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跪求化學高手解此普化題目....
Aballoonistispreparingtomakeatripinahelium-filledballoon.Thetripbeginsinearlymorningatatemperatureof15度C.Bymidafternoon,thetemperaturehasincreasedto30度C.Assumingthepressureremainsconstantat1.00atm,foreachmoleofheliumcalculate:(a)theinitialandfinalvolumes(ans:...顯示更多Aballoonistispreparingtomakeatripinahelium-filledballoon.Thetripbeginsinearlymorningatatemperatureof15度C.Bymidafternoon,thetemperaturehasincreasedto30度C.Assumingthepressureremainsconstantat1.00atm,foreachmoleofheliumcalculate:(a)theinitialandfinalvolumes(ans:initial=23.65Lfinal=24.89L)(b)thechangeininternalenergy.{Hint:heliumbehaveslikeanidealgas.SoE=3/2nRT.NotetheunitofRareconsistentwiththoseofE.}(ans:187J)(c)Thework(w)donebythehelium{inJ}(ans:-120J)(d)Theheat(q)transferred{inJ}(ans:310J)P.S.以上題目附了答案...原因是希望各位化學高手可以將過程告訴敝人...尤其是(c)(d)兩題更是不解...希望各位大大可以幫幫忙!!更新:可否請問bdref43這位大大...為什麼1atm.L=10333.23kg.L/m2*9.8m/s2*0.001m3/L?這三樣數字分別代表的意義是?
(a) 假設氦為理想氣體, V=nRT/P, R=0.08206 atm.L/mol.oK, Vi=0.08206*(273.15+15)=23.65 L, Vf=0.08206*(273.15+30)=24.88 L.(b) dE=(3/2)nRdT, R=8.31431 J/mol.oK, dE=1.5*8.31431*15=187 J(c) 1 atm.L = 10333.23 kg.L/m2 * 9.8 m/s2 * 0.001 m3/L = 101.3?kg.m2/s2 = 101.3 J, w=PdV=23.65-24.88=-1.23 atm.L=-1.23*101.3 J=-125 J(d) q=dE-w=187-(-125)=312 J 2006-11-12 01:26:00 補充： 一大氣壓=每cm^2的1033.323*cm水柱,質量=1033.323gcm=1.033323gcc/cm^2=10333.23gcc/m^2=10333.23kgL/m^2.壓力=質量*重力加速度 2006-11-12 07:40:38 補充： 上面補充再拷貝時電腦出了點問題,有些數字及單位不正確.重新更正.一大氣壓=每cm^2的1033.323*cm水柱,質量=1.033323Kg, 壓力=質量*重力加速度/面積. 重力加速度=9.8m/s^2, 1m=100cm, 1m^3=1000000cm^3=1000L,故1L=m^3, 1atm=1.033323Kg*9.8m/s^2/1cm^2=10.13 N/cm^2, 1atmL=10.13 NL/cm^2=10130Ncm=101.3Nm=101.3J
(a)如題,定壓下溫度從15度升到30度由PV=nRT可個別求出其初體積與末體積1xV(初)=1x0.082x(273+15),V(初)=23.616L(因各數值有效位數不同,故與解答不盡相同)同理得V(末)=24.846L(b)能量變化=末能減初能=3/2nR(T末-T初)=3/2x1x8.314x15=187.065J(c)W=-[P(末)V(末)-P(初)V(初)]=-[1x24.846-1x23.616]=-1.23J(d)E=q+W,故q=187.065+1.23=188.353(J)參考資料：ME
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