±ÖмÓÈë0.01 mol NaCl£¬ÈܽâÍê±Ïºó£¬Á½Ö»ÉÕ±°´Í¬ÑùµÄËÙ¶ÈÀäÈ´½µÎ£¬Ôò ( )
(A) A±ÏȽá±ù (B) B±ÏȽá±ù
(C) Á½±Í¬Ê±½á±ù (D) ²»ÄÜÔ¤²âÆä½á±ùµÄ´ÎÐò
´ð£º(A)¡£Ï¡ÈÜÒºµÄÒÀÊýÐÔÖ»ÓëÁ£×ÓµÄÊýÁ¿Óйأ¬¶øÓëÁ£×ÓµÄÐÔÖÊÎ޹ء£B±ÄÚÈÜÈëµÄÊÇNaCl£¬NaClÔÚË®ÖнâÀ룬ÆäÁ£×ÓÊý¼¸ºõÊÇA±ÖеÄÁ½±¶£¬B±ÖÐÈÜÒºµÄÄý¹ÌµãϽµµÃ¶à£¬ËùÒÔA±ÏȽá±ù¡£
13£®ÔÚºãεIJ£Á§ÕÖÖУ¬·âÈëÒ»±ÌÇË®£¨A±£©ºÍÒ»±´¿Ë®(B±)£¬Ê¹Á½±µÄÒºÃæÏàͬ£¬½«²£Á§ÕÖ³é³ÉÕæ¿Õ¡£¾ÀúÈô¸Éʱ¼äºó£¬Á½±ÒºÃæµÄ¸ß¶È½«ÊÇ ( )
(A) A ±¸ßÓÚ B ± (B) A ±µÈÓÚ B ± (C) A ±µÍÓÚ B ± (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»¯ÎÂ¶È (D) ÎüÊÕ»ìÄýÍÁÖеÄË®·Ý
´ð£º(C)¡£»ìÄýÍÁÖмÓÈëÉÙÁ¿ÑÎÀàºó£¬Ê¹Äý¹ÌµãϽµ£¬·ÀÖ¹»ìÄýÍÁ½á±ù¶øÓ°Ï콨ÖþÎïµÄÇ¿¶È¡£
15£®ÑμîµØµÄÅ©×÷ÎﳤÊƲ»Á¼£¬ÉõÖÁ¿Ýή£¬ÆäÖ÷ÒªÔÒòÊÇ ( )
(A) ÌìÆøÌ«ÈÈ (B) ºÜÉÙÏÂÓê (C) ·ÊÁϲ»×ã (D) Ë®·Ö´ÓÖ²Îïϸ°ûÏòÍÁÈÀµ¹Á÷
´ð£º(D)¡£ ÑμîµØÖк¬ÑÎÁ¿¸ß£¬Ë®ÔÚÖ²Îïϸ°ûÖеĻ¯Ñ§ÊÆ´óÓÚÔÚÑμîµØÖеĻ¯Ñ§ÊÆ£¬Ë®·Ö»á´ÓÖ²Îïϸ°ûÏòÍÁÈÀÉø͸£¬Ê¹Å©×÷ÎﳤÊƲ»Á¼¡£ Î壮ϰÌâ½âÎö
1£® ÔÚ298 Kʱ£¬ÓÐÒ»¸öH2SO4(B)µÄÏ¡Ë®ÈÜÒº£¬ÆäÃܶÈΪ1.060 3?103 kg?m?3£¬
H2SO4µÄÖÊÁ¿·ÖÊýwB?0.094 7¡£ÒÑÖªÔÚ¸ÃζÈÏ£¬´¿Ë®µÄÃܶÈΪ997.1 kg?m?3¡£ÊÔ¼ÆËãH2SO4µÄ£º¢Ù ÖÊÁ¿Ä¦¶ûŨ¶ÈmB£¬¢Ú ÎïÖʵÄÁ¿Å¨¶ÈcBºÍ¢Û ÎïÖʵÄÁ¿·ÖÊý
xB¡£
½â£º¢Ù ÖÊÁ¿Ä¦¶ûŨ¶ÈÊÇÖ¸£¬ÔÚ1 kgÈܼÁÖк¬ÈÜÖʵÄÎïÖʵÄÁ¿£¬²éÔ×ÓÁ¿±íµÃH2SO4µÄĦ¶ûÖÊÁ¿MB?0.098 08 kg?mol?1¡£ÉèÈÜÒºµÄÖÊÁ¿Îª1.0 kg£¬
mB?nBm(B)/MB? m(A)m(A)0.0947?1.0 kg/0.098 08 kg?mol?1??1.067mol?kg?1
(1?0.0947)?1.0 kg ¢Ú ÎïÖʵÄÁ¿Å¨¶ÈÊÇÖ¸£¬ÔÚ1 dm3 ÈÜÒºÖк¬ÈÜÖʵÄÎïÖʵÄÁ¿£¬ÉèÈÜÒºÖÊÁ¿Îª1 kg
cB?nBm(B)/MB ?Vm£¨ÈÜÒº£©/?£¨ÈÜÒº£©0.094 7?1.0 kg/0.098 08 kg?mol?1 ?
1.0kg/1.060 3?103kg?m?3?3 ?1.02?4310?mol?m3 1.0?24?moldm £¨3£©ÉèÈÜÒºÖÊÁ¿Îª1.0 kg£¬xB?nB
nB?nAnB?0.094 7?1.0 kg/0.098 08 kg?mol?1?0.966 mol nA?(1?0.094 7)?1.0 kg/0.018 kg?mol?1?50.29 mol
xB?nB0.966 mol??0.018 8
nB?nA(0.966?50.29)mol2£®½«0.6 molµÄÒÒ´¼£¨B£©ºÍ0.4 molµÄË®£¨A£©»ìºÏµÃÒÒ´¼µÄË®ÈÜÒº£¬ÈÜÒºµÄÃܶÈΪ849.4 kg?m?3¡£ÒÑÖªÈÜÒºÖÐÒÒ´¼µÄƫĦ¶ûÌå»ýVB?57.5?10?6 m3?mol?1£¬ÊÔÇóÈÜÒºÖÐË®µÄƫĦ¶ûÌå»ýVA¡£ÒÑ֪ˮºÍÒÒ´¼µÄĦ¶ûÖÊÁ¿·Ö±ðΪ
MA?18 g?mol?1£¬MB?46 g?mol?1¡£
½â£º¸ù¾ÝƫĦ¶ûÁ¿µÄ¼ÓºÍ¹«Ê½£¬ÓÐ
V?n B £¨1£© nAVA?BV V?m?m(A)?m(B)AnMA??B???nMB £¨2£©
ÉÏÃæÁ½Ê½Ó¦¸ÃÏàµÈ£¬ËùÒÔÓÐ
(18?0.4?46?0.6)?10?3kg 0.4 mol?VA?(0.6?57.5?10)m?849.4 kg?m?3?635½âµÃ VA?1.61?8?103?m?1 olm3£®ÔÚ298 Kʱ£¬ÓдóÁ¿µÄ¼×±½£¨A£©ºÍ±½£¨B£©µÄҺ̬»ìºÏÎÆäÖб½µÄĦ¶û·ÖÊýxB?0.20¡£Èç¹û½«1 mol´¿±½¼ÓÈëÕâ»ìºÏÎïÖУ¬¼ÆËãÕâ¸ö¹ý³ÌµÄ?G¡£
*½â£ºÉè1 mol´¿±½µÄGibbs×ÔÓÉÄÜΪGm,B£¬ÔÚҺ̬»ìºÏÎïÖÐ1 mol±½µÄƫĦ
¶ûGibbs×ÔÓÉÄÜΪGB£¬ËùÇóµÄ?G¾ÍÊÇÕâ1 mol±½£¬ÔÚ¼ÓÈë»ìºÏÎïµÄÇ°ºó£¬Gibbs×ÔÓÉÄܵIJîÖµ£¬
?G?GB?G*m,?B??B??BRTlnx
B ?(8.314?298?ln0.20)J?mol?1??3.99 kJ?mol?1
4£®ÔÚ263 KºÍ100 kPaÏ£¬ÓÐ1 mol¹ýÀäË®Äý¹Ì³ÉͬΡ¢Í¬Ñ¹µÄ±ù¡£ÇëÓû¯Ñ§ÊƼÆËãÕâ¹ý³ÌµÄ?G¡£ÒÑÖªÔÚ263 K ʱ£¬H2O(l)µÄ±¥ºÍÕôÆøѹ
p*(H2O,l)?287 Pa£¬H2O(s)µÄ±¥ºÍÕôÆøѹp*(H2O,s)?259 Pa¡£
½â£º¹ýÀäË®½á±ùÊǸö²»¿ÉÄæ¹ý³Ì£¬¿ÉÒÔÉè¼ÆÒ»¸öʼ¡¢ÖÕ̬ÏàͬµÄ¿ÉÄæ¹ý³Ì¡£ÔÚ±£³ÖζȲ»±äµÄÇé¿öÏ£¬·ÖÈçÏÂÎå²½µÈοÉÄæ±äѹ¹ý³Ì½øÐУº
£¨1£©´Ó100 kPaµÈοÉÄ潵ѹÖÁH2O(l)µÄ±¥ºÍÕôÆøѹ£» £¨2£©ÔÚH2O(l)µÄ±¥ºÍÕôÆøѹÏ´ï³ÉÆø-ÒºÁ½Ïàƽºâ£» £¨3£©H2O(g)µÄµÈοÉÄæ±äѹ¹ý³Ì£»
£¨4£©ÔÚH2O(s)µÄ±¥ºÍÕôÆøѹÏ´ï³ÉÆø-¹ÌÁ½Ïàƽºâ £¨5£©µÈοÉÄæÉýѹÖÁ100 kPa¡£¼´
1) H2O(l,100 ?k(P?a?)2(4)HO(l,287 P2a)25)HO(s,?2(5?9? P2a)(2) HO(g,287 Pa)3) ?(??H2O(g,259 Pa)H O(s,100 kPa) µÚ£¨1£©£¬£¨5£©Á½²½ÊÇÄý¾ÛÏàµÄµÈοÉÄæ±äѹ¹ý³Ì£¬ÓÉÓÚÄý¾ÛÏàµÄ¿ÉѹËõÐÔºÜС£¬Ä¦¶ûÌå»ý²»´ó£¬ÊÜѹÁ¦µÄÓ°ÏìºÜС£¬ËùÒÔÕâÁ½²½µÄGibbs ×ÔÓÉÄܵı仯ֵ¿ÉÒÔºöÂÔ²»¼Æ¡££¨2£©£¬£¨4£©Á½²½ÊǵÈΡ¢µÈѹ¿ÉÄæÏà±ä£¬Gibbs ×ÔÓÉÄܵı仯ֵµÈÓÚÁã¡£ËùÒÔ£¬×ܵÄGibbs ×ÔÓÉÄܵı仯ֵ¾ÍµÈÓÚµÚÈý²½µÄGibbs ×ÔÓÉÄܵı仯ֵ£¬ ?G??G(3)??*pl*psps* Vmdp?RTlnpl*259 Pa???1?1 ??8.314?263?ln? J?mol??224.46 J?mol
287 Pa??ÏÖÔÚÌâÄ¿ÒªÇóÓû¯Ñ§ÊÆÀ´¼ÆË㣬ÆäʵµÀÀíÊÇÒ»ÑùµÄ£¬¼ÆËã¸ü¼òµ¥¡£ÒòΪ´¿×é
·ÖµÄ»¯Ñ§Êƾ͵ÈÓÚĦ¶ûGibbs ×ÔÓÉÄÜ£¬Ë®ÕôÆø»¯Ñ§ÊƵıê׼̬¿ÉÒÔÏàÏû£¬ËùÒÔ
?G??G(3)??(H2O,g,259 Pa)??(H2O,g,287 Pa)
259 Pa??224.4?6 J?1 mol ?RT?ln287 Pa5£® ÒºÌåAÓëÒºÌåB¿ÉÒÔÐγÉÀíÏëµÄҺ̬»ìºÏÎï¡£ÔÚ343 Kʱ£¬1 mol AºÍ2 mol BËùÐγɵĻìºÏÎïµÄÕôÆøѹΪ50.663 kPa£¬ÈôÔÚÈÜÒºÖÐÔÙ¼ÓÈë3 mol A£¬ÔòÈÜÒºµÄÕôÆøѹÔö¼Óµ½70.928 kPa£¬ÊÔÇó£º
**£¨1£©AºÍBÔÚ343 KʱµÄ±¥ºÍÕôÆøѹpAºÍpB¡£
£¨2£©¶ÔÓÚµÚÒ»ÖÖ»ìºÏÎÔÚÆøÏàÖÐA£¬BµÄĦ¶û·ÖÊýyAºÍyB¡£ ½â£º £¨1£©ÒºÌ¬»ìºÏÎïÉϵÄ×ÜÕôÆøѹµÈÓÚAºÍBµÄÕôÆøѹµÄ¼ÓºÍ£¬
?? p?pAxA?pBxB
12????pB? £¨1£© 50.663 kPa?pA3321??70.928 kPa?pA??pB? £¨2£©
33 ÁªÁ¢(1)£¬(2)ʽ£¬½âµÃ
**pA?91.19 kPa pB?30.40 kPa
191.19 k?PappxA3?0.6?(2) yA?A? pp50.663 kPa?A yB?1?yA?0.4
6£®ÔÚ293 Kʱ£¬±½(1)µÄÕôÆøѹÊÇ13.332 kPa£¬ÐÁÍé(2)µÄÕôÆøѹΪ2.6664 kPa£¬ÏÖ½«1 molÐÁÍéÈÜÓÚ4 mol±½ÖУ¬ÐγÉÀíÏëµÄҺ̬»ìºÏÎï¡£ÊÔ¼ÆË㣺 £¨1£©ÏµÍ³µÄ×ÜÕôÆøѹ¡£ £¨2£©ÏµÍ³µÄÆøÏà×é³É¡£
£¨3£©½«£¨2£©ÖеÄÆøÏàÍêÈ«ÀäÄýÖÁÒºÏ࣬ÔÙ´ïµ½ÆøҺƽºâʱ£¬ÆøÏàµÄ×é³É¡£
??½â£º £¨1£©p?p1?p2?p1x1?p2x2
41?? ??13.332??2.6664??kPa?11.199 kPa
55?? £¨2£©y1?p113.332?0.8kPa??0.952 4 p11.199kPa y2?1?y1?1?0.9524?0.047 6
£¨3£©½«ÉÏÊöÆøÏàÍêÈ«ÀäÄýÖÁÆø-Һƽºâʱ£¬ÐÂÒºÏàµÄ×é³ÉÓëÉÏÊöÆøÏàµÄ×é
³ÉÏàͬ¡£
??y1 x2??y2 x1????p1??p2??p1??p2? p×Üx1x2