物理化学课件32

,1,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,3.4,熵变的计算,Q,系 ：,系统发生某变化时与环境交换的热。,(,实际交换的热,),1.,环境熵变的计算,系统,环境,T,，,p,pVT,变化,相变化,化学变化,恒压变温,恒容变温,可逆相变,不可逆相变,系统熵变的计算,液体、固体变温过程,气体变温过程,理想气体变温过程,2.,系统熵变的计算,封闭系统、,W,=0,热力学第三定律,液体、固体变温过程：,(1),单纯,pVT,变化过程熵变的计算,恒压变温过程：,变压变温过程：,压力变化不大时,忽略压力对熵的影响,理想气体,1,2,某过程,可逆过程,对混合物中的某一组分来说，公式中的,p,是该组分的,分压力,，体积为该气体实际占有的体积,(,混合气体的,总体积,，而不是分体积,),。,纯,pg,混合,pg,混合,pg,中某组分,(1) 有系统如以下图,容器及隔板绝热良好，抽去隔板后气体混合到达平衡，试求：混合过程的,例,1:,pg,A,p,1,=150kPa,T,1,=300K,V,A,1,=10m,3,pg,A,p,2,=150kPa,T,2,=300K,V,A,2,=5.0m,3,pg,A,p,=150kPa,T,=300K,V,=15m,3,同温同压的同种气体,，混合后，气体的状态没有改变，则系统的熵也没有改变。所以,mix,S,=,解：,(2) 有系统如以下图,pg,A,p,1,=150kPa,T,1,=300K,V,A,=10m,3,pg,B,p,2,=150kPa,T,2,=300K,V,B,=5.0m,3,容器及隔板绝热良好，抽去隔板后A、B混合到达平衡，试求：混合过程的,pg,A+B,p,=150kPa,T,=300K,V,=15m,3,解：,有系统如以下图,单原子,pg,A,p,A1,=150kPa,T,A1,=300K,V,A1,=10m,3,双原子,pg,B,p,B1,=300kPa,T,B1,=400K,V,B1,=5.0m,3,容器及隔板绝热良好，抽去隔板后A、B混合到达平衡，试求：混合过程的,(3),单原子,pg,A,p,A1,=150kPa,T,A1,=300K,V,A1,=10m,3,双原子,pg,B,p,B1,=300kPa,T,B1,=400K,V,B1,=5.0m,3,单原子,pg,A,+,双原子,pg,B,p,T,2,V,=15m,3,Q,=0,d,V,=0,单原子,pg,A,p,A1,=150kPa,T,A1,=300K,V,A1,=10m,3,双原子,pg,B,p,B1,=300kPa,T,B1,=400K,V,B1,=5.0m,3,单原子,pg,A,+,双原子,pg,B,p,T,2,V,=15m,3,Q,=0,d,V,=0,例,2,：,1mol,理想气体在恒温下体积增加,10,倍，求系统的熵变，环境的熵变。,设此过程为可逆过程,设为真空膨胀过程,pg,n,=1mol,p,1,V,1,pg,n,=1mol,p,2,V,2,=10,V,1,恒温,解：,pg,n=,1mol,p,1,V,1,pg,n=,1mol,p,2,V,2,=10,V,1,恒温可逆,恒温可逆过程,pg,n=,1mol,p,1,V,1,pg,n=,1mol,p,2,V,2,=10,V,1,恒温向真空膨胀过程,p,外,=0,d,T,=0,例,3,：,4mol,某双原子理想气体，由始态,600K,，,100kPa,绝热可逆压缩至压力为,1000kPa,，,问需要多少功？,Q,r,=0,pg,n=4mol,T,2,=,p,2,=1000kPa,pg,n=4mol,T,1,=600K,p,1,=100kPa,例,4,：某双原子理想气体,4mol,，由,600K,，,100kPa,的,始态经绝热压缩到,1000kPa,需消耗功,60kJ,求过程的,pg,n=4mol,T,1,=600K,p,1,=100kPa,pg,n=4mol,T,2,=,p,2,=1000kPa,Q,=0,(2),相变过程熵变的计算,可逆相变,可逆相变,: d,T,=0,d,p,=0,不可逆相变,设计过程,途径中的每一步必须可逆；,途径中每步,S,的计算有相应的公式可利用；,有计算每步,S,所需的热数据。,C,p,m,(l)=126.80J,mol,-1,K,-1,C,p,m,(s)=122.6J,mol,-1,K,-1,n=1mol C6H6l p1 =101.325kPa,T1=268.2K,n=1mol C6H6s p1 =101.325kPa,T1=268.2K,例5：求101.325kPa、-5时1mol液态苯凝固过程的H 和 S。苯的正常熔点为5 。,n=1mol C6H6l p1 =101.325kPa,T1=268.2K,n=1mol C6H6s p1 =101.325kPa,T1=268.2K,n=1mol C6H6l p2 =101.325kPa,T2=278.7K,n=1mol C6H6s p2 =101.325kPa,T2=278.7K,n=1mol C6H6l p1 =101.325kPa,T1=268.2K,n=1mol C6H6s p1 =101.325kPa,T1=268.2K,n=1mol C6H6l p2 =101.325kPa,T2=278.7K,n=1mol C6H6s p2 =101.325kPa,T2=278.7K,C,p,m,(l)=,111.46,J,mol,-1,K,-1,C,p,m,(g)=65.44J,mol,-1,K,-1,求,10mol,乙醇在正常沸点,(78.37 ),气化时的,S,。,n=10mol l p1=101.325kPa,T1=351.52K,n=10mol g p1=101.325kPa,T1=351.52K,例6：乙醇在25的饱和蒸气压为7.955kPa,摩,尔蒸发焓为42.59kJ mol-1,n=10mol l p1=101.325kPa,T1=351.52K,n=10mol g p1=101.325kPa,T1=351.52K,n=10mol l p2=7.955kPa,T2=298.15K,n=10mol g p2=7.955kPa,T2=298.15K,解：,例7: 在-10 ,101.325kPa下，1mol水结冰，计算该过,程的 (系)、 ，并判断该过程能否自发进展。,：水在-10, 101.325kPa下结冰放热5628J mol-1,-10水的饱和蒸汽压为285.7Pa,-10冰的饱和蒸汽压为260.0Pa,n,=1mol,H,2,O,(l),p,2,=285.7Pa,T,=263.15K,n,=1mol,H,2,O,(g),p,2,=285.7Pa,T,=263.15K,n,=1mol,H,2,O,(g),p,3,=260.0Pa,T,=263.15K,n,=1mol,H,2,O,(s),p,3,=260.0Pa,T,=263.15K,n,=1mol H,2,O(l),p,1,=101.325kPa,T,=263.15K,n,=1mol,H,2,O(,s),p,1,=101.325kPa,T,=263.15K,n,=1mol,H,2,O,(l),p,2,=285.7Pa,T,=263.15K,n,=1mol,H,2,O,(g),p,2,=285.7Pa,T,=263.15K,n,=1mol,H,2,O,(g),p,3,=260.0Pa,T,=263.15K,n,=1mol,H,2,O,(s),p,3,=260.0Pa,T,=263.15K,n=1mol H,2,O(l) p,1,=101.325kPa,T =263.15K,n=1mol,H,2,O(,s) p,1,=101.325kPa,T,1,=263.15K,n=1mol,H,2,O,(l),p,2,=285.7Pa,T =263.15K,n=1mol,H,2,O,(g),p,2,=285.7Pa,T =263.15K,n=1mol,H,2,O,(g),p,3,=260.0Pa,T =263.15K,n=1mol,H,2,O,(s),p,3,=260.0Pa,T =263.15K,水在,-10,101.325kPa,下,结冰放热,5628J,mol,-1,多步,混合,例8： 4mol某双原子理想气体，由始态600K，1000kPa依次经历以下过程：,绝热对抗600kPa恒定的环境压力，膨胀至平衡态再恒容加热至压力为800kPa,最后再绝热可逆膨胀至压力为500kPa的末态。,试求整个过程的,pg,n,=4mol,T,1,=600K,p,1,=1000kPa,pg,n,=4mol,T,3,=,p,3,=800kPa,pg,n,=4mol,T,2,=,p,2,=600kPa,d,V,=0,pg,n,=4mol,T,4,=,p,4,= 500kPa,Q,r,=0,Q,=0,p,外,p,2,/T,2,= p,3,/T,3,T2=,pg,n=4mol,T,1,=600K,p,1,=1000kPa,pg,n=4mol,T,3,=,p,3,=800kPa,pg,n=4mol,T,2,=,p,2,=600kPa,d,V,=0,pg, n=4mol,T,4,=,p,4,= 500kPa,Q,r,=0,Q,=0,p,外,T2=,V,2,=V,3,p,2,/T,2,= p,3,/T,3,pg,n=4mol,T,1,=600K,p,1,=1000kPa,pg,n=4mol,T,3,=,708.57K,p,3,=800kPa,pg,n=4mol,T,2,=,531.43K,p,2,=600kPa,d,V,=0,pg, n=4mol,T,4,=,619.53K,p,4,= 500kPa,Q,r,=0,Q,=0,p,外,Q,=0,p,外,d,V,=0,Q,r,=0,状态,1,状态,2,状态,3,状态,4,例9：有系统如以下图,单原子,pg,2mol,T,A1,=273.15K,双原子,pg,5mol,T,B1,=373.15K,容器及隔板绝热良好，抽去隔板后A、B混合到达平衡，试求：混合后的温度及混合过程的,100kPa,单原子,pg,A,p,A1,=,T,A1,=273.15K,双原子,pg,B,p,B1,=,T,B1,=373.15K,单原子,pg,A,+,双原子,pg,B,p,T,2,Q,=0,d,p,=0,100kPa,100kPa,100kPa,100kPa,单原子,pg,A,p,A1,=,T,A1,=273.15K,双原子,pg,B,p,B1,=,T,B1,=373.15K,单原子,pg,A,+,双原子,pg,B,p,T,2,Q,=0,d,p,=0,100kPa,100kPa,100kPa,100kPa,单原子,pg,A,p,A1,=,T,A1,=273.15K,双原子,pg,B,p,B1,=,T,B1,=373.15K,单原子,pg,A,+,双原子,pg,B,p,T,2,Q,=0,d,p,=0,100kPa,100kPa,100kPa,100kPa,*,C,p,m,(l)=126.80J,mol,-1,K,-1,C,p,m,(s)=122.6J,mol,-1,K,-1,n=1mol C6H6l p1 =101.325kPa,T1=268.2K,n=1mol C6H6s p1 =101.325kPa,T1=268.2K,例5：求101.325kPa、-5时1mol液态苯凝固过程的H 和 S,判断该过程能否自发进展。苯的正常熔点为5 ，摩尔融化焓,n=1mol C6H6l p1 =101.325kPa,T1=268.2K,n=1mol C6H6s p1 =101.325kPa,T1=268.2K,n=1mol C6H6l p2 =101.325kPa,T2=278.7K,n=1mol C6H6s p2 =101.325kPa,T2=278.7K,根据熵判据，该过程可以自发进展。,