µÚ ËÄ Õ ¶à×é·ÖÈÈÁ¦Ñ§
ºÏGibbs×ÔÓÉÄÜϽµ,¼´£º
?mixV?0, ?mixU?0, ?mixH?0, ?mixS>0, ?mixG<012.±±·½È˶¬Ìì³Ô¶³Àæ
Ç°£¬½«¶³Àæ·ÅÈëÁ¹Ë®ÖнþÅÝ£¬¹ýÒ»¶Îʱ¼äºó¶³ÀæÄÚ²¿½â¶³ÁË£¬µ«±íÃæ½áÁËÒ»²ã±¡±ù¡£ÊÔ½âÊÍÔÒò£¿´ð£ºÁ¹Ë®Î¶ȱȶ³Àæζȸߣ¬¿Éʹ¶³Àæ½â¶³¡£¶³
À溬ÓÐÌÇ·Ö£¬¹Ê¶³ÀæÄÚ²¿µÄÄý¹ÌµãµÍÓÚË®µÄ±ùµã¡£µ±¶³ÀæÄÚ²¿½â¶³Ê±£¬ÒªÎüÊÕÈÈÁ¿£¬¶ø½â¶³ºóµÄ¶³ÀæÄÚ²¿Î¶ÈÈÔÂÔµÍÓÚË®µÄ±ùµã£¬ËùÒÔ¶³ÀæÄÚ²¿½â
¶³ÁË£¬¶ø¶³Àæ±íÃæÉÏÈÔÄý½áÒ»²ã±¡±ù¡£¶þ¡¢¸ÅÄîÌâ
ÌâºÅ Ñ¡Ïî ÌâºÅ Ñ¡Ïî 1 C 9 B 2 A 10 D 3 B 11 C 4 C 12 A 5 B 13 A 6 D 14 C 7 B 15 D 8 B 1.2 mol AÎïÖʺÍ3 mol BÎïÖÊÔÚµÈΡ¢µÈѹÏ£¬»ìºÏÐγÉÀíÏëҺ̬»ìºÏÎ¸ÃϵͳÖÐAºÍBµÄƫĦ¶ûÌå»ý·Ö±ðΪ1.79¡Á10-5 m3?mol-1£¬2.15¡Á10-5 m3?mol-1£¬
Ôò»ìºÏÎïµÄ×ÜÌå»ýΪ£¨£©¡£
£¨A£©9.67¡Á10-5 m3£¨B£©9.85¡Á10-5 m3£¨C£©1.003¡Á10-4 m3£¨D£©8.95¡Á10-5 m3´ð£º£¨C£©ÔËÓÃƫĦ¶ûÁ¿µÄ¼¯ºÏ¹«Ê½£¨
V?nAVA?nBVB£©¡£
2.ÏÂÁÐƫ΢·ÖÖУ¬ÄܳÆΪƫĦ¶ûÁ¿µÄÊÇ£¨£©¡£
£¨A£©
??V???H?£¨B£© ??????nB?T£¬p£¬nC?C?B???nB?S£¬p£¬nC?C?B???G???S???X?X?£¨D£©´ð£º£¨A£©Æ«Ä¦¶ûÁ¿¶¨Òå¡£3.???????n?n?n?B?p£¬V£¬nC?C?B??B?T£¬H£¬nC?C?B??B?T£¬p£¬nC?C?B?£¨C£©
ÏÂÁÐƫ΢·ÖÖУ¬²»ÊÇ»¯Ñ§ÊƵÄÊÇ£¨£©¡£
£¨A£©
(??H??U)S,V,nC(C?B)£¨B£©???n?nB?B?T£¬p£¬n
C?C?B?£¨C£©
(?G?A£¨B£©¹ãÒåµÄ»¯Ñ§Ê½£º)T,p,nC(C?B)£¨D£©()T,V,nC(C?B)´ð£º
?nB?nB?U?H?A?G?B?()S,V,nC(C?B)?()S,p,nC(C?B)?()T,V,nC(C?B)?()T,p,nC(C?B)?nB?nB?nB?nB4.ÒÑÖª373 Kʱ£¬ÒºÌåAµÄ±¥ºÍÕôÆøѹΪ133.24 kPa£¬ÒºÌåBµÄ±¥ºÍÕôÆøѹΪ66.62 kPa¡£ÉèAºÍBÐγÉÀíÏëҺ̬»ìºÏÎµ±AÔÚÈÜÒºÖеÄĦ¶û·ÖÊýΪ0.5ʱ£¬ÔÚÆøÏàÖÐAµÄĦ¶û·ÖÊýΪ£¨£©¡£
£¨A£©1£¨B£©
12£¨C£©
23£¨D£©
1´ð£º£¨C£©ÓÃRoult¶¨ÂÉËã³ö×ÜÕôÆøѹ£¬ÔÙ°ÑAµÄÕôÆøѹ³ýÒÔ×ÜÕôÆøѹ¡£3?????1?xA??133.24?0.5?66.62??1?0.5??99.93kPap?pAxA?pBxB?pAxA?pB?pApAxA133.24?0.52yA????5.298 K£¬±ê׼ѹÁ¦Ï£¬±½ºÍ¼×±½ÐγÉÀíÏëҺ̬»ìºÏÎï¡£µÚÒ»·Ý»ìºÏÎïÌå»ýΪ2 dm£¬
pp99.9333
µÚ 29 Ò³ ¹² 90 Ò³ 2015-04-05£¨Óàѵˬ£©
µÚ ËÄ Õ ¶à×é·ÖÈÈÁ¦Ñ§
±½µÄĦ¶û·ÖÊýΪ0.25£¬±½µÄ»¯Ñ§ÊÆΪ?1£¬µÚ¶þ·Ý»ìºÏÎïµÄÌå»ýΪ1 dm3£¬±½µÄĦ¶û·ÖÊýΪ0.5£¬»¯Ñ§ÊÆΪ¦Ì2£¬Ôò£¨£©¡£
£¨A£©?1>¦Ì2£¨B£©?1<¦Ì2£¨C£©?1=¦Ì2£¨D£©²»È·¶¨´ð£º£¨B£©»¯Ñ§ÊÆÊÇƫĦ¶ûGibbs×ÔÓÉÄÜ£¬ÊÇÇ¿¶ÈÐÔÖÊ£¬Óë»ìºÏÎïµÄ×ÜÌå»ýÎ޹أ¬¶øÓë»ìºÏ
ÎïµÄŨ¶ÈÓйء£µÚÒ»·ÝµÄŨ¶ÈµÍÓÚµÚ¶þ·ÝµÄŨ¶È£¬¹Ê»¯Ñ§ÊÆС¡£6.ÔÚζÈTʱ£¬´¿ÒºÌåAµÄ±¥ºÍÕôÆøѹΪ
**£¬»¯Ñ§ÊÆΪ?A£¬²¢ÇÒÒÑÖªÔÚ´ópAÆøѹÁ¦ÏµÄÄý¹ÌµãΪ
Tf*£¬µ±AÖÐÈÜÈëÉÙÁ¿ÓëA²»ÐγɹÌ̬ÈÜÒºµÄÈÜÖʶøÐγÉΪϡÈÜҺʱ£¬ÉÏÊöÈýÎïÀíÁ¿·Ö±ðΪpA£¬?AºÍTf£¬Ôò£¨£©¡£
£¨A£©
******pA **£¨D£©´¿ÒºÌåAµÄ±¥ºÍÕôÆøѹºÍ»¯Ñ§ÊƱÈÏ¡ÈÜÒºÖеĴ󣬼ÓÈëÈÜÖʺó£¬Ï¡ÈÜTf*>Tf£¨D£©pA>pA£¬?A>?A£¬Tf*>Tf´ð£º ÒºµÄÄý¹Ìµã»áϽµ¡£7.ÔÚ298 Kʱ£¬AºÍBÁ½ÖÖÆøÌåµ¥¶ÀÔÚijһÈܼÁÖÐÈܽ⣬×ñÊØHenry¶¨ÂÉ£¬Henry³£Êý·Ö±ðΪkAºÍkB£¬ÇÒÖªkA>kB£¬Ôòµ±AºÍBѹÁ¦£¨Æ½ºâʱµÄ£©Ïàͬʱ£¬ÔÚÒ»¶¨Á¿µÄ¸ÃÈܼÁÖÐËùÈܽâµÄ¹ØϵΪ£¨£©¡££¨A£©AµÄÁ¿´óÓÚBµÄÁ¿£¨B£©AµÄÁ¿Ð¡ÓÚBµÄÁ¿£¨C£©AµÄÁ¿µÈÓÚBµÄ Á¿£¨D£©AµÄÁ¿ÓëBµÄÁ¿ÎÞ·¨±È½Ï´ð£º£¨B£©¸ù¾ÝHenry¶¨ÂÉ£¬µ±Æ½ºâѹÁ¦Ïàͬʱ£¬Henry³£Êý´óµÄÈܽâÁ¿·´¶øС¡£8.ÔÚ400 Kʱ£¬ÒºÌåAµÄÕôÆøѹΪ4¡Á104Pa£¬ÒºÌåBµÄÕôÆøѹΪ6¡Á104Pa£¬Á½Õß×é³ÉÀíÏëҺ̬»ìºÏÎƽºâʱÈÜÒºÖÐAµÄĦ¶û·ÖÊýΪ0.6£¬ÔòÆøÏàÖÐBµÄĦ¶û·ÖÊýΪ£¨£©¡£ £¨A£©0.60£¨B£©0.50£¨C£©0.40£¨D£©0.31´ð£º£¨B£©ÓÃRoult¶¨ÂÉËã³ö×ÜÕôÆøѹ£¬ÔÙ°ÑBµÄÕôÆøѹ³ýÒÔ×ÜÕôÆøѹ¡£9.ÔÚ50¡æʱ£¬ÒºÌåAµÄ±¥ ºÍÕôÆøѹÊÇÒºÌåB±¥ºÍÕôÆøѹµÄ3±¶£¬AºÍBÁ½ÒºÌåÐγÉÀíÏëҺ̬»ìºÏÎï¡£ÆøҺƽºâʱ£¬ÔÚÒºÏàÖÐAµÄĦ¶û·ÖÊýΪ0.5£¬ÔòÔÚÆøÏàÖÐBµÄĦ¶û·Ö ÊýΪ£¨£©¡£ £¨A£©0.15£¨B£©0.25£¨C£©0.50£¨D£©0.65´ð£º£¨B£©ÓÃRoult¶¨ÂÉËã³ö×ÜÕôÆøѹ£¬ÔÙ°ÑBµÄÕôÆøѹ³ýÒÔ×ÜÕôÆøѹ¡£10.298 K£¬±ê׼ѹÁ¦Ï£¬ Á½Æ¿º¬ÝÁµÄ±½ÈÜÒº£¬µÚһƿΪ2 dm3£¨ÈÜÓÐ0.5 molÝÁ£©£¬µÚ¶þƿΪ1 dm3£¨ÈÜÓÐ0.25 molÝÁ£©£¬ÈôÒÔ¦Ì1ºÍ¦Ì2·Ö±ð±íʾÁ½Æ¿ÖÐÝÁµÄ»¯Ñ§ÊÆ£¬Ôò£¨£©¡£ £¨A£©¦Ì1=10¦Ì2£¨B£©¦Ì1=2¦Ì2£¨C£©¦Ì1= 12¦Ì2£¨D£©¦Ì1=¦Ì2´ð£º£¨D£©»¯Ñ§ÊÆÊÇƫĦ¶ûGibbs×ÔÓÉÄÜ£¬ÊÇÇ¿¶ÈÐÔÖÊ£¬Óë»ìºÏÎïµÄ×ÜÌå»ýÎ޹أ¬¶øÓë »ìºÏÎïµÄŨ¶ÈÓйء£Á½·ÝµÄŨ¶ÈÏàͬ£¬¹Ê»¯Ñ§ÊÆÏàµÈ¡£11.ÔÚ273K£¬200kPaʱ£¬H2O£¨l£©µÄ»¯Ñ§ÊÆΪ¦Ì£¨H2O£¬l£©£¬H2O£¨s£©µÄ»¯Ñ§ÊÆΪ¦Ì£¨H2O£¬s£©£¬Á½ÕߵĴóС¹ØϵΪ£¨£©¡£ £¨A£©¦Ì£¨H2O£¬l£©>¦Ì£¨H2O£¬s£©£¨B£©¦Ì£¨H2O£¬l£©=¦Ì£¨H2O£¬s£© £¨C£©¦Ì£¨H2O£¬l£©<¦Ì£¨H2O£¬s£©£¨D£©ÎÞ·¨±È½Ï´ð£º£¨C£©Ñ¹Á¦´óÓÚÕý³£Ïà±ä»¯µÄѹÁ¦£¬Ñ¹Á¦Ôö´óÓÐÀûÓÚ±ùµÄÈÚ»¯£¬±ùµÄ»¯Ñ§ÊÆ´óÓÚË®µÄ»¯ ѧÊÆ¡£12.Á½Ö»ÉÕ±¸÷ÓÐ1 kgË®£¬ÏòA±ÖмÓÈë0.01 molÕáÌÇ£¬ÏòB±ÄÚÈÜÈë0.01 mol NaCl£¬Á½Ö»ÉÕ±°´Í¬ÑùËÙ¶ÈÀäÈ´½µÎ£¬ÔòÓУ¨£©¡££¨A£©A± ÏȽá±ù£¨B£©B±ÏȽá±ù£¨C£©Á½±Í¬Ê±½á±ù£¨D£©²»ÄÜÔ¤²âÆä½á±ùµÄÏȺó´ÎÐò´ð£º£¨A£©Ï¡ÈÜÒºµÄÒÀÊýÐÔÖ»ÓëÁ£×ÓÊýÓйأ¬¶øÓëÁ£×ÓµÄÐÔÖÊÎ޹ء£B ±ÄÚÈÜÈëNaCl£¬NaCl½âÀ룬ÆäÁ£×ÓÊý¼¸ºõÊÇA±ÖеÄÁ½±¶£¬B±µÄÄý¹ÌµãϽµµÃ¶à£¬ËùÒÔA±ÏȽá±ù¡£13.ÔÚºãγé¿ÕµÄ²£Á§ÕÖÖУ¬·âÈëÁ½±Òº ÃæÏàͬµÄÌÇË®£¨A±£©ºÍ´¿Ë®(B±)¡£¾ÀúÈô¸Éʱ¼äºó£¬Á½±ÒºÃæµÄ¸ß¶È½«ÊÇ£¨£©¡££¨A£©A±¸ßÓÚB±£¨B£©A±µÈÓÚB±£¨C£©A±µÍÓÚB±£¨D£© ÊÓζȶø¶¨´ð£º(A)´¿Ë®µÄ±¥ºÍÕôÆøѹ´óÓÚÌÇË®£¬´¿Ë®²»¶ÏÕô·¢£¬ÕôÆøÔÚº¬ÌÇË®µÄA±Ö⻶ÏÄý¾Û£¬ËùÒÔA±ÒºÃæ¸ßÓÚB±¡£ 14.¶¬¼¾½¨ÖþÊ©¹¤ÖУ¬ÎªÁ˱£Ö¤Ê©¹¤ÖÊÁ¿£¬³£ÔÚ½½×¢»ìÄýÍÁʱ¼ÓÈëÉÙÁ¿ÑÎÀ࣬ÆäÖ÷Òª×÷ÓÃÊÇ£¨£©¡££¨A£©Ôö¼Ó»ìÄýÍÁµÄÇ¿¶È£¨B£©·ÀÖ¹½¨ÖþÎï±»¸¯Ê´£¨C£© ½µµÍ»ìÄýÍÁµÄ¹Ì»¯Î¶ȣ¨D£©ÎüÊÕ»ìÄýÍÁÖеÄË®·Ý´ð£º£¨C£©»ìÄýÍÁÖÐ ¼ÓÈëÉÙÁ¿ÑÎÀàºó£¬Äý¹ÌµãϽµ£¬·ÀÖ¹»ìÄýÍÁ½á±ù¶øÓ°ÏìÖÊÁ¿¡£15.ÑÎ µÚ 30 Ò³ ¹² 90 Ò³ 2015-04-05£¨Óàѵˬ£© µÚ ËÄ Õ ¶à×é·ÖÈÈÁ¦Ñ§ ¼îµØµÄÅ©×÷ÎﳤÊƲ»Á¼£¬ÉõÖÁ¿Ýή£¬ÆäÖ÷ÒªÔÒòÊÇʲô£¨£©¡££¨A£©Ìì ÆøÌ«ÈÈ£¨B£©ºÜÉÙÏÂÓ꣨C£©·ÊÁϲ»×㣨D£©Ë®·Ö´ÓÖ²ÎïÏòÍÁÈÀµ¹Á÷´ð£º £¨D£©ÑμîµØÖк¬ÑÎÁ¿¸ß£¬Ë®ÔÚÖ²ÎïÖеĻ¯Ñ§ÊÆ´óÓÚÔÚÑμîµØÖеĻ¯ ѧÊÆ£¬Ë®·Ö»á´ÓÖ²ÎïÏòÍÁÈÀÉø͸£¬Ê¹Å©×÷ÎﳤÊƲ»Á¼¡£Èý¡¢Ï°Ìâ1. ÔÚ298 Kʱ£¬ÓÐH2SO4£¨B£©µÄÏ¡Ë®ÈÜÒº£¬ÆäÃܶÈΪ1.0603 ?10 3 kg¡¤m-3£¬£¬H2SO4£¨B£©µÄÖÊÁ¿·ÖÊýΪ0.0947¡£ÔÚ¸ÃζÈÏ´¿Ë®µÄÃܶÈΪ997.1 kg¡¤m-3¡£ÊÔ¼ÆËãH2SO4µÄ£¨1£©ÖÊÁ¿Ä¦¶ûŨ¶È£¨mB£©£»£¨2£©ÎïÖʵÄÁ¿Å¨¶È£¨cB£©£»£¨3£©ÎïÖʵÄÁ¿·ÖÊý£¨xB£©¡£½â£º£¨1£©ÖÊÁ¿Ä¦¶ûŨ¶ÈÊÇÖ¸1 kgÈܼÁÖк¬ÈÜÖʵÄÎïÖʵÄÁ¿£¬ÉèÈÜÒºÖÊÁ¿Îª100 g mB?0.0947?0.1 kg/MH2SO4(100?9.47)?10?3kg0.00947kg/0.09808kg?mol?1??1.067mol?kg?1£¨2£©ÎïÖʵÄÁ¿Å¨¶ÈÊÇÖ¸1 dmÈÜÒºÖк¬ÈÜÖʵÄÎïÖ浀 0.09053kg3 ??Á¿£¬ÉèÈÜÒºÖÊÁ¿Îª1 kg nm/MBcB?B?BVmsln/?1£¨3£©ÉèÈÜÒºÖÊÁ¿Îª1 kg 0.0947?1kg/0.09808kg?mol?1?3??1.024mol?dm1kg/1.0603?103kg?m?3 xB?nH2SO4nH2SO4?nH2OnH2SO4?0.0947?1 kg/0.09808 kg?mol?1?0.966 mol nH2O?(1?0.0947) kg/0.018 kg?mol?1?50.29 mol2.1 molË®£ÒÒ´¼ÈÜÒºÖУ¬Ë®µÄÎïÖʵÄÁ¿Îª 0.4mol£¬ÒÒ´¼µÄƫĦ¶ûÌå»ýΪ57.5¡Á10-6 m3?mol-1£¬ÈÜÒºµÄÃܶÈΪ849.4 kg?m-3£¬ÊÔÇóÈÜÒºÖÐË®µÄƫĦ¶ûÌå»ý¡£ÒÑ֪ˮºÍÒÒ´¼µÄĦ¶ûÖÊÁ¿·Ö±ðΪ18¡Á10-3 kg?mol-1ºÍ 46¡Á10-3 kg?mol-1¡£½â£ºÉèˮΪ A£¬ÒÒ´¼ÎªB£¬¸ù¾ÝƫĦ¶ûÁ¿µÄ¼¯ºÏ¹«Ê½£¬ÓÐV=nAVA+nBVB£¨1£© V?m??m??m???nAMA?nBMB??63£¨2£©Ôò£¨1£©=£¨2£©¼´ÓУº (18?0.4?46?0.6)?10?3kg0.4 mol?V??0.6 mol?57.5?10m?mol?849.4 kg?m?3?1B ½âµÃ V??1.618?10?5m3?mol?13.ÔÚ298Kʱ£¬ÓдóÁ¿µÄ¼×±½£¨A£©ºÍ±½£¨B£©µÄҺ̬»ìºÏÎÆäÖб½µÄĦ¶û·ÖÊýx=0.20¡£Èç¹û½«1mol ´¿±½¼ÓÈë´Ë»ìºÏÎïÖУ¬¼ÆËãÕâ¸ö¹ý³ÌµÄ¦¤G¡£½â£º¦¤G=G2-G1=¦Ì2(A£¬B)¨C[¦Ì1(A£¬B)+¦Ì*(B)]=[n¦ÌA+n(n+1)¦ÌB]-(n¦ÌA+n¦ÌB+ ???)=¦Ì-?B=£¨?B+RT lnx?BB µÚ 31 Ò³ ¹² 90 Ò³ 2015-04-05£¨Óàѵˬ£© µÚ ËÄ Õ ¶à×é·ÖÈÈÁ¦Ñ§ ?=8.314¡Á298¡ÁRT ln0.2=-3.99 kJ4.ÔÚ263KºÍ100kPaÏ£¬ÓÐ1mol¹ýÀäË®Äý¹Ì³ÉͬΡ¢Í¬Ñ¹µÄ±ù¡£ÇëÓû¯Ñ§ÊƼÆËã´Ë¹ý³ÌµÄ¦¤G¡£ÒÑÖªÔÚ263K?BB £©- ʱ£¬H2O£¨l£©µÄ±¥ºÍÕôÆûѹp*£¨H2O£¬l£©=278Pa£¬H2O£¨s£©µÄ±¥ºÍÕôÆûѹp*£¨H2O£¬l£©=259Pa¡£ ½â£ºÉèϵͳ¾5²½¿ÉÄæ¹ý³ÌÍê³É¸Ã±ä»¯£¬±£³Öζȶ¼Îª-10¡æ£¬ ¦¤G H2O£¨s£¬p¦È£© ¦¤G5 H2O£¨l£¬p¦È£© ¦¤G1 H2O£¨l£¬287Pa£© ¦¤G2 ¦¤G3 H2O£¨g£¬287kPa£© H2O£¨s£¬259kPa£© ¦¤G4 H2O£¨g£¬259kPa£© µÚ1£¬2£¬4£¬5²½µÄGibbs×ÔÓÉÄܵı仯ֵ¶¼¿ÉÒÔ²»¼Æ£¬Ö»¼ÆËãµÚÈý²½µÄGibbs×ÔÓÉÄܵı仯¡£ÏÖÔÚÌâÄ¿ÒªÇóÓû¯Ñ§ÊƼÆË㣬Æäʵ¸ü¼òµ¥£¬´¿×é·Ö µÄ»¯Ñ§Êƾ͵ÈÓÚĦ¶ûGibbs×ÔÓÉÄÜ£¬ËùÒÔ£º ?fusGm??(s)??(l)?nRTln*p2 *p1?1?8.314 J?K?1?mol?1?263 K?ln??224.46 J259 Pa287 Pa5.ÒºÌåAÓëÒºÌåBÐγÉÀíÏëҺ̬»ìºÏÎï¡£ÔÚ343 Kʱ£¬1 mol AºÍ2 mol BËù³É»ìºÏÎïµÄÕôÆøѹΪ50.663 kPa£¬ÈôÔÚÈÜÒºÖÐÔÙ¼ÓÈë3 mol A£¬ÔòÈÜÒºµÄÕôÆøѹÔö¼Óµ½70.928 kPa£¬ÊÔÇó£º £¨1£©pAºÍpB£» ??£¨2£©¶ÔµÚÒ»ÖÖ»ìºÏÎÆøÏàÖÐA£¬BµÄĦ¶û·ÖÊý¡£½â£º£¨1£© ??p?pAxA?pBxB 1221????50.663 kPa?pA??pB?70.928 kPa?pA??pB?ÁªÁ¢£¬Ê½£¬½âµÃp?A=91.19 kPa 3333pB=30.40 kPa ?1?pApAxA3?0.6y?1?y?0.46.293 Kʱ,±½£¨A£©µÄÕôÆøѹÊÇ13.332 ??£¨2£©yA?BApp50.663 kPa91.19 kPa?kPa£¬ÐÁÍ飨B£©µÄÕôÆøѹΪ2.6664 kPa£¬ÏÖ½«1 molÐÁÍéÈÜÓÚ4 mol±½ÖУ¬ÐγÉÀíÏëҺ̬»ìºÏÎï¡£ÊÔ¼ÆË㣺 £¨1£©ÏµÍ³µÄ×ÜÕôÆøѹ£» £¨2£©ÏµÍ³µÄÆøÏà×é³É£» µÚ 32 Ò³ ¹² 90 Ò³ 2015-04-05£¨Óàѵˬ£©