②合工大物理化学内部习题与详解第8章08-习题

1、1, RT/2. 1mol CO0, 101.325kPa ,:(1) CO c ;(2) c CO (x2 + y2 +z2).(1) 3, , C O c = 3 x 1 RT2/ L = 3 kT2= 3 x1.381 x10-23 J K -12x273.15K = 5.657 x10-21 J(2)m(CO) = M/L = 28.01 x103kg/(6.022x1023) = 4.651x1026kgV = RT/p = (8.314 x 273.15/101325)m3 = 22.41 x 103 m3(n 2 + n 2 + n 2 ) = cx 8mV2 / 3 / h 2

2、=3.811 x 10 2000-7-121 仁 仁 ZZZ.H-VWXGVN.FRP4426966701263CO I = 1.45 x1046 kgm2 , J 4 3Ac , T = 300K Ac/kT.cr = J (J + 1)h2 / 8 2 IrAc= 4(4 + 1) - 3(3 + 1)h2 / 8 2 I= h2 / 2 I= 3.068 x 10-22 JT=300K, Ac r /kT = 7.407 x 10200-7-122 仁 仁ZZZ.H-VWXGVN.FRP442696670126, 2 (2 -1)00-7-123 仁 仁 ZZZ.H-VWXGVN.FRP

3、442696670126 (1) HCl: i, gi = 1.nu +1=gu +1e - cu +1 / kT = e- Acu / kTnugue -cu / kT= exp -5.94 x 10 - 20 /(1.381 x 10 - 23 x 298.15)= exp( -14.43) = 5.41 x 10 -7 0(2) I2:nu +1nu= e -Acu / kT= exp -0.426 x 10 - 20 /(1.381 x 10 -23 x 298.15)00-7-12exp(-1.035)= 0.355 411 ,HlI2C 5.94 x 1020 J 0.426 x

4、1020 J. 25,= 仁仁ZZZ. H- VWXGV N.FRP44269667012615 0, 101.325 kPa ea, ni gi.(T / K)5/ 2 ( p /44 (m / kg)-3/ 2x 103 kg/L, m10-8 00-7-12 ni ea ,gni 106 n5g i 仁a 仁iiZZZ.H-VWXGVN.FRP44269667012620Cl2CO 810K 3070K, 300KCV,m, x2C=R(lnx ), x= exp( / T )-V , m , v(1 -x )2v( 2.7 2CV ,m,v (Cl2 ) = R x 6.72 x10-

5、2 J1- 6.72x10-2 ?( 10.233 2= 4.681J mol-1 K -1CV ,m ,v(CO) = R x 3.595 x10-5 J 1- 3.595x10-5?= 0.0313J mol-1 K -1CV ,m,t(Cl) = 3 R,C22V ,m ,r(Cl2 ) = R3 R +CV ,m (Cl 2 )= 2R + 4.681J mol -1 K -1= 25.467 J mol -1 K -100-7-126 仁 仁ZZZ.H-VWXGVN.FRP44269667012626, N2(g)+3H2(g)=2NH3(g)1000K . :( G 0- UH 0

6、- UTm,- J0, mJ mol-1 K -1 m, 2980, m T?kJ mol -11000KN2(g)198.0548.669H2(g)137.0938.468NH3(g)203.5579.916A f H 0(NH 3 ,298.15K ) = -46.11kJ mol-1m1000K( G 0- U( G 0- UJAJ m, T0, m=v B m, T0, m T?BT? B= 2 x (-203.557 ) - (-198.054) -3 x (-137.093)J mol -1 K -1 = 0200-72-12.179J -1-1 仁 仁 7ZZZ.H- VWXGVN.FRP442696670126AH0298= 2A f H0= -92 .22 kJ mol -1m0KAU 0,m =0AH-2980A( Hm, 298- U 0, m )= -92.22 - (2 x9.916 - 8.669 - 3 x 8.468)kJ mol -1= -77.979kJ mol -1AG0( G 0-UAUJ1000 = AJ1000AGm, T0T0, m+丿0,