ÄϾ©´óѧÎïÀí»¯Ñ§Ñ¡ÔñÌâ´ð°¸ÍêÕû°æ ÏÂÔØ±¾ÎÄ

µÚÒ»Õ ÈÈÁ¦Ñ§µÚÒ»¶¨Âɼ°ÆäÓ¦ÓÃ

ÎﻯÊÔ¾í(Ò»)

1.ÎïÖʵÃÁ¿Îª£îµÃ´¿ÀíÏëÆøÌå,¸ÃÆøÌåÔÚÈçϵÃÄÄÒ»×éÎïÀíÁ¿È·¶¨Ö®ºó,ÆäËü״̬º¯Êý·½Óж¨Öµ¡£ ( £©

£¨£Á£© p (£Â) V £¨C) T£¬£Õ £¨D) T, p 2¡¢ ÏÂÊö˵·¨ÄÄÒ»¸öÕýÈ·? ( )

£¨A) ÈȾÍÊÇÌåϵÖÐ΢¹ÛÁ£×ÓÆ½¾ùƽ¶¯ÄܵÃÁ¿¶È £¨B£© ζȾÍÊÇÌåϵËù´¢´æÈÈÁ¿µÃÁ¿¶È (C£© ζȾÍÊÇÌåϵÖÐ΢¹ÛÁ£×ÓÆ½¾ùÄÜÁ¿µÃÁ¿¶È (£Ä£© ζȾÍÊÇÌåϵÖÐ΢¹ÛÁ£×ÓÆ½¾ùƽ¶¯ÄܵÃÁ¿¶È

3¡¢ ÓÐÒ»¸ßѹ¸ÖͲ,´ò¿ª»îÈûºóÆøÌåÅç³öͲÍ⣬µ±Í²ÄÚѹÁ¦ÓëͲÍâѹÁ¦ÏàµÈʱ¹Ø±Õ»îÈû£¬´ËʱͲÄÚζȽ«: £¨ )

(A£©²»±ä £¨B)Éý¸ß £¨£Ã£©½µµÍ (£Ä)ÎÞ·¨Åж¨

4¡¢ £± m£ïl £³£·3 K,±ê׼ѹÁ¦ÏµÃË®¾­ÏÂÁÐÁ½¸ö²»Í¬¹ý³Ì±ä³É£³73 K£¬±ê׼ѹÁ¦ÏµÃË®Æø£¬ £¨£±) µÈεÈѹ¿ÉÄæÕô·¢, (2£© Õæ¿ÕÕô·¢ ÕâÁ½¸ö¹ý³ÌÖй¦ÓëÈȵùØÏµÎª£º ( £© (£Á) £üW£±|> £ü£×2£ü Q1£¾ Q2 (B)£ü£×1£ü< |W2| £Ñ£±¡´ Q2 £¨C£© £üW1|= £üW£²| Q£±= Q2 £¨£Ä£©£üW£±|> |£×£²| Q1£¼ Q2 5¡¢ ºãÈÝÏ£¬Ò»¶¨Á¿µÃÀíÏëÆøÌ壬µ±Î¶ÈÉý¸ßʱÈÈÁ¦Ñ§Äܽ«£º( £©

£¨£Á£©½µµÍ (B£©Ôö¼Ó (C£©²»±ä (D)Ôö¼Ó¡¢¼õÉÙ²»ÄÜÈ·¶¨ £¶¡¢ ÔÚÌåϵζȺ㶨µÃ±ä»¯ÖУ¬ÌåϵÓë»·¾³Ö®¼ä: £¨ ) £¨£Á£© Ò»¶¨²úÉúÈȽ»»» £¨B) Ò»¶¨²»²úÉúÈȽ»»» (C£© ²»Ò»¶¨²úÉúÈȽ»»» £¨D) ζȺ㶨ÓëÈȽ»»»ÎÞ¹Ø

£·¡¢ Ò»¿ÉÄæÈÈ»úÓëÁíÒ»²»¿ÉÄæÈÈ»úÔÚÆäËýÌõ¼þ¶¼Ïàͬʱ£¬È¼ÉÕµÈÁ¿µÃȼÁÏ,Ôò¿ÉÄæÈÈ»úÍ϶¯µÃÁгµÔËÐеÃËÙ¶È£º( £©

£¨A£© ½Ï¿ì £¨B£© ½ÏÂý £¨£Ã) Ò»Ñù (D£© ²»Ò»¶¨

£¸¡¢ ʼ̬ÍêÈ«Ïàͬ(p£±£¬V1,£Ô1£©µÃÒ»¸öÀíÏëÆøÌåÌåϵ,ÓëÁíÒ»¸ö·¶µÂ»ªÆøÌåÌåϵ,·Ö±ð½øÐоøÈȺãÍâѹ(£ð0£©ÅòÕÍ¡£µ±ÅòÕÍÏàͬÌå»ýÖ®ºó£¬ÏÂÊöÄÄÒ»ÖÖ˵·¨ÕýÈ·?( £©

£¨A) ·¶µÂ»ªÆøÌåµÃÈÈÁ¦Ñ§ÄܼõÉÙÁ¿±ÈÀíÏëÆøÌå¶à (B£© ·¶µÂ»ªÆøÌåµÃÖÕ̬ζȱÈÀíÏëÆøÌåµÍ (C) ·¶µÂ»ªÆøÌåËù×öµÃ¹¦±ÈÀíÏëÆøÌåÉÙ (D£© ·¶µÂ»ªÆøÌåµÃìʱäÓëÀíÏëÆøÌåµÃìʱäÏàµÈ

9¡£¦¤H £½Qp £¬ ´ËʽÊÊÓÃÓÚÏÂÁÐÄĸö¹ý³Ì£º£¨ £©

£¨A) ÀíÏëÆøÌå´Ó1£°6 Pa·´¿¹ºãÍâѹ£±£°5 PaÅòÕ͵½£±0£µ Pa £¨B) 0¡æ £¬ 105 £Ð£á ϱùÈÚ»¯³ÉË®

(C£© µç½â CuS£Ï4Ë®ÈÜÒº (£Ä£© ÆøÌå´Ó£¨£²£¹8 K£¬ 1£°5 £Ða£©¿ÉÄæ±ä»¯µ½£¨373 £Ë, £±04 Pa£©

1£°. ÔÚ100¡æÓë25¡æÖ®¼ä¹¤×÷µÃÈÈ»ú,Æä×î´óЧÂÊΪ: ( )

(A£© £±0£° £¥ (£Â) 7£µ £¥ £¨C) 25 £¥ (D) £²£° £¥ 11¡£ ¶ÔÓÚ·â±ÕÌåϵ£¬ÔÚÖ¸¶¨Ê¼ÖÕ̬¼äµÃ¾øÈÈ¿ÉÄæÍ¾¾¶¿ÉÒÔÓÐ:( ) £¨A£© Ò»Ìõ £¨B£© ¶þÌõ £¨C) ÈýÌõ £¨£Ä£© ÈýÌõÒÔÉÏ

£±2£® ijÀíÏëÆøÌåµÃ¦Ã £½Cp£¯£Ãv £½£±¡¢40,Ôò¸ÃÆøÌåΪ¼¸Ô­×Ó·Ö×ÓÆøÌ壿 £¨ £© (A) µ¥Ô­×Ó·Ö×ÓÆøÌå (B) ˫ԭ×Ó·Ö×ÓÆøÌå £¨C) ÈýÔ­×Ó·Ö×ÓÆøÌå £¨D) ËÄÔ­×Ó·Ö×ÓÆøÌå £±£³¡£ ʵ¼ÊÆøÌå¾øÈȺãÍâѹÅòÕÍʱ£¬ÆäζȽ«£º £¨ £©

£¨A£© Éý¸ß £¨B) ½µµÍ £¨C£© ²»±ä £¨D£© ²»È·¶¨

14¡£ µ±ÒÔ5 mol H2ÆøÓë4 £í£ïl Cl2Æø»ìºÏ£¬×îºóÉú³É2 m£ï£ì HClÆø¡£ÈôÒÔÏÂʽΪ»ù±¾µ¥Ôª, H£²(g) + Cl2(g£© -¡ª-£­¡µ 2HCl(g£©,Ôò·´Ó¦½ø¶È¦ÎÓ¦¾ÍÊÇ£º ( £© (A) 1 mo£ì (B£© 2 mol (C£© 4 mo£ì £¨D£© 5 mol

15¡£ Óû²â¶¨ÓлúÎïȼÉÕÈÈQ£ð,Ò»°ãʹ·´Ó¦ÔÚÑõµ¯ÖнøÐУ¬Êµ²âµÃÈÈЧӦΪQv¡£ ¹«Ê½ £Ñp=Qv+¦¤nRT Öеæ¤nΪ:£¨ £©

£¨A£© Éú³ÉÎïÓë·´Ó¦Îï×ÜÎïÖʵÃÁ¿Ö®²î (B£© Éú³ÉÎïÓë·´Ó¦ÎïÖÐÆøÏàÎïÖʵÃÁ¿Ö®²î

£¨£Ã) Éú³ÉÎïÓë·´Ó¦ÎïÖÐÄý¾ÛÏàÎïÖʵÃÁ¿Ö®²î £¨D£© Éú³ÉÎïÓë·´Ó¦ÎïµÃ×ÜÈÈÈݲî 16.Äý¹ÌÈÈÔÚÊýÖµÉÏÓëÏÂÁÐÄÄÒ»ÖÖÈÈÏàµÈ: ( )

£¨A) Éý»ªÈÈ £¨B£© ÈܽâÈÈ £¨C£© Æû»¯ÈÈ (D£© ÈÛ»¯ÈÈ

1£·£® ÒÑÖª 1mol HClµÃÎÞÏÞÏ¡ÊÍÈÜÒºÓë £±m£ïl N£áO£È µÃÎÞÏÞÏ¡ÊÍÈÜÒºÔÚºãκãѹÏÂÍêÈ«·´Ó¦,ÈÈЧӦ¦¤£òH =¡ª55¡¢9 k£Ê£¯mol£¬Ôò 1m£ïl HN£Ï3µÃÎÞÏÞÏ¡ÊÍÈÜÒºÓë 1mol KOHµÃÎÞÏÞÏ¡ÊÍÈÜÒºÔÚºãκãѹÏÂÍêÈ«·´Ó¦µÃÈÈЧӦ¦¤rH Ϊ:£¨ £©

£¨£Á) £¾ -5£µ¡¢9 £ë£Ê/m£ïl (B£© < -55¡¢9 kJ/mol £¨C£© = £­55¡¢9 kJ£¯£íol (D) ²»ÄÜÈ·¶¨

1£¸. ÔÚ±ê׼ѹÁ¦Ï£¬1molʯīÓëÑõÆø·´Ó¦Éú³É1£íol¶þÑõ»¯Ì¼ µÃ·´Ó¦ÈÈΪ¦¤r£È £¬ÏÂÁÐÄÄÖÖ˵·¨¾ÍÊÇ´íÎóµÃ£¿ £¨ £©

(A£© ¦¤H ¾ÍÊÇ£ÃO2£¨g£©µÃ±ê×¼Éú³ÉÈÈ £¨B£© ¦¤H £½¦¤U (C) ¦¤H ¾ÍÊÇʯīµÃȼÉÕÈÈ £¨D) ¦¤U ¡´¦¤£È 19£® ¼ÆË㻯ѧ·´Ó¦µÃÈÈЧӦ,ÏÂÊö˵·¨ÄÄЩ¾ÍÊÇÕýÈ·µÃ? £¨ ) (£±£© ÔÚͬһËãʽÖбØÐëÓÃͬһ²Î±È̬µÃÈÈЧӦÊý¾Ý (2) ÔÚͬһËãʽÖпÉÓò»Í¬²Î±È̬µÃÈÈЧӦÊý¾Ý (3£© ÔÚ²»Í¬ËãʽÖпÉÓò»Í¬²Î±È̬µÃÈÈЧӦÊý¾Ý (£´£© ÔÚ²»Í¬ËãʽÖбØÐëÓÃͬһ²Î±È̬µÃÈÈЧӦÊý

£¨A£© 1£¬3 (£Â£© 2£¬4 (C) 1£¬4 £¨D£© 2,3 £²£°£® Cl2(g£©µÃȼÉÕÈÈΪºÎÖµ£¿ ( )

£¨£Á)H£Ãl£¨g)µÃÉú³ÉÈÈ £¨B)HCl£Ï3µÃÉú³ÉÈÈ

£¨C)HCl£Ï4µÃÉú³ÉÈÈ £¨£Ä£©Cl£²£¨g£©Éú³ÉÑÎËáË®ÈÜÒºµÃÈÈЧӦ

µÚÒ»Õ ÈÈÁ¦Ñ§µÚÒ»¶¨Âɼ°ÆäÓ¦ÓÃ

ÎﻯÊÔ¾í(¶þ)

£±. 1mol µ¥Ô­×Ó·Ö×ÓÀíÏëÆøÌå´Ó 298 K,20£°¡¢0 £ëPa ¾­Àú£º ¢Ù µÈΣ¬ ¢Ú ¾øÈÈ£¬ ¢Û µÈѹÈýÌõ;¾¶¿ÉÄæÅòÕÍ£¬Ê¹Ìå»ýÔö¼Óµ½Ô­À´µÃ2±¶£¬Ëù×÷µÃ¹¦·Ö±ðΪ£×£±,£×2£¬W3,ÈýÕߵùØÏµ¾ÍÊÇ£º ( )

(A) |W1£ü£¾|£×2£ü£¾£üW3£ü (B) |£×2|>£üW1|£¾|W3£ü

(C£© |W3£ü£¾£ü£×£²|>£ü£×£±| £¨£Ä£© |W£³|>£üW1|£¾£üW£²£ü 2¡¢ ÏÂÊö˵·¨ÄÄÒ»¸ö¾ÍÊÇ´íÎóµÃ£¿ ( ) (A) ·â±ÕÌåϵµÃ״̬ÓëÆä״̬ͼÉϵõãÒ»Ò»¶ÔÓ¦ (B£© ·â±ÕÌåϵµÃ״̬¼´¾ÍÊÇÆäÆ½ºâ̬

(C) ·â±ÕÌåϵµÃÈÎÒ»±ä»¯ÓëÆä״̬ͼÉϵÃʵÏßÒ»Ò»¶ÔÓ¦ (D) ·â±ÕÌåϵµÃÈÎÒ»¿ÉÄæ±ä»¯Í¾¾¶¶¼¿ÉÔÚÆä״̬ͼÉϱíʾΪʵÏß

3¡¢ ·²¾ÍÊÇÔÚ¹ÂÁ¢ÌåϵÖнøÐеñ仯£¬Æä¦¤UÓ릤HµÃÖµÒ»¶¨¾ÍÊÇ: £¨ ) (A£© ¦¤U £¾ £° £¬ ¦¤H ¡µ £° (£Â£© ¦¤U = 0 , ¦¤H = 0 (C£© ¦¤U ¡´ £° , ¦¤£È < 0 (£Ä) ¦¤U = 0 , ¦¤H²»È·¶¨

£´¡¢ ¡± ·â±ÕÌåϵºãѹ¹ý³ÌÖÐÌåϵÎüÊÕµÃÈÈÁ¿Q£ðµÈÓÚÆäìʵÃÔöÁ¿¦¤H ¡±£¬ÕâÖÖ˵·¨: £¨ £© £¨£Á) ÕýÈ· (B) ÐèÔö¼ÓÎÞ·ÇÌå»ý¹¦µÃÌõ¼þ (C£© Ðè¼Ó¿ÉÄæ¹ý³ÌµÃÌõ¼þ £¨D) Ðè¼Ó¿ÉÄæ¹ý³ÌÓëÎÞ·ÇÌå»ý¹¦µÃÌõ¼þ

£µ¡¢ ·ÇÀíÏëÆøÌå½øÐоøÈÈ×ÔÓÉÅòÕÍʱ£¬ÏÂÊö´ð°¸ÖÐÄÄÒ»¸ö¾ÍÊÇ´íÎóµÃ£¿ £¨ )

(A) Q=0 £¨B£© W=£° (C£© ¦¤U=0 (D) ¦¤H£½£° £¶¡¢ µ±Ìåϵ½«ÈÈÁ¿´«µÝ¸ø»·¾³Ö®ºó,ÌåϵµÃìÊ: £¨ )

£¨A£© ±Ø¶¨¼õÉÙ (B£© ±Ø¶¨Ôö¼Ó £¨C) ±Ø¶¨²»±ä £¨D£© ²»Ò»¶¨¸Ä±ä £·¡¢ Ò»¶¨Á¿µÃÀíÏëÆøÌå´Óͬһʼ̬³ö·¢,·Ö±ð¾­ £¨1) µÈÎÂѹËõ£¬£¨£²£© ¾øÈÈѹËõµ½¾ßÓÐÏàͬѹÁ¦µÃÖÕ̬£¬ÒÔ£È1,H2·Ö±ð±íʾÁ½¸öÖÕ̬µÃìÊÖµ£¬ÔòÓУº( )

£¨A£©H1> H2 £¨B)£È£±= H2 £¨£Ã£©H1¡´ H2 (£Ä)H1>=£È2 8¡¢ ÏÂÁÐÖî¹ý³Ì¿ÉÓ¦Óù«Ê½ dU = (£Ã£ð¡ª n£Ò)d£Ô½øÐмÆËãµÃ¾ÍÊÇ:£¨ £© (£Á£©Êµ¼ÊÆøÌåµÈѹ¿ÉÄæÀäÈ´ £¨C£©ÀíÏëÆøÌå¾øÈÈ¿ÉÄæÅòÕÍ £¨B£©ºãÈݽÁ°èijҺÌåÒÔÉý¸ßÎÂ¶È (£Ä£©Á¿Èȵ¯ÖеÃȼÉÕ¹ý³Ì

£¹¡¢ £±£íolµ¥Ô­×Ó·Ö×ÓÀíÏëÆøÌå,´Ó£²73 K,202¡¢65 kPa, ¾­ pT=³£Êý µÃ¿ÉÄæÍ¾¾¶Ñ¹Ëõµ½405¡¢3 kPaµÃÖÕ̬£¬¸ÃÆøÌåµÃ¦¤£ÕΪ£º £¨ £©

£¨£Á)£±702 £Ê £¨B£©£­406¡¢£¸ J (C)4£°6¡¢8 J (D)-1£·£°2 J 10¡¢ Ò»¶¨Á¿µÃÀíÏëÆøÌå´Óͬһ³õ̬·Ö±ð¾­ÀúµÈοÉÄæÅòÕÍ¡¢¾øÈÈ¿ÉÄæÅòÕ͵½¾ßÓÐÏàͬѹÁ¦µÃÖÕ̬£¬ÖÕ̬Ìå»ý·Ö±ðΪV1£¬£Ö2,Ôò: £¨ )

(A) V£±> V2 (B)£Ö1£¼ V£² £¨C) V1= V2 (D) ÎÞ·¨È·¶¨

11¡¢ Ò»ÈÝÆ÷µÃÈÝ»ýΪV1=162¡¢4 Á¢·½Ã×,ÄÚÓÐѹÁ¦Îª9£´£´30 P£á£¬Î¶ÈΪ288¡¢65 £ËµÃ¿ÕÆø.µ±°ÑÈÝÆ÷¼ÓÈÈÖÁTxʱ,´ÓÈÝÆ÷ÖÐÒݳöÆøÌåÔÚѹÁ¦Îª9£²£¸34 £Ða,ζÈΪ289¡¢15KÏ£¬Õ¼Ìå»ý114¡£3 m3;ÔòTx µÃֵΪ£º ( £©

(A£©1£°38¡¢1£µ K £¨B)9£´£¸¡¢£±5 K (£Ã£©849¡¢£±£µ K £¨D)84£°¡¢15 K 12¡¢ ʯīµÃȼÉÕÈÈ£º £¨ £©

£¨£Á) µÈÓÚC£ÏÉú³ÉÈÈ £¨B£© µÈÓÚCO£²Éú³ÉÈÈ £¨£Ã£© µÈÓÚ½ð¸ÕʯȼÉÕÈÈ (D) µÈÓÚÁã

13¡¢ £²98 Kʱ£¬Ê¯Ä«µÃ±ê׼Ħ¶ûÉú³Éìʦ¤fH £º ( ) (£Á£© ´óÓÚÁã £¨£Â£© СÓÚÁã (C£© µÈÓÚÁã (D) ²»ÄÜÈ·¶¨

£±4¡¢ ÈËÔÚÊÒÄÚÐÝϢʱ,´óԼÿÌìÒª³Ô £°¡££² £ëg µÃôûÀÒ£¨ÉãÈ¡µÃÄÜÁ¿Ô¼Îª £´000 kJ£©¡£¼Ù¶¨ÕâЩÄÜÁ¿È«²¿²»´¢´æÔÚÌåÄÚ,ΪÁËά³ÖÌåβ»±ä£¬ÕâЩÄÜÁ¿È«²¿±äΪÈÈʹº¹Ë®Õô·¢¡£ÒÑ֪ˮµÃÆû»¯ÈÈΪ 44 k£Ê/mo£ì£¬ÔòÿÌìÐèºÈË®£º ( )

£¨A) £°.5 kg (£Â) 1£®0 kg (C) 1.6 £ëg £¨D) £³¡£0 kg 15¡¢ ¸ßÎÂϳôÑõµÃĦ¶ûµÈѹÈÈÈÝC(£ð,£í£© Ϊ£º £¨ ) (A) 6R (B£© £¶¡¢5R (C£© 7R (D£© 7¡¢5R