Ìì½ò´óѧµÚÎå°æ-Áõ¿¡¼ª-ÎïÀí»¯Ñ§¿ÎºóϰÌâ´ð°¸(È«) ÏÂÔØ±¾ÎÄ

ÎïÀí»¯Ñ§ÉϲáϰÌâ½â£¨Ìì½ò´óѧµÚÎå°æ£©

µÚÒ»ÕÂ ÆøÌåµÄpVT¹ØÏµ

1-1ÎïÖʵÄÌåÅòÕÍϵÊý?VÓëµÈÎÂѹËõϵÊý?TµÄ¶¨ÒåÈçÏ£º

?V?1V1??V???V?? ?T??? ?? ??V??p?T??T?pÊÔµ¼³öÀíÏëÆøÌåµÄ?V¡¢?TÓëѹÁ¦¡¢Î¶ȵĹØÏµ£¿ ½â£º¶ÔÓÚÀíÏëÆøÌ壬pV=nRT

?V?1??V?1??(nRT/p)?1nR1V???T?1 ? ??????V??T?pV??T?pVpVT?T??1??V?1??(nRT/p)?1nRT1V?????? ??2???p?1 ????V??p?TV??pVp?TVp1-2 Æø¹ñÄÚÓÐ121.6kPa¡¢27¡æµÄÂÈÒÒÏ©£¨C2H3Cl£©ÆøÌå300m3£¬ÈôÒÔÿСʱ90kgµÄÁ÷Á¿ÊäÍùʹÓóµ¼ä£¬ÊÔÎÊÖü´æµÄÆøÌåÄÜÓöàÉÙСʱ£¿

½â£ºÉèÂÈÒÒϩΪÀíÏëÆøÌå£¬Æø¹ñÄÚÂÈÒÒÏ©µÄÎïÖʵÄÁ¿Îª

pV121.6?103?300n???14618.623mol RT8.314?300.15ÿСʱ90kgµÄÁ÷Á¿ÕÛºÏpĦ¶ûÊýΪ

90?10390?103v???1441.153mol?h?1 MC2H3Cl62.45n/v=£¨14618.623¡Â1441.153£©=10.144Сʱ

1-3 0¡æ¡¢101.325kPaµÄÌõ¼þ³£³ÆÎªÆøÌåµÄ±ê×¼×´¿ö¡£ÊÔÇó¼×ÍéÔÚ±ê×¼×´¿öϵÄÃܶȡ£

½â£º?CH4pn101325?16?10?3??MCH4??MCH4??0.714kg?m?3 VRT8.314?273.151-4 Ò»³é³ÉÕæ¿ÕµÄÇòÐÎÈÝÆ÷£¬ÖÊÁ¿Îª25.0000g¡£³äÒÔ4¡æË®Ö®ºó£¬

1

ÎïÀí»¯Ñ§ÉϲáϰÌâ½â£¨Ìì½ò´óѧµÚÎå°æ£©

×ÜÖÊÁ¿Îª125.0000g¡£Èô¸ÄÓóäÒÔ25¡æ¡¢13.33kPaµÄij̼Ç⻯ºÏÎïÆøÌ壬Ôò×ÜÖÊÁ¿Îª25.0163g¡£ÊÔ¹ÀËã¸ÃÆøÌåµÄĦ¶ûÖÊÁ¿¡£

½â£ºÏÈÇóÈÝÆ÷µÄÈÝ»ýV?125.0000?25.000?100.0000cm3?100.0000cm3

?HO(l)21n=m/M=pV/RT

M?RTm8.314?298.15?(25.0163?25.0000)??30.31g?mol ?4pV13330?101-5 Á½¸öÌå»ý¾ùΪVµÄ²£Á§ÇòÅÝÖ®¼äÓÃϸ¹ÜÁ¬½Ó£¬ÅÝÄÚÃÜ·â×űê×¼×´¿öÌõ¼þÏÂµÄ¿ÕÆø¡£Èô½«ÆäÖÐÒ»¸öÇò¼ÓÈȵ½100¡æ£¬ÁíÒ»¸öÇòÔòά³Ö0¡æ£¬ºöÂÔÁ¬½Ó¹ÜÖÐÆøÌåÌå»ý£¬ÊÔÇó¸ÃÈÝÆ÷ÄÚ¿ÕÆøµÄѹÁ¦¡£ ½â£º·½·¨Ò»£ºÔÚÌâÄ¿Ëù¸ø³öµÄÌõ¼þÏ£¬ÆøÌåµÄÁ¿²»±ä¡£²¢ÇÒÉè²£Á§ÅݵÄÌå»ý²»ËæÎ¶ȶø±ä»¯£¬Ôòʼ̬Ϊ n?n1,i?n2,i?2piV/(RTi) ÖÕ̬£¨f£©Ê± n?n1,fT1,fT2,fn??pf?VR??T1,f?T2,f?n2,f?pf?VV??R??T1,fT2,f?pfV???R??T2,f?T1,f??TT?1,f2,f?? ???2pi?T1,fT2,f??????T?T?T?i?1,f2,f??

2?101.325?373.15?273.15 ??117.00kPa273.15(373.15?273.15)1-6 0¡æÊ±Âȼ×Í飨CH3Cl£©ÆøÌåµÄÃܶȦÑËæÑ¹Á¦µÄ±ä»¯ÈçÏ¡£ÊÔ×÷¦Ñ/p¡ªpͼ£¬ÓÃÍâÍÆ·¨ÇóÂȼ×ÍéµÄÏà¶Ô·Ö×ÓÖÊÁ¿¡£

P/kPa ¦Ñ/2.3074 1.5263 £¨g¡¤dm£© ½â£º½«Êý¾Ý´¦ÀíÈçÏ£º

-3101.325 67.550 50.663 33.775 25.331 1.1401 0.75713 0.56660 2

ÎïÀí»¯Ñ§ÉϲáϰÌâ½â£¨Ìì½ò´óѧµÚÎå°æ£©

101.32

P/kPa

5

(¦Ñ/p)/£¨g¡¤dm-3¡¤kPa£© ×÷(¦Ñ/p)¶Ôpͼ

0.02290.02280.02270.02260.02250.02240.02230.02220204060p8050.66

67.550

3

0.02242 0.02237 33.775 25.331

0.02270.02260.0227

0

50

¦Ñ/pÏßÐÔ (¦Ñ/p)¦Ñ/p100120 µ±p¡ú0ʱ£¬(¦Ñ/p)=0.02225£¬ÔòÂȼ×ÍéµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª

M???/p?p?0RT?0.02225?8.314?273.15?50.529g?mol?1

1-7 ½ñÓÐ20¡æµÄÒÒÍé-¶¡Íé»ìºÏÆøÌ壬³äÈëÒ»³éÕæ¿ÕµÄ200 cm3

ÈÝÆ÷ÖУ¬Ö±ÖÁѹÁ¦´ï101.325kPa£¬²âµÃÈÝÆ÷ÖлìºÏÆøÌåµÄÖÊÁ¿Îª0.3879g¡£ÊÔÇó¸Ã»ìºÏÆøÌåÖÐÁ½ÖÖ×é·ÖµÄĦ¶û·ÖÊý¼°·ÖѹÁ¦¡£ ½â£ºÉèAΪÒÒÍ飬BΪ¶¡Íé¡£

pV101325?200?10?6n???0.008315mol RT8.314?293.15m0.3897?yAMA?yBMB??46.867g?mol?1 £¨1£© n0.008315 ?30.0694yA?58.123yBM?yA?yB?1 £¨2£©

ÁªÁ¢·½³Ì£¨1£©Ó루2£©Çó½âµÃyB?0.599,yB?0.401

pA?yAp?0.401?101.325?40.63kPapB?yBp?0.599?101.325?60.69kPa

1-8 ÈçͼËùʾһ´ø¸ô°åµÄÈÝÆ÷ÖУ¬Á½²à·Ö±ðÓÐͬÎÂͬѹµÄÇâÆøÓë

3

ÎïÀí»¯Ñ§ÉϲáϰÌâ½â£¨Ìì½ò´óѧµÚÎå°æ£©

µªÆø£¬¶þÕß¾ù¿ËÊÓΪÀíÏëÆøÌå¡£

N2 H2 3dm3 p T 1dm3 p T £¨1£©±£³ÖÈÝÆ÷ÄÚζȺ㶨ʱ³éÈ¥¸ô°å£¬ÇÒ¸ô°å±¾ÉíµÄÌå»ý¿ÉºöÂÔ²»¼Æ£¬ÊÔÇóÁ½ÖÖÆøÌå»ìºÏºóµÄѹÁ¦¡£

£¨2£©¸ô°å³éȥǰºó£¬H2¼°N2µÄĦ¶ûÌå»ýÊÇ·ñÏàͬ£¿

£¨3£©¸ô°å³éÈ¥ºó£¬»ìºÏÆøÌåÖÐH2¼°N2µÄ·ÖѹÁ¦Ö®±ÈÒÔ¼°ËüÃǵķÖÌå»ý¸÷ΪÈô¸É£¿

½â£º£¨1£©³é¸ô°åǰÁ½²àѹÁ¦¾ùΪp£¬Î¶ȾùΪT¡£

pH2?nH2RT3dm32?pN2?nN2RT1dm3?p £¨1£©

µÃ£ºnH?3nN2

¶ø³éÈ¥¸ô°åºó£¬Ìå»ýΪ4dm3£¬Î¶ÈΪ£¬ËùÒÔѹÁ¦Îª

4nN2RTnN2RTnRTRTp??(nN2?3nN2)??V4dm34dm31dm3 £¨2£©

±È½Ïʽ£¨1£©¡¢£¨2£©£¬¿É¼û³éÈ¥¸ô°åºóÁ½ÖÖÆøÌå»ìºÏºóµÄѹÁ¦ÈÔΪp¡£ £¨2£©³é¸ô°åǰ£¬H2µÄĦ¶ûÌå»ýΪVm,HVm,N2?RT/p

2?RT/p£¬N2µÄĦ¶ûÌå»ý

³éÈ¥¸ô°åºó

V×Ü?nH2Vm,H2?nN2Vm,N2?nRT/p?(3nN2?nN2)RT/p ?? nH23nN2RTp?3nN2?nN2RTp

4