物理化学第三章练习题.ppt

单击以编辑母版标题样式,,单击以编辑母版文本样式,,第二级,,第三级,,第四级,,第五级,,*,*,若某可逆热机分别从(a)600K, (b)1000K的高温热源吸热, 向300K的冷却水放热, 问每吸100kJ热各能作多少功?,Q,2,W,可,Q,1,热源,T,1,热源,T,2,可,(a),T,1,= 600K,,T,2,= 300K,,Q,1,=100kJ,,,= ,W,r,/,Q,1,= (,T,1,,,T,2,)/,T,1,,,=(600K,,300K,,)/ 600K = 0.5,,,W,r,=,,,Q,1,=,,0.5,,100kJ =,5,0kJ,(b),T,1,= 1000K,,T,2,= 300K,,Q,1,= 100kJ,,,= ,W,r,/,Q,1,= (,T,1,,,T,2,)/,T,1,,,= (1000K,,300K,,) / 1000K = 0.7,,,W,r,=,,,Q,1,=,,0.7,,100kJ =,7,0kJ,1,10/24/2024,1,,5,这是一个自发过程可设想在高温热源,T,1,和低温热源,T,2,之间有无数个热源，每个热源的温度仅相差,,d,T,，将系统逐个与每个热源接触并达到热平衡，最终的结果是热从高温热源可逆地传到了低温热源。

整个过程的熵变,,,△S= △S,1,+ △S,2,=5.76J·K,-1,10/24/2024,2,,7,10/24/2024,3,,9,P,1,T,1,V,1,P,2,T,2,= T,1,V,2,①恒温可逆,P,2,V′ T,1,′,②恒容,恒压,P,2,V,1,T′,恒压,③绝热,①△U=0 Q= -W=,=5.76J·K,-1,②W,1,=0,10/24/2024,4,,Q,1,=△U,1,=nC,V,m,(T,1,′-T,1,)= - 3.118KJ,Q,2,=△H,1,=nC,p,m,(T,2,- T,1,′)= 4.356KJ,Q=Q,1,+Q,2,=1.247KJ,△S= △S,1,+ △S,2,=5.76J·K,-1,③ Q,1,=0 △S,1,=0,10/24/2024,5,,Q = Q,1,+Q,2,P,1,T,1,V,1,P,2,T,2,= T,1,V,2,Q,P,2,V′ T,1,′,②恒容,,Q,1,恒压,,Q,2,恒温可逆,10/24/2024,6,,,在恒熵条件下, 将3.45mol理想气体从15,℃,, 100kPa压缩到700kPa, 然后保持容积不变, 降温至15,℃,. 求过程之,Q,,,W,,,,U,,,,H,及,,S,. 已知,C,p,m,=20.785 J,,mol,－1,,,K,－1,.,n,=3.45mol, pg,,T,1,=288.15K,,p,1,=100kPa,n,=3.45mol, pg,,T,3,=288.15K,,V,3,=,V,2,d,S,=0,n,=3.45mol, pg,,T,2,=？,,p,2,=700kPa,压缩,d,V,=0,降温,恒熵过程指绝热可逆过程,10,′,10/24/2024,7,,,一个两端封闭的绝热气缸中, 装有一无摩擦的导热活塞, 将气缸分成两部分. 最初, 活塞被固定于气缸中央, 一边是1dm,3,, 300K, 200kPa的空气; 另一边是1dm,3,, 300K, 100kPa的空气. 把固定活塞的销钉取走, 于是活塞就移动至平衡位置.试求最终的温度, 压力及隔离系统总熵变.,返回,12,′,气缸绝热,导热活塞,p,a,=200kPa,,T,a,=300K,,V,a,=1dm,3,p,b,=100kPa,,T,b,=300K,,V,b,=1dm,3,p,,T,,V,1,p,,T,,V,2,绝热恒容,空气视为理想气体, 则由,,U =,0,可知,T,=,T,a,=,T,b,=300K.,( 亦可由,,U =n,(左),C,V,m,(,T,-,T,a,) +,n,(右),C,V,m,(,T,-,T,b,) =,,0,求得 ),左边:,pV,1,=,p,a,V,a,右边:,pV,2,=,p,b,V,b,10/24/2024,8,,返回,10/24/2024,9,,12,PT,= c =100×298 =29800 kPa·k,T,2,= 29800/200 = 149k,△U=nC,V,m,△T=2×2.5×8.314×(149-298)= - 6.194kJ,△H=nC,p,m,△T=2×3.5×8.314×(149-298)= - 8.672kJ,= - 51.87J·K,-1,Q=△U – w = - 11.149KJ,D,D(,-,=,D,T,S,) = - 8.672-[149,×(-51.87+2×205.1)-,,(298×2×205.1)] ×10,-3,,=60.177kJ,H,G,10/24/2024,10,,13,恒温可逆,4mol,,T,1,= 750K,,p,1,＝150kPa,,V,1,4mol,,T,3,＝,T,2,,,p,3,＝100kPa,,V,3,,4mol,,T,2,,,p,2,＝50kPa,,V,2,＝,V,1,恒容,10/24/2024,11,,,已知－5,℃,固态苯的饱和蒸气压为2.28kPa, 1mol, －5,℃,过冷液体苯在,p,=101.325kPa下凝固时,,,S,m,=－35.46 J,,K,－1,,mol,－1,, 放热9860 J,,mol,－1,. 求－5,℃,时, 液态苯的饱和蒸气压. 设苯蒸气为理想气体.,16,′,1mol过冷液体苯在恒温268.15K凝固. 设过程经由如下图所示的5个步骤完成:,l,,,p,0,,,T,s,,,p,0,,,T,l,,,p,l,,,T,g,,,p,l,,,T,g,,,p,s,,,T,s,,,p,s,,,T,,G,,G,1,,G,2,,G,3,,G,4,,G,5,,H,,S,10/24/2024,12,,p,l,= 2.669kPa,l,,,p,0,,,T,s,,,p,0,,,T,l,,,p,l,,,T,g,,,p,l,,,T,g,,,p,s,,,T,s,,,p,s,,,T,,G,,G,1,,G,2,,G,3,,G,4,,G,5,,H,,S,10/24/2024,13,,16,△U=W,nC,V,m,(T,2,-T,1,)= - P(V,2,-V,1,),△U=W=nC,V,m,(T,2,-T,1,)= - 2.394KJ,△H=nC,p,m,(T,2,-T,1,)= - 3.991KJ,=10.73J·K,-1,10/24/2024,14,,,(1)1kg,温度为273K的水与373K的恒温热源接触, 当水温升至373K时, 求水的熵变, 热源的熵变及隔离系统总熵变.,,(2)倘若水是先与保持323K的恒温热源接触, 达到平衡后再与保持373K的恒温热源接触, 并使水温最终升至373K, 求总熵变.,,(3)根据(1), (2)计算结果, 说明用何种加热方式既能使水温由273K升至373K, 又能使总熵变接近于零? 设水的比热容为 4.184 J K,－1,g,－1,.,17,′,m,=1kg H,2,O(,l,),T,1,=273K,热源,T,=373K,m,=1kg H,2,O(,l,),T,=373K,热源,T,=373K,(1) 水的比热容,c,= 4.184J K,－1,g,－1,10/24/2024,15,,(3) 计算表明, 采用温度递增的不同热源加热以缩小热源与系统间的温差,,,S,(总)将减小并趋于零. 当采用可逆加热时, 系统每次升温d,T,, 则需无穷多个热源,,,S,(总) = 0.,10/24/2024,16,,,在100kPa下有10g 27,℃,的水与20g 72,℃,的水在绝热器中混合, 求最终水温及过程的总熵变. 水的定压比热容为,c,p,,=4.184 J g,－1,K,－1,.,,绝热混合,恒压100kPa,m,a,=10g,, t,a,=27,℃,,水,m,b,=20g,, t,b,=72,℃,,水,m,=30g,, t,,,水,19,10/24/2024,17,,20,△S= △S(H,2,)+ △S(CH,4,),=31.83J·K,-1,H,2,,300K,,100dm,3,CH,4,,,300K,,50dm,3,H,2,CH,4,,300K,,(100+50)dm,3,恒温,,恒压,10/24/2024,18,,,今有1mol氧气从900,℃,, 700kPa绝热可逆膨胀到140kPa, 求此过程的,,H,及,,G. 已知标准熵S,,1173,=248.67 J,,mol,－1,,K,－1,. 定压摩尔热容为:,,C,p,m,/(J,,mol,－1,,K,－1,),,= 28.17 + 6.297,,10,－3,(,T/K,)－0.7494,,10,－6,(,T/K,),2,22,′,n,=1mol, O,2,,p,1,=700kPa,,T,1,=1173.15K,n,=1mol, O,2,,p,2,=140kPa,,T,2,=？,绝热可逆,,,S =,0,因题给,C,p,,m,是温度的函数, 故不宜用,绝热过程方程,求,T,2,.,10/24/2024,19,,始态的规定熵,10/24/2024,20,,22,△U= △U,1,+ △U,2,=0,n,1,C,V,m,(T,2,-T,1,)+ n,2,C,V,m,(T,2,-T,1,′,)=0,2× (T,2,-200)+ 4×(T,2,-500)=0,T,2,=400K,2mol N,2,,200K,,50dm,3,4mol N,2,,500K,,75dm,3,(2+4)mol N,2,,T,2,,(50+75)dm,3,2mol N,2,,T,2,,dm,3,4mol N,2,,T,2,,dm,3,恒温恒压,,△S′,△S,1_,+ △S,2,10/24/2024,21,,△S= △S,1,+ △S,2,=10.73J·K,-1,△S′= 0,10/24/2024,22,,,在一绝热容器中有1kg 25,℃,的水，现向容器中加入0.5kg 0,℃,的冰，这是系统的始态。

求系统达到平衡态后，过程的△S.,绝热混合,恒压,m,a,=1kg,, t,a,=25,℃,,水,m,b,=0.5kg,, t,b,=0,℃, 冰,m,=1.5kg,, t,,,水,24,′,假设冰完全融化，则,代入数据求得 t＝－9.887,℃,＜0 ℃，不合理,故终态温度为0,℃,设融化的冰为m，则1000×4.184×（0－25）＋333.3m＝0,,m＝313.8g,10/24/2024,23,,,1mol水在373.15K, 101.325kPa下恒温恒压气化为水蒸气, 并继续升温降压为473.15K, 50.66kPa. 求整个过程的,,G,. 设水蒸气为理想气体, 水气的定压热容,C,p,m,/(J,,mol,－1,,K,－1,) = 30.54 + 10.29,,10,－3,(,T/K,), 其它数据可查附录.,24,′,H,2,O(l),,T,1,=373.15K,,p,1,=101.325kPa,H,2,O(g),,T,2,=,T,1,,p,2,=,p,1,H,2,O(g),,T,3,=473.15K,,p,3,= 50.66kPa,(1),,,G,1,=0,(2),,,G,2,,H,2,,S,2,,10/24/2024,24,,10/24/2024,25,,,298.15K, 101.325kPa,下1mol过冷水蒸气变为298.15K,,101.325kPa,的液态水. 求此过程的,,S,及,,G,. 已知,298.15K,下水的饱和蒸气压为3.1674,kPa,, 气化热为2217J,,g,－1,. 此过程能否自发进行?,24,1mol, H,2,O(g),,T,,=298.15K,,p,1,=101.325kPa,1mol, H,2,O(l),,T,=298.15K,,,p,1,=101.325kPa,恒温恒压,1mol, H,2,O(g),,T,,=298.15K,,p,2,=3.1674kPa,1mol, H,2,O(l),,T,,=298.15K,,p,2,=3.1674kPa,(1),(2),(3),10/24/2024,26,,10/24/2024,27,,一定量某物质的恒压过程,T,一定, 将上式对,p,微分, 可得,将麦克斯韦关系式,代入上式, 得,30,′,10/24/2024,28,,1mol某物质,,S,m,=,f,(,T,,,V,m,),上式在恒压条件下除以d,T,, 可得,由麦克斯韦关系式, 可知,将式(2), (3), (4)式代入(1)式, 得,32,′,10/24/2024,29,,,今有两个用绝热外套围着的容器, 均处于压力,p,=101.325 kPa下. 在一个容器中有0.5mol液态苯与0.5mol固态苯成平衡; 在另一容器中有0.8mol冰与0.2mol水成平衡. 求两容器互相接触达平衡后的,,,S,.已知常压下苯的熔点为5,℃,, 冰的熔点为0,℃,, 固态苯的热容为122.59 J,,mol,－1,,,K,－1,, 苯的熔化热为9916 J,,mol,－1,, 冰的熔化热为6004 J,,mol,－1,.,,(应知水的定压比热容为,c,p,,=4.184 J g,－1,K,－1,.),27,′,,假设冰全部熔化, 苯全部凝固, 末态温度均为,t,. 在,恒压,101.325kPa和与外部,绝热,条件下进行内部的相变和变温过程:,0.5mol苯,(,l,),,0.5mol苯,(,s,),t,1,=5,℃,0.8mol冰,(,s,),,0.2mol水,(,l,),t,2,=0,℃,可逆相变,1mol苯(,s,),,t,1,=5,℃,1mol水(,l,),,t,2,=0,℃,1mol苯(,s,),,t,1mol水(,l,),,t,变温,10/24/2024,30,,10/24/2024,31,,28,① nRT=PV,P,2,=,②,△H= △H,1,+ △H,2,=0.1×25.104+0= 2.5104KJ,③ W=0 Q= △U= △H - △(PV)≈ △H –P,2,V,2,,,= (2.5104+25.662×10×10,-3,)KJ=2.2538KJ,<101.325kPa,0.1mol 液,,35.51℃,,101.325KPa,0.1mol 气,,35.51℃,,P,2,不可逆,,,相变,0.1mol 气,,35.51℃,,101.325KPa,①,②,10/24/2024,32,,,已知,298.15K,时液态水的标准摩尔生成吉布斯函数,,f,G,,m,(H,2,O,l),=－237.129 kJ,,mol,－1,.,298.15K,时水的饱和蒸气压为3.1663kPa . 求298.15K时水蒸汽的标准摩尔生成吉布斯函数,m,.,H,2,O,,(l),,100kPa,,f,G,,m,,′,H,2,O,,(l),,3.1663kPa,,G,1,H,2,O,,(g),,3.1663kPa,,G,2,,G,3,(1/2)O,2,(g),,100kPa,H,2,O,,(g),,100kPa,H,2,(g),,100kPa,,f,G,,m,恒温298.15K下,,,f,G,,m,,′,,=－237.129 kJ,,mol,－1,,G,2,= 0,,f,G,,m,,=,,G,1,+,,G,2,+,,G,3,+,,f,G,,m,,′,,,= －228.57 kJ,,mol,－1,33,H,2,(g) + (1/2)O,2,(g), H,2,O(g),10/24/2024,33,,,f,G,,m,(H,2,O,g) = ,f,G,,m,(H,2,O,l) +,G,,,G,1,H,2,O,,(l),,100kPa,H,2,O,,(l),,3.1663kPa,H,2,O,,(g),,3.1663kPa,,G,2,,G,3,H,2,O,,(g),,100kPa,,G,10/24/2024,34,,34,假设终态N,2,分压 =(120 – 101.325)kPa=18.675kPa,H,2,O,,(l),,101.325kPa,H,2,O,,(g),,72kPa,H,2,O,,(g),,101.325kPa,②,①,W= - P △V = -RT△n,,= -8.314×373.15×3 = -9.308kJ,△U = Q +W = 112.696kJ,N,2,,,120kPa,N,2,,48kPa,3,10/24/2024,35,,△A = △U - T△S = 112.696 -373.15×350.72×10,-3,=-18.176kJ,△G = △H - T△S= 122.004 -373.15×350.72×10,-3,=-8.868kJ,dA = -SdT - PdV,10/24/2024,36,,nH,2,O,,(g),,P,2,=101.325kPa,,t,,,, V,1,≠V,n H,2,O,,(g),,V=100dm,3,,P,1,=120kPa,,t=100 ℃,n,1,H,2,O,,(g),,n,2,H,2,O,,(l),,P,2,=101.325kPa,,V , t,②,①,不可逆,,相变,35,,n=P,1,V/RT=3.868mol n,1,=P,2,V/RT=3.266mol,,n,1,= 0.602mol,①理想气体恒温可逆变化,△H,1,＝0,△S,1,＝nRlnP,1,/P,2,-1,10/24/2024,37,,②可逆相变过程,△H,2,＝－n △,Vap,H＝-24.479 kJ,△S,2,= △H,2,-1,△H= △H,1,+ △H,2,= -24. 479 kJ,△S= △S,1,+ △S,2,= -60. 161J. K,-1,W=0 Q= △U= △H- △(PV)= △H -V△P=-22.449kJ,△G = △H- T△S=-2.030kJ,或 △G= △G,1,+ △G,2,= △G,1,＝ nRTlnP,1,/P,2,=-2.030kJ,△A = △U- T△S=-0.162kJ,错误：Q=Q,1,+Q,2,,,△S= Q/T=△U/T,10/24/2024,38,,36,100℃ H,2,O(g),,101.325 kPa,100℃ H,2,O(l),,101.325 kPa,120℃ H,2,O(l),,101.325 kPa,120℃ H,2,O(g),,101.325 kPa,不可逆相变,(2),可逆相变,(1),,(3),,dG = -SdT + VdP,△G = △H - T△S = -119.77kJ,10/24/2024,39,,37,H,2,O(l),,100kPa,,- 5℃,H,2,O(S),,100kPa,,- 5℃,△G,△S,H,2,O(g),,P,l,s,,- 5℃,△G,2,H,2,O(l),,P,l,s,,- 5℃,△G,1,H,2,O(g),,P,s,s,,- 5℃,△G,3,△G,4,H,2,O(S),,P,s,s,,- 5℃,△G,5,△G,1,≈0,△G,5,≈0,(凝聚态物质恒温变压),△G,1,= △G,4,=0,(可逆相变),,0℃ H,2,O(s),,100 kPa,0℃ H,2,O(l),,100 kPa,－5℃ H,2,O(l),,100 kPa,－5℃ H,2,O(s),,100 kPa,不可逆相变,,G,2,＝0,可逆相变,,,G,1,,,G,3,,10/24/2024,40,,,－5℃ H,2,O (s),,59.8 MPa,－5℃ H,2,O(l),,59.8 MPa,－5℃ H,2,O(l),,100 kPa,－5℃ H,2,O (s),,100 kPa,不可逆相变,,G,2,可逆相变,,,G,1,,G,3,,38,,G,=,,G,1,+,,G,2,+,,G,3,= -5.377k J,dG = -SdT + VdP,dG = VdP,,G,1,= V(l)(P,2,– P,1,)=(59.8×10,3,-100) ×10,3,/999.2=59748J,,G,3,= V(S)(P,1,– P,2,)= (100 -59.8×10,3,) ×10,3,/916.7=-65125J,,G,2,= 0,10/24/2024,41,,40,,r,G,m,=,,G,m,1,+,,G,m,2,+,,r,G,m,⊙,= 162.86k J ·mol,-1,,r,S,m,=(,,r,H,m,-,,r,G,m,)/T=,286.46J·k,-1,,mol,-1,,G,m,1,,G,m,2,本章完,10/24/2024,42,,在d,V,= 0的条件下, 式 d,U,=,T,d,S,－,p,d,V,除以d,p,可得:,在d,p,= 0的条件下, 式 d,H,=,T,d,S,+,V,d,p,除以d,V,可得:,44,10/24/2024,43,,由理想气体状态方程,pV = nRT,可知：,(2),对理想气体:,,10/24/2024,44,,,48,,已知水在77,℃,时的饱和蒸气压为41.847,kPa,, 试求:,,(1)表示蒸气压与温度关系的方程式中的常数,A,和,B,;,,,(2)水的摩尔蒸发热;,,(3)在多在压力下水的沸点为105,℃,.,,P,=121.041kPa,10/24/2024,45,,,水与氯仿的正常沸点分别为100,℃,和61.5,℃,, 摩尔蒸发热分别为40.67kJ,,mol,－1,和29.50kJ,,mol,－1,. 求二者具有相同饱和蒸气压时的温度.,假设二者的蒸发焓都与温度无关, 蒸气为理想气体, 正常沸点即,p,=101.325kPa下液体的沸点.,将,p,=101325Pa,,T,=373.15K及,,evp,H,m,(水) = 40670J,,mol,－1,代入上式得,C,1,= 24.6354,; 将,p,=101325Pa,,T,= 334.65K及,,evp,H,m,(氯仿)=29500J,,mol,－1,代入上式得,C,2,=22.1289,.,,两式相减, 可得,49,10/24/2024,46,,。