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5 mol single atomic perfect gas£¬from the initial state of 300K, 50kPa is firstly adiabatical reversible compressed to 100 kPa, then isobaricly cooled to decrease the volume to 85dm3 .Calculate W, Q,¡÷U, ¡÷H and ¡÷S of the whole process.

´ð°¸£ºQ£½-19.892kJ,W=13.935kJ,¡÷U=-5.958kJ£¬¡÷H=-9.930kJ,¡÷S=-68.66J/K 5. ¼×´¼ÔÚ101.325kPaϵķе㣨Õý³£·Ðµã£©Îª64.65¡æ£¬ÔÚ´ËÌõ¼þϵÄĦ¶ûÕô·¢ìÊ

?vapHm?35.32kJ?mol?1¡£ÇóÔÚÉÏÊöζÈѹÁ¦Ï£¬1kg¼×´¼È«²¿³ÉΪ¼×´¼ÕôÆøÊ±µÄ

Q,W,¡÷U£¬¡÷H£¬¡÷S,¡÷A¼°¡÷G

At 101.325kPa, the boiling point of CH3OH is 64.65¡æ, ?vapHm?35.32kJ?mol at this condition. Calculate Q,W,¡÷U,¡÷H£¬¡÷S,¡÷A and ¡÷G of the following process: 1kg CH3OH (l, 64.65¡æ,101.325kPa)¡úCH3OH (g, 64.65¡æ,101.325kPa)

´ð°¸£ºQ£½¡÷H£½1102.65kJ,W=-87.65kJ,¡÷U=1014.65kJ£¬¡÷S=3.263kJ/K£¬¡÷A=-86.58kJ£¬¡÷G=0

6. ÒÑ֪ˮµÄ·ÐµãÊÇ100¡æ£¬Ä¦¶û¶¨Ñ¹ÈÈÈÝCp,m?H2O,l??75.20J?mol?K,Æû»¯

?1?1?1ìÊ?vapHm?40.67kJ?mol£¬Ë®ÆøÄ¦¶û¶¨Ñ¹ÈÈÈÝ

?1Cp,m?H2O,g??33.57J?mol?1?K?1(Ħ¶û¶¨Ñ¹ÈÈÈÝ¿ÉÊÓΪÓëζÈÎÞ¹Ø)£¬Çó¹ý³Ì£º

1molH2O(l,60¡æ,101325Pa) ¡ú1molH2O(g,60¡æ,101325Pa)µÄQ,W,¡÷U£¬¡÷H£¬¡÷S,¡÷A¼°¡÷G Given the boiling point of water is 100¡æ, Cp,m?H2O,l??75.20J?mol?K,

?1?1?vapHm?40.67kJ?mol?1, Cp,m?H2O,g??33.57J?mol?1?K?1( don¡¯t change with

temperature),

Calculate Q,W,¡÷U£¬¡÷H£¬¡÷S,¡÷A and ¡÷G of the following process: 1molH2O(l,60¡æ,101325Pa) ¡ú1molH2O(g,60¡æ,101325Pa)

´ð°¸£ºQ£½¡÷H£½42.34kJ,W=-2.77kJ,¡÷U=39.57kJ£¬¡÷S=113.7J/K,¡÷A=1.69kJ£¬¡÷G=4.06kJ

7. ÒÑÖª1mol¡¢-5¡æ¡¢100kPaµÄ¹ýÀäÒºÌå±½ÍêÈ«Äý¹ÌΪ£­5¡æ¡¢100kPa¹Ì̬±½µÄìØ±ä»¯Îª-35.5J/K,¹Ì̬±½ÔÚ£­5¡æÊ±µÄÕôÆøÑ¹Îª2280Pa,Ħ¶ûÈÛ»¯ìÊΪ9874J/mol¡£¼ÆËã¹ýÀäҺ̬±½ÔÚ£­5¡æÊ±µÄÕôÆøÑ¹¡£

Given that ¡÷S of the process 1molC6H6(l,-5¡æ,100kPa) ¡ú1molC6H6 (s,-5¡æ, 100kPa), ´ð°¸£º2680Pa

8. 4molÀíÏëÆøÌå´Ó300K¡¢p?ºãѹ¼ÓÈȵ½600K,Çó´Ë¹ý³ÌµÄQ¡¢W¡¢¡÷U¡¢¡÷H¡¢¡÷S¡¢

?¡÷A¡¢¡÷G¡£ÒÑÖª´ËÀíÏëÆøÌåµÄSm(300K)=150.0J¡¤K-1¡¤mol-1£¬Cp,m =30.00J¡¤K-1¡¤mol-1

4 mol perfect gas from the initial state of 300K, p? isobaricly heated to 600K. Calculate Q,W,¡÷U,¡÷H,¡÷S,¡÷A and ¡÷G of the process. Given for the perfect gas ?(300K)=150.0J¡¤K-1¡¤mol-1£¬Cp,m =30.00J¡¤K-1¡¤mol-1.Sm´ð°¸£º¡÷U=26.02kJ; ¡÷H=Q=36.0kJ;W=-9.978kJ; ¡÷S=83.18J/K;¡÷A=-203.9kJ; ¡÷G=-193.9kJ

9. 8molij˫ԭ×ÓÀíÏëÆøÌåÓÉʼ̬(400K,0.20MPa)·Ö±ð¾­ÏÂÁÐÈý¸ö²»Í¬¹ý³Ì±äµ½¸Ã¹ý³ÌÖ¸¶¨µÄÖÕ̬£¬·Ö±ð¼ÆËã¸÷¹ý³ÌµÄQ,W,¡÷U;¡÷H;¡÷S;¡÷A;¡÷G. ¹ý³Ì¢ñ£ººãοÉÄæÅòÕ͵½0.10MPa; ¢ò: ×ÔÓÉÅòÕ͵½0.10MPa;

¢ó: ºãÎÂ϶Կ¹ºãÍâѹ0.10MPaÅòÕ͵½Æ½ºâ

8 mol double atomic perfect gas from the initial state of 400K,0.20MPa through the following three different paths to given final state. Calculate Q,W,¡÷U;¡÷H;¡÷S;¡÷A;¡÷Gof each paths.

(1) Isothermal reversible expands to 0.10MPa. (2) Free expands to 0.10MPa.

(3) Isothermal expands against an external constant pressure of 0.10MPa to equilibrium. ´ð°¸£º ¹ý³Ì ¢ñ ¢ò ¢ó W/kJ -18.41 0 -6.65 Q/kJ 18.41 0 6.65 ¡÷U/kJ 0 0 0 ¡÷H/kJ 0 0 0 ¡÷S/kJ 46. 10 46.10 46.10 ¡÷A/kJ -18.44 -18.44 -18.44 ¡÷G/kJ -18.44 -18.44 -18.44 3

10. ½«×°ÓÐ0.1molÒÒÃÑÒºÌåµÄ΢С²£Á§Æ¿·ÅÈëÈÝ»ýΪ10dmµÄºãÈÝÃܱյÄÕæ¿ÕÈÝÆ÷ÖÐ,²¢ÔÚ35.51¡æµÄºãβÛÖкãΡ£35.51¡æÎªÔÚ101.325kPaÏÂÒÒÃѵķе㡣ÒÑÖªÔÚ´ËÌõ¼þÏÂÒÒÃѵÄĦ¶ûÕô·¢ìÊΪ25.10kJ/mol¡£½ñ½«Ð¡²£Á§Æ¿´òÆÆ£¬ÒÒÃÑÕô·¢ÖÁƽºâ̬£¬Ç󣺣¨1£©ÒÒÃÑÕôÆøµÄѹÁ¦£»£¨2£©¹ý³ÌµÄW,Q,¡÷U,¡÷H,¡÷S,¡÷G,¡÷A¡£

´ð°¸£ºP=25.664kPa;W=0J;Q=¡÷U=2.254kJ;¡÷H=2.51kJ;¡÷S=9.275J/K;

¡÷A=-0.61kJ;¡÷G=-0.353kJ

3

11.½«×°ÓÐ0.1molÒÒÃÑÒºÌåµÄ΢С²£Á§ÅÝ·ÅÈë35¡æ¡¢101325Pa¡¢10dmµÄ±£ÎÂÆ¿ÖÐ,ÆäÖгäÂúN2(g),½«Ð¡²£Á§ÅÝ´òËéºó,ÒÒÃÑÈ«²¿Æû»¯,ÐγɵĻìºÏÆøÌå¿ÉÊÓΪÀíÏëÆøÌå¡£ÒÑÖªÒÒÃÑÔÚ101325PaʱµÄÕý³£·ÐµãΪ35¡æ,ÆäÆû»¯ìÊΪ25.10kJ/mol¡£¼ÆË㣨1£©»ìºÏÆøÖÐÒÒÃѵķÖѹ£»£¨2£©µªÆøµÄ¡÷H,¡÷S,¡÷G£»£¨3£©ÒÒÃѵġ÷H,¡÷S,¡÷G ´ð°¸£º(1)P=25.664kPa;(2)¡÷U=0kJ;¡÷H=0kJ;¡÷G=0kJ£»

(3)¡÷S=9.275J/K;¡÷H=2510kJ;¡÷G=-0.353kJ

12.ÒÑÖª298KʱʯīºÍ½ð¸ÕʯµÄ±ê׼Ħ¶ûȼÉÕìÊ·Ö±ðΪ-393.511kJ/molºÍ-395.407kJ/mol£¬±ê׼Ħ¶ûìØ·Ö±ðΪ5.694J?K?1?mol?1ºÍ2.439J?K?1?mol?1£¬Ìå»ý

?ÖÊÁ¿·Ö±ðΪ2.260ºÍ3.520g/cm3¡££¨1£©¼ÆËãC(ʯī) ¡úC(½ð¸Õʯ)µÄ?rGm?298K?; (2)25¡æÊ±Ðè¶à´óѹÁ¦ÄÜʹÉÏÊöת±ä³ÉΪ¿ÉÄÜ£¨Ê¯Ä«ºÍ½ð¸ÕʯµÄѹËõϵÊý¾ù¿É½üËÆÊÓΪÁ㣩

??Given at 298K, ?cHm(C, graphite)= -393.51kJ/mol, ?cHm (C, diamond)= ?-395.407kJ/mol, Sm( C, graphite)= 5.694J?K?1??mol?1,Sm( C, diamond)=

2.439J?K?1?mol?1,

¦Ñ( C, graphite)= 2.260g/cm3, ¦Ñ( C, diamond)= 3.520g/cm3.

(1) Calculate ?rGm?298K? of this process: C(graphite) ¡úC(diamond);

?(2) How to control pressure to make the above change realized.

´ð°¸£º(1) 2.867kJ/mol; (2) p>1.51¡Á109Pa 13£® ÒÑ֪ˮÔÚ77¡æÊ±µÄ±¥ºÍÕôÆøÑ¹Îª41.891kPa,Ë®ÔÚ101.325kPaϵÄÕý³£·ÐµãΪ100¡æ,Çó:

(1) ÏÂÃæ±íʾˮµÄÕôÆøÑ¹ÓëζȹØÏµµÄ·½³ÌʽÖеÄAºÍBÖµ;

lg?pPa???AT?B

(2) ÔÚ´Ëζȷ¶Î§ÄÚË®µÄĦ¶ûÕô·¢ìÊ; (3) ÔÚ¶à´óѹÁ¦ÏÂË®µÄ·ÐµãΪ105¡æ£®

Given that saturated vapor pressure of water is 41.891kPa at 77¡æ.Normal boiling point of water is 100¡æ under 101.325kPa. Please calculate:

(1) The value A and B below expressing the relationship of water vapor pressure with

temperature.

lg?pPa???AT?B

(2) Molar vapor enthalpy in this temperature scope.

(3) How much the pressure when boiling point of water is 105¡æ£® ´ð°¸ (1) A=2179.133K,B=10.8455; (2) 41.719kJ/mol;(3) 121.042kPa

14.Ë®ºÍÂÈ·ÂÔÚ101.325kPaϵÄÕý³£·Ðµã·Ö±ðΪ100¡æºÍ61.5¡æ,Ħ¶ûÕô·¢ìÊ·Ö±ðΪ

?vapHm?H2O??40.668kJ?mol?1ºÍ?vapHm?CHCl3??29.50kJ?mol?1,ÇóÁ½ÒºÌå

¾ßÓÐÏàͬ±¥ºÍÕôÆøÑ¹Ê±µÄζÈ.

The normal boiling point of H2O and CH3Cl are 100¡æ and 61.5¡æ respectively,

?vapHm?H2O??40.668kJ?mol?1 and

?vapHm?CHCl3??29.50kJ?mol?1,

Calculate the temperature when the two liquid has the same saturated pressure. ´ð°¸:262.9¡æ

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(A) > (B) < (C) = (D)ÎÞ·¨È·¶¨

2.ÔÚ?¡¢?Á½ÏàÖж¼º¬ÓÐAºÍBÁ½ÖÖÎïÖÊ£¬µ±´ïµ½ÏàÆ½ºâʱ£¬ÏÂÁÐÈýÖÖÇé¿öÕýÈ·µÄÊÇ£º £¨ £©

¦Á (A?)A??¦ÁB ? ¦Á(B)¦Â?AA? C B ) (D) ÎÞ·¨È·¶¨ ¦Á ? A ? ¦Â(?3.ÔÚºãΡ¢ºãѹÏÂ,ÓÉAºÍBÐγÉÀíÏëҺ̬»¯ºÏÎïµÄ¹ý³ÌµÄ?mixUºÍ?mixA£¨ £©

(A)?mixU=0£¬?mixA<0 (B)?mixU<0£¬?mixA=0 (C) ?mixU>0£¬ ?mixA>0 (D) ?mixU=0£¬?mixA>0 4.150¡æ£¬101 325PaµÄҺ̬H2O(l)µÄ»¯Ñ§ÊÆ?l£¬ 150¡æ£¬101325PaµÄÆøÌ¬H2O(g)µÄ»¯Ñ§ÊÆ?g £¬¶þÕߵĹØÏµÎª£¨ £©

(A) ?l > ?g (B) ?l < ?g (C) ?l = ?g (D) ÎÞ·¨È·¶¨

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