物理化学电子教案第二章(下) 热力学第一定律及其应用 环境 surroundings 无物质交换 封闭本系 closed system △U=Q+W 能量交换 4上一内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 物理化学电子教案—第二章（下） = + U Q W
Adiabatic process In Fig Pv T In reversible process the work A() done (below the line ab) is B/p: V,) big than the work done(below the line ac)in adiabatic Clp: v,) reversible process Slope in ab: 图1.6绝热可逆过程(AC)与 等温可逆过程(AB)功的图解(示意图) Slope in ac Os==yp yy>I 4上一内容下一内容◇回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 Adiabatic process In Fig.PV: Slope in AB： ( )T p p V V = − Slope in AC： ( )S p p V V = − In reversible process the work done （below the line AB）is big than the work done (below the line AC) in adiabatic reversible process. 1 T
Fig. adiabatic and isothermal rev. process In an isothermal p expansion heat continuously flows B/: Vy) Into the system, and so the e pressure a C(p: V, does not fall as much as in a the ermally isolated adiabatic F 图1.6绝热可逆过程(AC)与 expansion 等温可逆过程(AB)功的图解(示意图) 4上一内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 Fig. adiabatic and isothermal rev. process In an isothermal expansion heat continuously flows into the system, and so the pressure does not fall as much as in a thermally isolated ,adiabatic expansion
Work in adiabatic process (1)work in pg. a. r. process K y (P=K) K 1 ecause P,li=p,l=K therefore W=p22-p, V nR(T-T) 4上一内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 Work in adiabatic process （1）work in pg. a. r. process. 2 1 = d V V K V V − 1 1 2 1 = 1 1 ( ) (1 ) K V V − − − − − therefore 2 2 1 1 = 1 p V pV W − − p V p V K 1 1 2 2 because = = 2 1 d V V W p V = − ( ) pV K = 2 1 ( ) 1 nR T T − = −
a better calculation method (2)another way to calculate W=△U=2CdT C(2-7)(设G与7无关) This equation is true for all adiabatic expansions or contractions involving a perfect gas or not, reversible or not 4上-内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 A better calculation method （2）another way to calculate W U = This equation is true for all adiabatic expansions or contractions involving a perfect gas or not, reversible or not. 2 1 = ( ( ) C T T V C T V − 设 与 无关） 2 1 d T V T = C T
2.7 real gas Joule- Thomson experiment Joule在1843年所做的气体自由膨胀实验是不够精 确的,1852年 Joule和 Thomson设计了新的实验,称为 throttling process 在这个实验中,使人们对实际气体的U和H的性质 有所了解,并且在获得低温和气体液化工业中有重要 应用。 4上一内容下一内容◇回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 2.7 real gas Joule-Thomson experiment Joule在1843年所做的气体自由膨胀实验是不够精 确的，1852年Joule和Thomson 设计了新的实验，称为 throttling process。 在这个实验中，使人们对实际气体的U和H的性质 有所了解，并且在获得低温和气体液化工业中有重要 应用
throttling process The Joule- Thomson expansion压缩区|多北参膨胀区 consists of allowing a gas to expand through a porous plug p, V,T. from a region of higher 焦耳一汤姆逊实验(1) pressure to a region of lower ressure as depicted in Fig 压缩区 多孔塞 膨胀区 pA The process is carried out steadily and adiabatically 焦耳一汤姆逊实验(2) 4上一内容下一内容◇回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 throttling process The Joule-Thomson expansion consists of allowing a gas to expand through a porous plug from a region of higher pressure to a region of lower pressure as depicted in Fig. The process is carried out steadily and adiabatically
throttling process 压缩区多孔塞膨胀区压缩区 多孔塞‖膨胀区 P 焦耳一汤姆逊实验(1) 焦耳一汤姆逊实验(2) When the flow is sufficiently slow, the gas has well defined pressure and temperature on both sides of the restriction 4上一内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 throttling process When the flow is sufficiently slow,the gas has well defined pressure and temperature on both sides of the restriction
U and in the throttling process adiabatic 2 =0, then U7,-U1=△=W Left, Surroundings compresses gaseous system work, system got W=-P△=n(△=0-1=-71) Right gaseous system expansion k surr. got 2=-2△=-P2V2(△=12-0=12) 4上一内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 U and H in the throttling process W p V 1 1 = − Left, Surroundings compresses gaseous system: work, system got： Adiabatic Q=0 ，then： U U U W 2 1 − = = Right, gaseous system expansion： work , surr. got: W p V 2 2 = − 1 1 1 1 = − = − pV V V V ( =0 ) 2 2 2 2 = − − = p V V V V ( = 0 )
U and in the throttling process the sum of work W=n+w2=p,i-p2V2 then C2-U1=n1-P2V2 72 tpV 2 Throttling process is const -enthalpy process It is isenthalpic 4上一内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 U and H in the throttling process the sum of work: W W W pV p V = + = − 1 2 1 1 2 2 then U U pV p V 2 1 1 1 2 2 − = − Throttling process is const.-enthalpy process! H H 2 1 = U p V U pV 2 2 2 1 1 1 + = + It is isenthalpic