淮阴师范学院《物理化学》课外习题
?H = nCp,m ?T =31.25 kJ
Q = ?U – W = 23.75 kJ
6 [解] 单原子理想气体CV,m,A = (3/2)R 双原子理想气体CV,m,B = (5/2)R
因过程绝热 Q = 0，所以 T1 = 400K
?U = W = -p?V
(n ACV,m,A + n BCV,m,B )(T2 – T1)
= -100?103[(n A+n B)RT2/100?103 - (n A+n B)RT1/200?103] 解得 T2 = 331.03K
?U = W = (n ACV,m,A + n BCV,m,B )(T2 – T1) = -5.448kJ ?H = ?U +? (pV) = ?U +(n A+n B)R?T = 8.315 kJ
7 [解] (1) ? = ?n/? = (-10/128)/(-1) = 78.02 mmol
(2)?cUm?=?U?(1000/78.02) = -401.727?(1000/78.02) = -5149.1 kJ?mol-1 (3) ?cHm? = ?cUm? + (??i,g)RT = ?Um + (-2)RT = -5154.1 kJ?mol-1
8 [解] 爆炸反应的最高温度与压力应按恒容绝热计算。以1molH2为计算基准： n(H2)=1mol；n(O2)={1+50%}/2）mol=0.75mol;n(N2)=(0.75×79/21)mol=2.8214mol. 可通过下列过程求算爆炸时的最高温度及最高压力：
终态温度也可由ΔrHm计算 ΔrHm=ΔrUm+Δr(pV)m; ΔrUm=0 Δr(pV)m=
=(33.85T/K-11331.67) J·mol-1 ΔH1=ΔfHm=(H2O,g,298.15K)=-241.82kJ·mol-1
ΔH2= =148.603(T/K-298.15)J·mol-1
ΔrHm=Δr(pV)m=(33.85T/K-11331.6)J·mol-1
=ΔH1+ΔH2
={-241.82×103+148.603(T/K-298.15)} J·mol-1 由上式可得：
T=
由理想气体状态方程可求得： p2 = 715.44 kPa
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