第一章习题及答案
8.1mol 理想气体,始态为2×101.325kPa 、11.2dm 3,经p T =常数的可逆过程压缩到终态为4×101.325kPa ,已知C V =3/2R 。求:
(1)终态的体积和温度。(2)ΔU 和ΔH (3)所作的功。
解:(1)T 1=p 1V 1/nR 273314.8/102.112026503
=××=?K 因pT =常数
故T 2=p 1T 1/p 2=202.65×273/405.3=136.5K
V 2=nRT -2/p 2=8.314×136.5/405.3=2.8dm 3
(2)单原子理想气体C V ,m =3/2R,C p ,m =5/2R
ΔU =C V (T 2-T 1)=3/2×8.314×(136.5-273)=-1702J ΔH =C p (T 2-T 1)=5/2×8.314×(136.5-273)=-2837J (3)pT =B,p =B/T V=RT/p=RT 2/B,d V=(2RT/B)d T
J
2270)2735.136(314.82d 2d B
2B d =?××?=?=?=?=∫∫∫T
R T RT
T V p W 9.1mol 理想气体从373.15K 、0.025m 3经下述四个过程变为373.15K 、0.1m 3:(1)等温可逆膨胀;(2)向真空膨胀;
(3)等外压为终态压力下膨胀;
(4)等温下先以等外压等于气体体积为0.05m 3时的压力膨胀至0.05m 3,再以等外压等于终态压力下膨胀至0.1m 3。
求诸过程系统所作的体积功。解:
(1)
∫?=?=1
2ln
d V V nRT V p W J 4301025
.01
.0ln
15.373314.81?=×××?=J (2))(0)(1212e V V V V p W ?×?=??==0
(3)
)
()(122
122V V V nRT
V V p W ?×?=??=J 2326)025.01.0(1
.015.373314.81?=?×××?=J
(4))]05.01.0(1
.0[)025.005.0(05.0?×?+??
=nRT
nRT W =-3102J 15.298.15K 的0.5g 正庚烷在等容条件下完全燃烧使热容为8175.5J·K -1的量热计温度上升了2.94℃,求正庚烷在298.15K 完全燃烧时的ΔH 。解:C 7H 16(1)+11O 2(g)=7CO 2(g)-8H 2O(1)
M=100
RT
n T C n nRT Q Q H V V P g ?+?=?+==?正庚烷kJ 1.4817298314.8)117(5
.0100
94.25.8175?=××?+×
×?=16.试求下列反应在298.15K 、101.325kPa 时的等压热效应。
(1)2H 2S (g )+SO 2(g )→2H 2O (l )+3S (斜方)Q V =-223.8kJ (2)2C (石墨)+O 2(g )→2CO (g )Q V =-231.3kJ (3)H 2(g )+Cl 2(g )→2HCl (g )
Q V =-184kJ
解:(1)298
314.8)30(108.2238××?+×?=?+=?nRT Q Q V p =-231.2kJ
(2)298314.8)12(103.2313
××?+×?=P Q =-228.8kJ (3)kJ 184298314.8)22(101843?=××?+×?=P Q 18.已知下列反应在298K 时的热效应(1)Na (s )+1/2Cl 2(g )→NaCl (s )kJ 411m r ?=?H (2)H 2(g )+S (s )+2O 2(g )→H 2SO 4(l )kJ 3.811m r ?=?H (3)2Na (s )+S (s )+2O 2(g )→Na 2SO 4(s )kJ 1383m r ?=?H (4)1/2H 2(g )+1/2Cl 2(g )→HCl (g )
kJ
3.92m r ?=?H 求反应2NaCl (s )+H 2SO 4(l )→Na 2SO 4(s )+2HCl (g )在298K 时的m
r H ?和m r U ?。
解:根据赫斯定律,所求反应=[(3)+(4)×2]-[(1)×2+(2)]
]
2[]2[2,m r l ,m r 4,m r 3,m r m r H H H H H ?+×??×?+?=?=)]
3.811(2)411[(]2)3.92(1383[?+×??×?+?7.65=kJ·mol -1nRT
H U m r m
r ΔΔΔ?==74.60298314.82107.653
=××?×kJ