1¡¢1molµ¥Ô×ÓÀíÏëÆøÌåʼ̬Ϊ273K¡¢pºãÎÂÏÂѹÁ¦¼Ó±¶£¬¼ÆËãÆäQ¡¢W¡¢¦¤U¡¢¦¤H¡¢¦¤S¡¢¦¤G¡¢¦¤A¡££¨ÒÑÖª273K¡¢pÏÂ¸ÃÆøÌåµÄĦ¶ûìØÎª100J¡¤K-1¡¤mol-1£©
2¡¢1molÀíÏëÆøÌå´Ó300K£¬100kPaϵÈѹ¼ÓÈȵ½600K£¬Çó´Ë¹ý³ÌµÄQ¡¢W¡¢¦¤U¡¢¦¤H¡¢¦¤S¡¢¦¤A¡¢¦¤G¡£ÒÑÖª´ËÀíÏëÆøÌå300KʱµÄSm?=150.0J¡¤K£1¡¤mol£1£¬Cp,m=30.0J¡¤K£1¡¤mol
£1
¡£
3¡¢ 1mol ÀíÏëÆøÌåʼ̬Ϊ27¡æ¡¢1MPa£¬ÁîÆä·´¿¹ºã¶¨µÄÍâѹ0.2MPaÅòÕ͵½Ìå»ýΪÔÀ´µÄ5±¶£¬Ñ¹Á¦ÓëÍâѹÏàͬ¡£ÊÔ¼ÆËã´Ë¹ý³ÌµÄQ¡¢W¡¢¦¤U¡¢¦¤H¡¢¦¤S¡¢¦¤A¡¢¦¤G¡£ÒÑÖªÀíÏëÆøÌåµÄºãÈÝĦ¶ûÈÈÈÝΪ12.471J¡¤mol-1¡¤K-1
4¡¢ÔÚ298.15Kʱ£¬½«1molO2´Ó101.325kPaµÈοÉÄæÑ¹Ëõµ½6.0¡Á101.325kPa£¬ÇóQ£¬ W£¬
?U£¬ ?H£¬ ?A£¬?SÌåϵ£¬?S¸ôÀë¡£
5¡¢273.2£Ë¡¢Ñ¹Á¦Îª500kPaµÄijÀíÏëÆøÌå2dm3£¬ÔÚÍâѹΪ100kPaϵÈÎÂÅòÕÍ£¬Ö±µ½ÆøÌåµÄѹÁ¦Ò²µÈÓÚ100kPaΪֹ¡£Çó¹ý³ÌÖеÄQ¡¢W¡¢¦¤U¡¢¦¤H¡¢¦¤S¡¢¦¤A¡¢¦¤G¡£
6¡¢2molË«Ô×ÓÀíÏëÆøÌåʼ̬Ϊ298K¡¢p?¾¹ýºãÈÝ¿ÉÄæ¹ý³ÌÖÁѹÁ¦¼Ó±¶£¬ÊÔ¼ÆËã¸Ã¹ý³ÌµÄQ¡¢W¡¢¦¤U¡¢¦¤H¡¢¦¤S¡¢¦¤A¡¢¦¤G¡£ÒÑÖª298K¡¢p?ÏÂ¸ÃÆøÌåµÄĦ¶ûìØÎª100 J¡¤K-1¡¤mol-1¡£
7¡¢3molË«Ô×ÓÀíÏëÆøÌå´Óʼ̬100kPa£¬75 dm3£¬ÏȺãοÉÄæÑ¹ËõʹÌå»ýËõСÖÁ50 dm3£¬ÔÙºãѹ¼ÓÈÈÖÁ100 dm3£¬ÇóÕû¸ö¹ý³ÌµÄQ£¬W£¬¦¤U£¬¦¤H¼° ¦¤S¡£
8¡¢5 molÀíÏëÆøÌ壨Cpm = 29.10 J¡¤K-1¡¤mol-1£©£¬ÓÉʼ̬400 K£¬200 kPa¶¨Ñ¹ÀäÈ´µ½300 K£¬ÊÔ¼ÆËã¹ý³ÌµÄQ¡¢W¡¢¦¤U¡¢¦¤H¡¢¦¤S¡¢¦¤A¡¢¦¤G¡£
9¡¢ÔÚÏÂÁÐÇé¿öÏ£¬1 molÀíÏëÆøÌåÔÚ27¡æ¶¨ÎÂÅòÕÍ£¬´Ó50 dm3ÖÁ100 dm3£¬Çó¹ý³ÌµÄQ¡¢W¡¢¦¤U¡¢¦¤H¡¢¦¤S¡£
£¨1£©¿ÉÄæÅòÕÍ£»£¨2£©ÅòÕ͹ý³ÌËù×÷µÄ¹¦µÈÓÚ×î´ó¹¦µÄ50 %£»£¨3£©ÏòÕæ¿ÕÅòÕÍ¡£
10¡¢2 molijÀíÏëÆøÌ壬Æä¶¨ÈÝĦ¶ûÈÈÈÝ Cv,m =3/2R£¬ÓÉ500 K£¬405.2 kPaµÄʼ̬£¬ÒÀ´Î¾ÀúÏÂÁйý³Ì£º£¨1£©ÔÚºãÍâѹ202.6 kPaÏ£¬¾øÈÈÅòÕÍÖÁƽºâ̬£¬£¨2£©ÔÙ¿ÉÄæ¾øÈÈÅòÕÍÖÁ101.3kPa£»£¨3£©×îºó¶¨ÈݼÓÈÈÖÁ500 KµÄÖÕ̬¡£ÊÔÇóÕû¸ö¹ý³ÌµÄQ¡¢W¡¢¦¤U¡¢¦¤H¡¢¦¤S¡£
11¡¢ÒÑÖª´¿B( l )ÔÚ100 kPaÏ£¬80¡æÊ±·ÐÌÚ£¬ÆäĦ¶ûÆû»¯ìÊvapHm = 30878 J¡¤mol-1¡£BÒºÌåµÄ¶¨Ñ¹Ä¦¶ûÈÈÈÝCpm=14.27 J¡¤K-1¡¤mol-1¡£½ñ½«1 mol£¬40 kPaµÄB( g )ÔÚ¶¨ÎÂ80¡æµÄÌõ¼þÏÂѹËõ³É100 kPaµÄB( l )¡£È»ºóÔÙ¶¨Ñ¹½µÎÂÖÁ60¡æ¡£Çó´Ë¹ý³ÌµÄ¦¤S¡£ÉèB( g )ΪÀíÏëÆøÌå¡£
12¡¢ÔÚ25¡æÊ±1 mol O2´Ó1000 kPa×ÔÓÉÅòÕ͵½100 kPa£¬Çó´Ë¹ý³ÌµÄ¦¤U¡¢¦¤H¡¢¦¤S£¬?¦¤A¡¢¦¤G£¨ÉèO2ΪÀíÏëÆøÌ壩¡£
13¡¢4 molijÀíÏëÆøÌ壬ÆäCVm = 2.5 R£¬ÓÉ600 K£¬100 kPaµÄʼ̬£¬¾¾øÈÈ¡¢·´¿¹Ñ¹Á¦ºã¶¨Îª600 kPaµÄ»·¾³Ñ¹Á¦ÅòÕÍÖÁƽºâ̬֮ºó£¬ÔÙ¶¨Ñ¹¼ÓÈȵ½600 KµÄÖÕ̬¡£ÊÔÇóÕû¸ö¹ý³ÌµÄ¦¤S£¬¦¤A¡¢¦¤G¸÷ΪÈô¸É£¿
14¡¢±½ÔÚÕý³£·Ðµã353 KʱĦ¶ûÆû»¯ìÊΪ30.75 kJ¡¤mol-1¡£½ñ½«353 K£¬101.326 kPaϵÄ1 molҺ̬±½ÏòÕæ¿ÕµÈÎÂÕô·¢±äΪͬÎÂͬѹµÄ±½ÕôÆø£¨ÉèΪÀíÏëÆøÌ壩¡£ £¨1£©Çó´Ë¹ý³ÌµÄQ£¬W£¬¦¤U¡¢¦¤H¡¢¦¤S£¬¦¤A¡¢¦¤G£» £¨2£©Ó¦ÓÃÓйØÔÀí£¬Åжϴ˹ý³ÌÊÇ·ñΪ²»¿ÉÄæ¹ý³Ì¡£
15¡¢½«Ò»Ð¡²£Á§Æ¿·ÅÈëÕæ¿ÕÈÝÆ÷ÖУ¬Æ¿ÖÐÒÑ·âÈë1 molҺ̬ˮ£¨100¡æ£¬101.3 kPa£©£¬Õæ¿ÕÈÝÆ÷Ç¡ºÃÄÜÈÝÄÉ1molË®ÕôÆø£¨100¡æ£¬101.3kPa£©¡£Èô±£³ÖÕû¸öϵͳµÄζÈΪ100¡æ£¬½«Æ¿»÷ÆÆºó£¬Ë®È«²¿Æø»¯ÎªË®ÕôÆø¡£ÊÔ¼ÆËã´Ë¹ý³ÌµÄQ£¬W£¬¦¤U¡¢¦¤H¡¢¦¤S£¬¦¤A¡¢¦¤G¡£¸ù¾Ý¼ÆËã½á¹û˵Ã÷´Ë¹ý³ÌÊÇ·ñ¿ÉÄæ£¿ÓÃÄÄÒ»¸öÈÈÁ¦Ñ§º¯Êý×÷ΪÅоݣ¿ ÒÑ֪ˮÔÚ100¡æ£¬101.3 kPaµÄĦ¶ûÆø»¯ìÊΪ40.64 kJ¡¤mol-1¡£ÉèÕôÆøÎªÀíÏëÆøÌå¡£