl) = p外Vg = nRT =18.314383.2 = 3186 ( J )H= Qp = (10.092 )362103=33.3103 ( J )U= QW=33.31033186= 30.1103 ( J )S = Q /T = (10.092 )362103 /383.2= 86.9 (JK -1)G = 0A = - W=UTS = -3186 ( J )11. 1mol O2于298.2K时: (1)由101.3 kPa等温可逆压缩到608.0 kPa,求Q、W、U、H、A、G、S和S孤立
13、;(2) 若自始至终用608.0 kPa的外压,等温压缩到终态,求上述各热力学量的变化。解: 等温过程,纯p V T 变化。设O2为理想气体。(1) U=H=0Q=W= pdV = nRT ln = nRT ln =18.314298.2ln(101.3/608.0) = -4443 ( J )S体 = nR ln = nR ln= 18.314ln(101.3/608.0) = -14.9 ( J )S环 = -Q /T环= -(-4443)/298.2= 14.9 (JK -1)S孤立=S体+S环= (-14.9)+ 14.9= 0 (可逆过程)G =Vdp = nRT ln = 18.3
14、14298.2ln(608.0/101.3) = 4443 ( J )A = -pdV = - W = 4443 ( J )(2) U=H=0Q=W= p外(V2 V1) = p外()= nRT(1)=18.314298.2(1) = 12.401103 ( J )S体 = nR ln = nR ln= 18.314ln(101.3/608.0) = -14.9 ( J )S环 = -Q /T环= -(12.401103)/(298.2)= 41.6 (JK -1)S孤立=S体+S环= (-14.9)+ 41.6= 26.7 (JK -1)S孤立 0,自发过程。G =Vdp = nRT ln
15、= 18.314298.2ln(608.0/101.3) = 4443 ( J )A = -pdV = - W = 4443 ( J )12. 25,1mol O2从101325 Pa绝热可逆压缩到6101325 Pa,求Q、W、U、H、G、S。已知25氧的规定熵为205.03 JK-1mol -1。(氧为双原子分子,若为理想气体,Cp,m = R,= )解: 设O2为理想气体。纯p V T 变化。= = 1.4,T1p11 -= T2p21 -T2= T1 (p1/ p2) (1) / =298(101325/ 6101325) (11.4) / 1.4 =497.2 ( K )Q= 0U
16、= W = nCV,mdT =n(Cp,m-R )dT = 1(8.3148.314 )(497.2298) =4140 ( J )W = 4140 ( J )H = nCp,mdT =1(8.314)(497.2298) =5796.5 ( J )S体 = Q /T = 0设计定压升温和定温加压两个可逆过程代替绝热可逆压缩(令始、终态p V T相同)来计算G。定压(101325 Pa)升温(298497.2K):规定熵: ST = S298 += 205.03 + 18.314ln(T /298)= 39.23 + 29.1lnT dG = -SdT + Vdp,定p下,GT = -SdT
17、= -39.23 + 29.1lnT dT= -39.23(497.2298)-29.1497.2(ln497.2-1)-298(ln298-1)= -7814.6 -34634.2= -42449 ( J )定温(497.2K)加压(1013256101325 Pa): Gp = Vdp = nRT ln=18.314497.2ln=7406.6 ( J )G =GT +Gp =(-42449) +7406.6= -35042 ( J )13. 0,1MPa,10 dm3的单原子分子理想气体,绝热膨胀至0.1MPa,计算Q、W、U、H、S 。(a) p外p;(b) p外0.1MPa ;(c)
18、 p外0。(单原子理想气体,CV,m = R,= )解:(a) p外p,可逆绝热膨胀。= ,T1p11 -= T2p21 -T2= T1 (p1/ p2) (1) / =273(1106/0.1106 ) 2/5 =108.7 ( K )n = 4.4 (mol )Q= 0U = W = nCV,mdT = 4.48.314(108.7273) = 9016 ( J )W = 9016 ( J )H = nCp,mdT =4.48.314(108.7273) = 15026 ( J )S体 = Q /T = 0(b) p外0.1MPa,不可逆绝热膨胀。由于U = W,则nCV,mdT = nC
19、V,m(T2T1) = p外(V2 V1) = p外()CV,m(T2T1) =R(T2)8.314(T2273) =8.314(T2)T2= 174.7 ( K )Q= 0U = W = nCV,mdT = 4.48.314(174.7273) = 5394 ( J )W = 5394 ( J )H = nCp,mdT =4.48.314(174.7273) = 8990 ( J )S体 = nCp,mdT/T + nR ln= nCp,m ln(T2/ T1) + nR ln=4.48.314ln(174.7/ 273)+ 4.48.314ln(1106/0.1106 )= 43.4 (J
20、K -1)(c) p外0,不可逆绝热膨胀。Q= 0W = p外(V2 V1) =0U = 0 ,对理想气体,则温度未变,所以H = 0S体 = nR ln=4.48.314ln(1106/0.1106 ) = 84.2 (JK -1)14. 在25、101.325 kPa下,1mol过冷水蒸气变为25、101.325 kPa的液态水,求此过程的S及G。已知25水的饱和蒸气压为3.1674 kPa,汽化热为2217 kJkg -1。上述过程能否自发进行?解: 设计可逆过程来计算S和G ,设蒸气为理想气体:S1 = pdV/ T= nR ln =nR ln=18.314ln(101.325/3.1
21、674 ) = 28.8 (JK -1)S2 = Q /T = (-H汽化 /T) = 118(-2217)/298 = -133.9 (JK -1)S3 = 0 (恒温下,S = pdV/ T,液、固的S随V、p变化很小)S体系 = S1 +S2 +S3 = -105.1 (JK -1)G1 =Vdp = nRT ln=18.314298ln(3.1674/ 101.325) = -8585.8 ( J )G2 = 0G3 = Vdp =(118/1)10-6(101.325103-3.1674103)= 1.82 ( J )(恒温下,液、固的V随p变化很小)G = G1 +G2 +G3 =
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