ÎïÀí»¯Ñ§¿¼ÊÔÌâ¿â

Òò

1*lVm,AºÜС£¬ËùÒÔ´¿AµÄÕôÆûѹpAËæÍâѹpµÄ±ä»¯ºÜС¡£½«ÉÏʽд³É RT? dln(pA/p)?1Vm,Adp RT*pA1?pA2£¬²¢ÊÓVm,lAΪ³£Êý£¬»ý·ÖÉÏʽ

µ±ÍâѹÓÉp1?p2ÒºÌåAµÄÕôÆûѹÓÉ

pA2 ?pA1dln(pA/µÃ

1P2p)?Vm,A?p1dp

RT?*lpA2Vm,Aln?(P2?P1)

pA1RTÓ¦Óøûý·Öʽ¿É½âÊ͵±ÍâѹΪpʱˮµÄ±¥ºÍÕôÆûѹÎÊÌâ¡£²éµÃË®µÄ

*Vml?18.016cm3?mol?1£¬ÎÞ¶èÐÔÆøÌåʱp1?pA1?3167Pa£¬ÓжèÐÔÆøÌåʱ

p2?p?105Pa£¬ÔòÉÏʽΪ

pA218.016?10?6ln?(105?3167) 31678.314?298pA2?3169Pa

?½á¹û˵Ã÷£º298Kʱ£¬ÍâѹΪpʱ´¿Ë®µÄ±¥ºÍÕôÆûѹÓëÎÞÍâѹʱ´¿Ë®µÄÕôÆûѹ»ù±¾Îޱ仯¡£

ÓÉÒÔÉÏÌÖÂÛ¿ÉÒÔÀí½â£¬µ±Íâѹp?pʱ£¬pB?pB£¬ËùÒÔÀíÏëÈÜÒºÖÐ×é·ÖBµÄ±ê׼̬»¯Ñ§ÊÆ

***??B(T,pB)??B(T,p)??B(T,p)

?***?B(T,pB)Ëù´¦µÄ״̬¿ÉÒÔÈÏΪÊÇζÈΪT¡¢Ñ¹Á¦Îªp?ʱ´¿ÒºÌåBµÄ״̬¡£

46. ÅжÏÏÂÁйý³ÌµÄQ¡¢W¡¢¡÷U¡¢¡÷H¡¢¡÷S¡¢¡÷GÖµµÄÕý¸º¡£ ( 1£©ÀíÏëÆøÌå×ÔÓÉÅòÕÍ¡£ ( 2£©Á½ÖÖÀíÏëÆøÌåÔÚ¾øÈÈÏäÖлìºÏ¡£

½â£º

¹ý³Ì Q W 0 0 ¡÷U 0 0 ¡÷H 0 0 ¡÷S > 0 >0 ¡÷G <0 <0 £¨1£© 0 £¨2£© 0

47¡¢ ˵Ã÷ÏÂÁи÷ʽµÄÊÊÓÃÌõ¼þ¡£ ( 1£© ¡÷G = ¡÷HÒ»T¡÷S£»£¨2£©dG £½Ò»SdT + Vdp £¨3£©-¡÷G = -W?

´ð£º¹«Ê½£¨1£©£º·â±ÕÌåϵµÄ¶¨Î¹ý³Ì

¹«Ê½£¨2£©£º²»×÷ÆäËû¹¦¡¢¾ùÏà¡¢´¿×é·Ö£¨»ò×é³É²»±äµÄ¶à×é·Ö£©µÄ·â±ÕÌåϵ ¹«Ê½£¨3£©£º·â±ÕÌåϵ¡¢¶¨Î¡¢¶¨Ñ¹µÄ¿ÉÄæ¹ý³Ì¡£

48¡¢ 298Kʱ£¬1mol ÀíÏëÆøÌå´ÓÌå»ý10dm3ÅòÕ͵½20dm3£¬¼ÆË㣨1£©¶¨Î¿ÉÄæÅòÕÍ£»

£¨2£©ÏòÕæ¿ÕÅòÕÍÁ½ÖÖÇé¿öÏ嵀 ¡÷G ½â£º £¨1£©

?G?nRTlnP2V10?nRTln1?1?8.314?298ln??1717.3J P1V220 £¨2£© ¡÷G = £­1717.3 J

49¡¢ ijµ°°×ÖÊÓÉÌìÈ»ÕÛµþ̬±äµ½ÕÅ¿ª×´Ì¬µÄ±äÐÔ¹ý³ÌµÄìʱä¡÷HºÍìØ±ä¡÷S·Ö±ðΪ

251.04kJ¡¤mol-1ºÍ753J¡¤K-1¡¤mol-1£¬¼ÆË㣨1£©298Kʱµ°°×±äÐÔ¹ý³ÌµÄ¡÷G£» (2) ·¢Éú±äÐÔ¹ý³ÌµÄ×îµÍζȡ£ ½â£º½«¡÷HºÍ¡÷S½üËÆ¿´×÷ÓëζÈÎÞ¹Ø

£¨1£©?G??H?T?S?251.04?298?753?10 £¨2£©T?

50¡¢ 298K ,P? Ï£¬1molǦÓëÒÒËáÍ­ÔÚÔ­µç³ØÄÚ·´Ó¦¿ÉµÃµç¹¦9183.87kJ£¬ÎüÈÈ216.35kJ,

¼ÆËã¡÷U¡¢¡÷H¡¢¡÷SºÍ¡÷G

½â£º ¡÷G = W? = £­ 9183.87kJ ¡÷S = Q / T = 216.35 / 298 = 726 J/K

¡÷U = Q + W = £­ 9183.87 + 216.35 = £­8967.52 kJ ¡÷H = ¡÷G + T¡÷S = £­8967.52 kJ 51¡¢ ¹ãÒå»¯Ñ§ÊÆ ?B?(²»ÊÇÆ«Ä¦¶ûÁ¿£¿ ´ð£º (

?3?26.646kJ

?H251040??333.4K ?S753?G?U?H?F)T,P,nZ?()S,V,nZ?()S,P,nZ?()T,V,nZʽÖÐÄļ¸Ïî?nB?nB?nB?nB?U?H?F)S,V,nZ¡¢()S,P,nZ¡¢()T,V,nZ²»ÊÇÆ«Ä¦¶ûÁ¿ ?nB?nB?nB52¡¢ ÓÉ 2.0 mol AºÍ1.5 mol B ×é³ÉµÄ¶þ×é·ÖÈÜÒºµÄÌå»ýΪ425cm-3£¬ÒÑÖªVB , m Ϊ

250.0cm-3¡¤mol-1£¬ÇóVA,m ¡£ ½â£º 425?2?VA,m?1.5?250.0 VA,m?25.0cm3?mol?1

53¡¢ 298K¼°P?Ï£¬½«1molҺ̬±½¼ÓÈ˵½x±½?0.2µÄ±½ºÍ¼×±½¹¹³ÉµÄÁ¿ºÜ´óµÄÈÜÒºÖУ¬

Çó¸Ã¹ý³ÌµÄ¡÷G ¡£

½â£º Éè±½ÓÃA±íʾ£¬¼×±½ÓÃB±íʾ

»ìºÏǰ G1?Gm,A?nAGA,m?nBGB,m »ìºÏºó G2?(nA?1)GA,m?nBGB,m

»ìºÏ¹ý³Ì ?G?G2?G1?GA,m?Gm,A??A??A?RTlnxA ?8.314?298?ln0.2??3.99kJ

54¡¢ 308K ʱ£¬±ûͪµÄ±¥ºÍÕôÆøÑ¹Îª4.3¡Á104 Pa£¬½ñ²âµÃx

ÂÈ·Â = 0.3

**?µÄÂÈ·ÂÒ»±ûͪÈÜÒºÕô

ÆøÖбûͪÕôÆøµÄ·ÖѹΪ2.7¡Á104 Pa £¬ÎÊ´ËÈÜÒºÊÇ·ñΪÀíÏëÈÜÒº£¿ ½â£º ÈôΪÀíÏëÈÜÒº£¬±Ø·ûºÏRaoult¶¨ÂÉ£¬ÔòÓÐ

*4 p±ûͪ?p±ûͪx±ûͪ?43000?(1?0.3)?3.01?10Pa

44 ÓÉÓÚp±ûͪ?3.01?10Pa?2.7?10Pa£¬Òò´Ë´ËÈÜÒº²»ÊÇÀíÏëÈÜÒº¡£

55¡¢ ÒÑÖª370.26K ´¿Ë®µÄÕôÆøÑ¹Îª91293.8Pa£¬ÔÚÖÊÁ¿·ÖÊýΪ0.03µÄÒÒ´¼Ë®ÈÜÒºÉÏ·½£¬Õô

Æø×ÜѹΪ101325Pa¡£¼ÆËãÏàͬζÈʱÒÒ´¼µÄÎïÖʵÄÁ¿·ÖÊýΪ0.02µÄË®ÈÜÒºÉÏ£º(1) Ë®µÄÕôÆø·Öѹ£» (2) ÒÒ´¼µÄÕôÆø·Öѹ¡££¨MH2O?18g?mol;MC2H5OH?46g?mol£© ½â£º °ÑÈÜÒº¿´×÷ÊÇÏ¡ÈÜÒº£¬ÓÃA±íʾˮ£¬ÓÃB±íʾÒÒ´¼

£¨1£© pA?pAxA,2?91293.8?(1?0.02)?89467.9Pa £¨2£© µ±

*?1?1mB3?0.03 £¬ ¼´ mB?mA ʱ

mA?mB97 xB,1?nBmB461???0.012

nA?nBmA?mB1?97?4618463?18 ËùÒÔ pA,1?p*AxA,1?91293.8?(1?0.012)?90202.2Pa pB,1?p×Ü?pA,1?101325?90202.2?11122.8Pa kx?pB,1xB,1?11122.8?926900

0.012 pB?kxxB?926900?0.02?18538Pa

56.ÒÑÖª370.26K ´¿Ë®µÄÕôÆøÑ¹Îª91293.8Pa£¬ÔÚĦ¶û·ÖÊýΪ0.03µÄÒÒ´¼Ë®ÈÜÒºÉÏ·½£¬ÕôÆø×ÜѹΪ101325Pa¡£¼ÆËãÏàͬζÈʱÒÒ´¼µÄĦ¶û·ÖÊýΪ0.02µÄË®ÈÜÒºÉÏ£º(1) Ë®µÄÕôÆø·Öѹ£» (2) ÒÒ´¼µÄÕôÆø·Öѹ¡££¨MH2O?18g?mol;MC2H5OH?46g?mol£© ½â£º £¨1£©

*pË®?pË®xË®?1?1 ?91293.8?£¨1-0.02£© ?89467.9£¨Pa£©£¨2£©

*p?pË®xË®?kxxÒÒ´¼?101325?91293.8??1?0.03??kx?0.03 ?kx?425667.1(Pa)pÒÒ´¼?kxxÒÒ´¼?425667.1?0.02 ?8513.3(Pa)57.ÒÑÖª370.26K ´¿Ë®µÄÕôÆøÑ¹Îª91293.8Pa£¬ÔÚÖÊÁ¿·ÖÊýΪ0.03µÄÒÒ´¼Ë®ÈÜÒºÉÏ·½£¬ÕôÆø×ÜѹΪ101325Pa¡£¼ÆËãÏàͬζÈʱÒÒ´¼µÄÖÊÁ¿·ÖÊýΪ0.02µÄË®ÈÜÒºÉÏ£º(1) Ë®µÄÕôÆø·Öѹ£» (2) ÒÒ´¼µÄÕôÆø·Öѹ¡££¨MH2O?18g?mol;MC2H5OH?46g?mol£© ½â£º°ÑÈÜÒº¿´×÷ÊÇÏ¡ÈÜÒº£¬ÓÃA±íʾˮ£¬ÓÃB±íʾÒÒ´¼ £¨1£©µ±

?1?1mB2?0.02£¬ ¼´ mB?mA ʱ

mA?mB98 xA?nAmA181???0.992

98?46nA?nBmAmB1??2?181846

ÁªÏµ¿Í·þ£º779662525#qq.com(#Ìæ»»Îª@)