BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY 第，、章镜计热力学和步美文司数及参考答素 this temperature and 1 atm pressure). m=1007852x10kgmo1=1673559×10-”ke 6.022045×1025mo -y7 -[2红L67359x109g038062x10J-KX30K™x02461705m (6.626176×10Js)7 =7.69115×100 Note that the molecular partition function is a dimensionless quantity and may be interpreted approximately as the number of energy levels accessible to the single hydrogen atom in this volume 唱cw2” For H(g) J厂20.67359x10-”kg1.38062x10-JKX300K)] (6.626176×10-4Js)2 ×1380662×10-KX300K=1636274 101325Nm2 =&31K.mr得+16362g=1368W.Kmr G=-(8.314J.K-.mol-)3000K16.36274)=-408.160 kJ.mol- For Ha(g) s=831Knm得+174024时-16s4Ka TEL010-64434903 1 BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL:010-64434903 1 第八章 统计热力学初步英文习题及参考答案 1.Calculate the molecular partition function for translational motion for a hydrogen atom at 3000 K in a volume of 0.2461705m3 (This is the volume that would contain 1 mol of perfect gas at this temperature and 1 atm pressure). 1.673559 10 kg 6.022045 10 mol 1.007852 10 kg mol 27 23 1 3 1 - - - - = ´ ´ ´ × m = V h mkT qt 3 / 2 2 2 ÷ ø ö ç è æ = p (0.2461705 ) (6.626176 10 Js) 2 (1.673559 10 kg)(1.380662 10 J K )(300K) 3 3/ 2 34 2 27 23 1 ú ´ m û ù ê ë é ´ ´ ´ × = - - - - p =7.69115×1030 Note that the molecular partition function is a dimensionless quantity and may be interpreted approximately as the number of energy levels accessible to the single hydrogen atom in this volume. 2.Calculate the translational contributions to S°and G°for H(g) and H2(g) at 3000K. ïþ ï ý ü ïî ï í ì ú û ù ê ë é = + θ 3 / 2 2 0 2 ln 2 5 p kT h mkT St R p ， ú ú û ù ê ê ë é ÷ ø ö ç è æ = - θ 3 / 2 2 0 2 ln p kT h mkT Gt RT p For H(g) ú ú û ù ê ê ë é ÷ ø ö ç è æ θ 3 / 2 2 2 ln p kT h pmkT 16.36274 101325Nm (1.380662 10 JK )(3000K) (6.626176 10 Js) 2 (1.673559 10 kg)(1.380662 10 JK )(3000K) ln -2 23 -1 3 / 2 34 2 27 23 -1 = þ ý ´ ü ´ ïî ï í ì ú û ù ê ë é ´ ´ ´ = - - - - p θ 1 1 1 1 16.36274 156.84 K mol 2 5 (8.314J K mol ) - - - - ÷ = × × ø ö ç è æ S = × × + J t ( ) θ 1 1 1 (8.314J K mol ) 3000 (16.36274) 408.160kJ mol - - - G = - × × K = - × t For H2(g) ú ú û ù ê ê ë é ÷ ø ö ç è æ θ 3/ 2 2 2 ln p kT h pmkT =17.40246 θ 1 1 1 1 17.40246 165.485J K mol 2 5 (8.314J K mol ) - - - - ÷ = × × ø ö ç è æ St = × × +