参考理工典型例题分析

到体积为始态2倍的终态。′=Q=Q=CV,mCp,mRCpCV

3.5R1.4 TTV2 300K(2)04227.41 11V 11pp1202kPa 76.542S1W1U1nCV,m(T2T1

222.58.314Jmol1K1(227.4K3003018理想气体绝热、反抗等外压过程，因为pp，V WU (TT)5R(TT V Wp(VV)p(VV nRT

(2V1V1)RT2所 5R(TT)RT T5300K250 WU (TT V 58.314Jmol1K(250K300K)2079SnRlnV2V

V

TlnT 250 22.58.314JmolKln 28.314JmolKln300

3.947J想气体的内能是温度的函数，所以必然存在T2′>T2，相应的p2′>p2。例2有两系统如下图所示，已知两系统中气体A，B皆是理想气体，且CV,,m,A=1.5R，销钉销钉 ∆S且随气体B温度的变化，气体A也将相应发生变化。解；(1)上述理想气体混合过程(a)

1mol3008.314Jmol1K1100kPa0.025ABVBnBRTB 2mol4008.314Jmol1K1200kPa0.033BVVAVB0.025m30.033m30.058 Q0，W0，U nACV,m,A(TTA)nBCV,m,B(TTB)11.5R(T300K)22.5R(T400K) T376.92

Rln

CV

lnT1Rln0.058

11.5Rln376.923009.89J

Rln

CV

ln1Rln0.058

11.5Rln376.924006.83JSSASB16.72J Q0，UWUnACV,m,A(T2TA)nBCV,m,B(T2TB11.5R(T2300K)22.5R(T23006.5R(T2300Wp(VV)nRT

nB

B2R(T20.5 6.5R(T2300K)2R(T20.5T2264.7

CV

ln

1.5Rln1.58.314Jmol1K1ln264.73001.56JV SBnBRlnB

BCV

ln

ln

Rln2(2.51)8.314Jmol1K1ln264.730028.314Jmol1K1ln2001004.24JSSASB2.68J500K400KW、∆S、∆A∆G。300KθSm20Jmol1K1θ∆S、∆A及∆G可用状态函数法直接根据始、终态计算。有关∆A∆G的计算方法归纳为两种：AU式 TST2S2

GHTSS2SS1AU GH A1dA1(SdTpdVATzp 11G2dG11

(SdTVd

Vdz

第法：过程（1）为理想气体等熵过程，即∆S1＝0即SnRlnV2 lnT2nRlnp1 lnT2 V

p 得13.58.314Jmol1K1ln 30018.314Jmol1K1ln100kPa400T2445.8方程式计算T2。CpCV

3.5R1.4 0p2 200kPa 0T2T1p 300K100kPa 1

445.8p

T3p500K400kPa448.6T2 445.8T2

nRln p4 T11mol8.314Jmol1K1ln10013.58.314Jmol1K1ln400300S4SS14.1JK120JK115.89JTST4S4T1S1400K15.89JK1300K20JK1356UnCV2079HnCp

T)1mol2.58.314Jmol1K1(400K30011T)1mol3.58.314Jmol1K1(400K300 291011AUTS2079J356J1723GHTS2910J356J2554WW1W2W3nCV,m(T1T2)nR(T4T31mol2.58.314Jmol1K1(300K445.81mol8.314Jmol1K1(400K500K)386240.1mol乙醚(C2H5)2O(l)35℃，101.325kPa10dm3的长颈瓶中瓶中nmolN2气，今将小玻璃瓶打醚完全知乙101.325kPa下的沸点为35℃，在此条件下乙醚的摩尔蒸发焓为25.104kJ·mol-1，试计算N2气的∆H、∆S分析：本题为等温下含有相变及理想气体混合过程中，N2气和乙醚各状态函数改变量的计解：(1)设乙醚全部气化，n乙醚p

n乙醚RT

0.1mol8.314JmolK1308K

25.62由于p乙醚101.325kPa乙醚101.325kPap乙醚101.325kPaT1=T1=皆未变，乙醚等温蒸发后，N2始态变为混合态，N2在混合态 等N

S

nRlnV2N 0N2

N NT2T2=T1T3=P乙醚

0.1mol25.104kJmol12.51041SH12.5104kJ8.147J1 308G1H2pS2n乙醚Rln 20.1mol8.314JmolK1p乙1.143J

101.32525.62G2H2T2S2T1S2308K1.143JK1352.2HH1H22.5104S乙醚S1S29.29JGG1G2352.2510mol理想气体在等温下通过：(1)可逆膨胀，(2)2倍，解：(1)可逆膨胀，系统的熵变

nRln

10mol8.314Jmol1K1ln57.6JRRS 57.6JKTSS系S(2)向真空膨胀，此为不可逆过程，系统的熵变不能用过程的热温商计算，但这两个过程的S2S157.6JSS隔S系S环57.6J设混合前后压力均为100kPa，氢气和氮气可视为理想气体。V倍V nR 0.5mol8.314Jmol1K1ln22.9J 1V nR 0.5mol8.314Jmol1K1ln22.9J 1Smix5.8J气体的温度从400K上升到600K，计算该过程的熵变。p、V、T均变的过程，计算熵变要分步进行，根据已知条件可以套用SnRlnV2

V

ln1mol8.314Jmol1K1

501001mol38.314Jmol1K1ln60020.706J

400例8求10mol水在373K及101.325kPa蒸发过程的熵变，以及系统的熵变。若保持压373K40.68kJ·mol1，水和水蒸气75.3J·mol1·K133.6J·mol1·K1。解：(1)在沸点时的气－液两相相变是可逆过程，系统的熵变：SvapT

373

1.09kJ QR10mol40.68kJmol1 373

1.09kJ S＝0，系统处于两相平衡状态H2O(g,300K,H2H2O(g,300K,H2O(l,300K, H2H2O(g,373K,H2O(l,373K,SS1S2

p

ln

10mol75.3Jmol1K1ln300164J

p,m

ln

73.18JS

vapT

373

1.09kJS91.91J 均在等压下进行的，所以QpH，系统与环境交换的热效应

HH1H2H

T2nCp1

(H

dT10mol75.3Jmol1K1(373K30054.97H2nvapH10mol40.68kJmol1400.68

T2nCp,m(HO,g)dT10mol33.6Jmol1K1(300K24.53HH1H2H3ST

431.12kJ300

1.44kJS ＝1.35kJS因系统的熵变S0，因此该过程不能自发进行91molH2O（l298