1.系统的同一状态具有相同的体积. ( ) 2.系统的不同状态可具有相同的体积. ( ) 3.状态改变,系统的所有状态函数都改变. ( )
4.系统的某状态函数改变了,其状态一定改变. ( )
5.在恒容下,一定量理想气体,当温度升高时,其焓值将增加.（ ） 6.隔离系统中发生的实际过程,?U?0,W?0。（ ）
7.一定量的理想气体从相同的始态分别经等温可逆膨胀,绝热可逆膨胀达到具有相同压力的终态,终态的体积分别为V2和V2’,则V2 >V2’ .（ ）
8.1mol纯理想气体,当其T和U确定之后,其它的状态函数才有定值。（ ） 9.某实际气体经历一不可逆循环,该过程的Q=400J,则过程的W=400J。（ ） 10.在一定的温度下,?fHm金刚石,s,T?0 kJ/mol。 ( )
???11. 1mol、100℃、101325Pa下的水变成同温同压的水蒸气，若水蒸气视为理想气体，因过程温度不变，则ΔU=0,ΔH=0.( )
12.反应 CO(g) + 0.5O2 (g) = CO2 (g)的标准摩尔反应焓?rHm?T? 即是CO2的标准摩
?尔生成焓变?fHm?T?。（ ）
?13.pV?常数适用于非体积功为零的理想气体绝热可逆过程.（ ）
14.实际气体的节流膨胀过程是等焓过程。 （ ） 15.绝热、恒压非体积功为零的过程是等焓过程。（ ）
16. 298.15K时,C(石墨)的标准摩尔燃烧焓于CO2 (g)的标准摩尔生成焓值相等。（ ） 17.若某一系统从一始态出发经一循环过程又回到始态，则系统的热力学能的增量为零。（ ） 选择题答案:
1.D;2.B;3.A;4.C;5.C;6.A;7.A;8.A;9.D;10.A;11.B;12.D;13.B;14.C;15.A 填空题答案:
1. W3>W1>W2,；2. △H=0；3. 1.9kJ/mol；4. 30 kJ, 50kJ；5. >0,>0,=0,>0；6. △T=0,ΣγB(g)=0；7. R;8.2.5V; 9. 0; 10. 150kPa; 11. 7.0×10-3m3; 12. 在一系列无限接近平衡条件下进行的过程,成为可逆过程; 13. CO2(g), H2O(l); 14. -6.197kJ/mol; 15. 0; 0; 0; 0; 16. -4.989kJ; 17. >; 18. -128.0Kj/mol , -120.6kJ/mol; 19. 52.25kJ/mol; 20. ?r??T??,??p?H低于; 21. = = = = 判断题答案:
1.√；2.√；3.×；4.√；5.√；6.√；7.√；8.×；9.√；10.√；11.×；12.×；13.√；14.√；15.√;16.√；17.√ 四、计算题
1. 10mol理想气体由25℃、106Pa膨胀到25℃、105Pa, 设过程由(1) 自由膨胀；(2)反抗恒外压105Pa膨胀；(3) 恒温可逆膨胀。分别计算上述各过程的W,Q,△U和△H。 10 mol pg whose initial state is T1=25℃,p1=106Pa,through the following different paths expands to final state T2=25℃,p2=106Pa . Calculate W,Q,△U and △H of each paths. (1) Free expansion process.
(2) Expends against an external constant pressure Psu=105Pa . (3) Expends isothermal reversible.
答案：(1)0,0,0,0;(2) -22.3kJ,22.3kJ,0,0;(3) -57.1kJ,57.1kJ,0,0
2. 2mol单原子理想气体，由600K,1.00Mpa反抗恒外压100kPa绝热膨胀到平衡。计算该过程的W,Q,△U和△H。
2mol single atomic perfect gas adiabatically Expends against an external constant pressure Psu=100kPa from T1=600K，p1=1.00Mpa to equilibrium. calculate W，Q,△H， △U of this process.
答案：-5.39kJ, 0,-5.39kJ, -8.98kJ 3. 在298.15K、6×101.3kPa压力下，1mol单原子理想气体进行绝热膨胀，最终压力为101.3kPa,若为(1)可逆膨胀；(2)反抗恒外压101.3kPa膨胀。求上述二绝热过程的气体的最终温度以及W,△U和△H。
1 mol single atomic perfect gas whose initial state is T1=298.15K,p1=6×101.3kPa, adiabatically expends through the following two different paths to final state p2=101.3kPa . Calculate T2,and W,△U, △H of each paths.
答案：(1)145.6K;-1.902kJ, -1.902kJ, -3.171kJ;(2) 198.8K;-1.239kJ, -1.239kJ, -2.065kJ 4. 2mol某理想气体，Cp,m?3.5R。由始态100kPa,50dm3,先恒容加热使压力升高至200kPa,再恒压冷却使体积缩小至25dm3,其整个过程的W,Q,△U和△H。
2 mol perfect gas，with Cp,m?3.5R. From the initial state of 100kPa,50dm3 is firstly isochoricly heated to increase the pressure to 200kPa ,then isobaricly cooled to decrease the volume to 25dm3 . Calculate W, Q,△U, △H of the whole process. 答案: Q=-5.00 kJ, W=5.00kJ;△U=0;△H=0
5. 4mol某理想气体，Cp,m?3.5R。由始态100kPa,100dm3,先恒压加热使体积增大到150dm3,再恒容加热使压力增大到150kPa,求整个过程的W,Q,△U和△H。
4 mol perfect gas，with Cp,m?3.5R. From the initial state of 100kPa,100dm3 is firstly isobaricly heated to increase the volume to 150dm3 ,then isochoricly heated to increase the pressure to 150kPa .Calculate W, Q,△U, △H of the whole process. 答案: Q=23.75 kJ, W=-5.00kJ;△U=18.75kJ;△H=31.25kJ
6. 5mol双原子理想气体从始态300K，200kPa,先恒温可逆膨胀到压力为50kPa,再绝热可逆压缩到末态压力200kPa。求末态温度T及整个过程的Q,W,△U和△H。
5 mol double atomic perfect gas， from the initial state of 300K，200kPa is firstly isothermal reversible expended to 50kPa ,then adiabatic reversible compressed to the final pressure of 200kPa .Calculate final temperature T2 and Q,W,△U, △H of the whole process. 答案：T=445.80K;Q=17.29kJ; W=-2.14kJ;△U=15.15kJ,△H=21.21kJ
7. 2mol单原子理想气体A,3mol双原子理想气体B形成的理想气体混合系统，由350K,72.75dm3的始态，分别经下列过程，到达各自的平衡末态： (1) 恒温可逆膨胀至120dm3
(2) 恒温、外压恒定为121.25kPa膨胀至120dm3 (3) 绝热可逆膨胀至120dm3
(4) 绝热、反抗121.25kPa的恒定外压至平衡。 分别计算各过程的Q,W,△U和△H。
2 mol single atomic perfect gas A and 3 mol double atomic perfect gas B are mixed to form a mixture system. The mixture system from the initial state of 350K, 72.75dm3 through the following different paths reaches to its equilibrium state. Calculate Q,W,△U, △H of each paths.
(1) Isothermal reversible expands to 120dm3
(2) Isothermal expands against an external constant pressure of 121.25kPa to 120dm3 (3) Adiabatic reversible expands to 120dm3
(4) Adiabatic expands against an external constant pressure of 121.25kPa to equilibrium. 答案：(1) Q=-W=7282J,△U=△H=0 (2) Q=-W=5729J,△U=△H=0
(3)Q=0, W=△U=-6479.6J,△H=-9565.1 J (4) Q=0, W=△U=-3881J,△H=-5729.8 J 8. 2mol单原子理想气体，从273K, 206.6kPa的始态沿pV?1?1?pV11的途径可逆加热
到405.2kPa的末态，求此过程的Q,W,△U和△H。
2mol single atomic perfect gas, from initial state T1=273K，p1=206.6kPa through
?1Q,△H， △U of pV?1?pV11 reversible heated to final state p2=405.2kPa. Calculate W，
this process.
答案：Q=27.24kJ, W=-6.81kJ, △U=20.429kJ, △H=34.048kJ
9. 已知水(H2O,l)在100℃的饱和蒸气压p*?101.325kPa，在此温度、压力下水的摩尔蒸发焓?vapHm?40.668kJ?mol?1。求在100℃，101.325kPa下使1kg水蒸气全部凝结成液体水时的Q,W,△U和△H。设水蒸气适用理想气体状态方程。 Calculate the Q,W, △U,△H of the following process: 1kg H2O(g, 100℃,101.325kPa)→H2O(l, 100℃,101.325kPa). Given that: p*(H2O,100℃)=101.325kPa, △vapHm(H2O,100℃)= 40.668kJ·mol-1, Supposing the vapor obeys the state equation of perfect gas 答案：Q=△H=-2257kJ, W=172.2kJ, △U=-2085kJ 10. (1)1mol水在100℃，101.325kPa恒压蒸发为同温同压下的蒸气(假设为理想气体)吸热为40.67kJ/mol,求：上述过程的Q,W,△U和△H？(2)始态同上，当外压恒为50kPa时将水恒温蒸发，然后将此1mol, 100℃，50kPa的水蒸气恒温可逆加压变为末态100℃，101.325kPa的水蒸气，求此过程的Q,W,△U和△H？(3)将1mol水(100℃，101.325kPa)在真空中蒸发为同温同压的水蒸气，求过程的Q,W,△U和△H？ (1) 1mol water under 100℃ and 101.325kPa is vaporized at an external constant pressure to vapor with the same temperature and same pressure( Assume as perfect gas), the heat absorbed in this process is 40.67kJ/mol. Calculate the Q,W, △U,△H of process.
(2)The initial state is same as (1), firstly the water is isothermal vaporized at an external constant pressure of 50kPa, then the vapor is isothermal reversible compressed from 100℃，50kPa to the 100℃，101.325kPa . Calculate the Q,W, △U,△H of the process （3）1mol water under 100℃and 101.325kPa is vaporized isothermally to vacuum and is changed to vapor with the same temperature and same pressure. Calculate the Q,W, △U,△H of process.
答案：(1)Q=△H=40.67kJ, W=-3.102kJ, △U=37.57kJ;(2) Q=38.48kJ;△H=40.67kJ, W=-0.911kJ, △U=37.57kJ;(3) Q=△U =37.57kJ, W=0kJ, △H =40.67kJ 11. 2mol,60℃,100kPa的液态苯在恒外压下全部变为60℃，24kPa的蒸气，请计算该过程的Q,W,△U和△H？(已知40℃时，苯的蒸气压为24kPa，汽化焓为33.43kJ/mol,假定苯(l)和(g)的摩尔定压热容可近似看做与温度无关，分别为141.5J?mol?K94.12J?mol?K，忽略液体的体积)
2mol liquid benzene under 60℃and 100kPa is vaporized completely at an external constant pressure to vapor of 60℃and 24kPa. Calculate the Q,W, △U,△H of process.
?1?1?1?1及
(Given p*(C6H6,40℃)=24kPa, △vapHm(C6H6,40℃)= 33.43kJ·mol-1,
Cp,m?C6H6,l??141.5J?mol?1?K?1, Cp,m?C6H6,g??94.12J?mol?1?K?1,
bothCp,m?C6H6,l? and Cp,m?C6H6,g? don’t change with temperature, the volume of liquid can be omitted).
答案：Q=△H=64.96kJ, W=-5.54kJ, △U=59.42kJ
12. 已知C(石墨)及H2(g)在25℃时的标准摩尔燃烧焓分别为?393.51kJ?mol?1及
?285.84kJ?mol?1；水在25℃时的汽化焓为44.0kJ?mol?1，反应：C(石墨)＋
2H2O(g)→2H2 (g)+ CO2 (g)在25℃时的标准摩尔反应焓?rHm?298.15K?为多少？
???Given at 25℃ , ?cHm (C, graphite)= -393.51kJ/mol, ?cHm (H2, g)= -285.84kJ/mol;
△vapHm(H2O,25℃)= 40.0kJ·mol-1, Calculate ?rHm?298.15K?of the following reaction
?at 25℃.: C(graphite)＋2H2O(g)→2H2 (g)+ CO2 (g) 答案：?rHm?298.15K?＝90.17kJ?mol
??113. 气相反应A(g) +B(g)→Y(g)在500℃，100kPa进行。已知数据： 物质 A(g) B(g) Y(g) ???fHm?298.15K?/kJ?mol?1 25℃~500℃的Cp,m/J?mol?K 19.1 4.2 30.0 ??1?1-235 52 -241 ?试求?rHm?298.15K?、?rHm?773.15K?、?rUm?773.15K?
?For gas reaction A(g) +B(g)→Y(g) proceeds at 500℃，100kPa. The date is shown in the
following table: substance A(g) B(g) Y(g) ???fHm?298.15K?/kJ?mol?1 25℃~500℃的Cp,m/J?mol?K 19.1 4.2 30.0 ??1?1-235 52 -241 ?Calculate ?rHm?298.15K?、?rHm?773.15K?、?rUm?773.15K? of the reaction.
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