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16. 298.15Kʱ,C(ʯī)µÄ±ê׼Ħ¶ûȼÉÕìÊÓÚCO2 (g)µÄ±ê׼Ħ¶ûÉú³ÉìÊÖµÏàµÈ¡££¨ £© 17.Èôijһϵͳ´Óһʼ̬³ö·¢¾Ò»Ñ»·¹ý³ÌÓֻص½Ê¼Ì¬£¬ÔòϵͳµÄÈÈÁ¦Ñ§ÄܵÄÔöÁ¿ÎªÁã¡££¨ £© Ñ¡ÔñÌâ´ð°¸:
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 Ìî¿ÕÌâ´ð°¸:
1. W3>W1>W2,£»2. ¡÷H=0£»3. 1.9kJ/mol£»4. 30 kJ, 50kJ£»5. >0,>0,=0,>0£»6. ¡÷T=0,¦²¦ÃB(g)=0£»7. R;8.2.5V; 9. 0; 10. 150kPa; 11. 7.0¡Á10-3m3; 12. ÔÚһϵÁÐÎÞÏÞ½Ó½üƽºâÌõ¼þϽøÐеĹý³Ì,³ÉΪ¿ÉÄæ¹ý³Ì; 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?HµÍÓÚ; 21. = = = = ÅжÏÌâ´ð°¸:
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1. 10molÀíÏëÆøÌåÓÉ25¡æ¡¢106PaÅòÕ͵½25¡æ¡¢105Pa, Éè¹ý³ÌÓÉ(1) ×ÔÓÉÅòÕÍ£»(2)·´¿¹ºãÍâѹ105PaÅòÕÍ£»(3) ºãοÉÄæÅòÕÍ¡£·Ö±ð¼ÆËãÉÏÊö¸÷¹ý³ÌµÄW,Q,¡÷UºÍ¡÷H¡£ 10 mol pg whose initial state is T1=25¡æ,p1=106Pa,through the following different paths expands to final state T2=25¡æ,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.
´ð°¸£º(1)0,0,0,0;(2) -22.3kJ,22.3kJ,0,0;(3) -57.1kJ,57.1kJ,0,0
2. 2molµ¥Ô×ÓÀíÏëÆøÌ壬ÓÉ600K,1.00Mpa·´¿¹ºãÍâѹ100kPa¾øÈÈÅòÕ͵½Æ½ºâ¡£¼ÆËã¸Ã¹ý³ÌµÄW,Q,¡÷UºÍ¡÷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.
´ð°¸£º-5.39kJ, 0,-5.39kJ, -8.98kJ 3. ÔÚ298.15K¡¢6¡Á101.3kPaѹÁ¦Ï£¬1molµ¥Ô×ÓÀíÏëÆøÌå½øÐоøÈÈÅòÕÍ£¬×îÖÕѹÁ¦Îª101.3kPa,ÈôΪ(1)¿ÉÄæÅòÕÍ£»(2)·´¿¹ºãÍâѹ101.3kPaÅòÕÍ¡£ÇóÉÏÊö¶þ¾øÈȹý³ÌµÄÆøÌåµÄ×îÖÕζÈÒÔ¼°W,¡÷UºÍ¡÷H¡£
1 mol single atomic perfect gas whose initial state is T1=298.15K,p1=6¡Á101.3kPa, adiabatically expends through the following two different paths to final state p2=101.3kPa . Calculate T2,and W,¡÷U, ¡÷H of each paths.
´ð°¸£º(1)145.6K;-1.902kJ, -1.902kJ, -3.171kJ;(2) 198.8K;-1.239kJ, -1.239kJ, -2.065kJ 4. 2molijÀíÏëÆøÌ壬Cp,m?3.5R¡£ÓÉʼ̬100kPa,50dm3,ÏȺãÈݼÓÈÈʹѹÁ¦Éý¸ßÖÁ200kPa,ÔÙºãѹÀäȴʹÌå»ýËõСÖÁ25dm3,ÆäÕû¸ö¹ý³ÌµÄW,Q,¡÷UºÍ¡÷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. ´ð°¸: Q=-5.00 kJ, W=5.00kJ;¡÷U=0;¡÷H=0
5. 4molijÀíÏëÆøÌ壬Cp,m?3.5R¡£ÓÉʼ̬100kPa,100dm3,ÏȺãѹ¼ÓÈÈʹÌå»ýÔö´óµ½150dm3,ÔÙºãÈݼÓÈÈʹѹÁ¦Ôö´óµ½150kPa,ÇóÕû¸ö¹ý³ÌµÄW,Q,¡÷UºÍ¡÷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. ´ð°¸: Q=23.75 kJ, W=-5.00kJ;¡÷U=18.75kJ;¡÷H=31.25kJ
6. 5molË«Ô×ÓÀíÏëÆøÌå´Óʼ̬300K£¬200kPa,ÏȺãοÉÄæÅòÕ͵½Ñ¹Á¦Îª50kPa,ÔÙ¾øÈÈ¿ÉÄæѹËõµ½Ä©Ì¬Ñ¹Á¦200kPa¡£Çóĩ̬ζÈT¼°Õû¸ö¹ý³ÌµÄQ,W,¡÷UºÍ¡÷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. ´ð°¸£ºT=445.80K;Q=17.29kJ; W=-2.14kJ;¡÷U=15.15kJ,¡÷H=21.21kJ
7. 2molµ¥Ô×ÓÀíÏëÆøÌåA,3molË«Ô×ÓÀíÏëÆøÌåBÐγɵÄÀíÏëÆøÌå»ìºÏϵͳ£¬ÓÉ350K,72.75dm3µÄʼ̬£¬·Ö±ð¾ÏÂÁйý³Ì£¬µ½´ï¸÷×ÔµÄƽºâĩ̬£º (1) ºãοÉÄæÅòÕÍÖÁ120dm3
(2) ºãΡ¢Íâѹºã¶¨Îª121.25kPaÅòÕÍÖÁ120dm3 (3) ¾øÈÈ¿ÉÄæÅòÕÍÖÁ120dm3
(4) ¾øÈÈ¡¢·´¿¹121.25kPaµÄºã¶¨ÍâѹÖÁƽºâ¡£ ·Ö±ð¼ÆËã¸÷¹ý³ÌµÄQ,W,¡÷UºÍ¡÷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. ´ð°¸£º(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. 2molµ¥Ô×ÓÀíÏëÆøÌ壬´Ó273K, 206.6kPaµÄʼ̬ÑØpV?1?1?pV11µÄ;¾¶¿ÉÄæ¼ÓÈÈ
µ½405.2kPaµÄĩ̬£¬Çó´Ë¹ý³ÌµÄQ,W,¡÷UºÍ¡÷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£¬