复旦物理化学第一章习题答案
第一章习题解答
1. 体系为隔离体系, ?U=0 W=Q=0
2. (1)
W=p ?V=p(V g -V l )≈pV g =nRT=1?8.314?373.15=3102 J (2) W=p ?V=p(V s –V l )J 16.0018.0100.111092.01101325M 11p 3
3l
s
=???? ???-?=???
? ?
?ρ-ρ=
3. (1)恒温可逆膨胀
J 4299025
.01
.0ln 2.373314.8V V ln
RT W 12=?==
(2)真空膨胀 W = 0
(3)恒外压膨胀 W = p
外
(V 2–V 3) =
()1
22
V V V
RT
-???
? ?
?-=2
1
V V 1RT ?
?
? ??-??=1.0025.012.373314.8= 2327 J
(4)二次膨胀 W=W 1 + W 2 ???
?
??-+???? ?
?-=3221
V V 1RT V V 1RT
J
31031.005.01RT 05.0025.01RT =??
? ??-+??? ??-=
4. ?H=n ??H m,汽化=40670 J
?U=?H –?(pV)=?H –p (V g -V l )=40670–101325(30200–1880)?10–6
=40670–3058=37611 J
5. C p,m =29.07–0.836?103
T+2.01?10–6T 2
(1) Q p =?H 1000
300
3623T T m ,p T 1001.231T 10836.02107.29dT C n 2
1
?
??
?????+??-==--?
=20349–380+625=20.62 kJ
(2) Q V =?U=?H –?(pV)=?H –(p 2V 1–
p 1V 1)???
?
?
?-?-?=1
12
2
nRT V V
nRT H V 2=V 1 ∴ Q V =?H –nR(T 2–T 1)=20.62–R(1000-300)?10–3
=14.80 kJ
(3) 1
-1
-
m
,p m
,p mol K J 46.29300
100020621
T Q C ??=-=?=
6.(1)等温可逆膨胀 ?U =?H = 0
Q =W J 16311
5
ln
102106.506p
p
ln
V p p p ln nRT 332
11
1
2
1
=???===-
(2)等温恒外压膨胀 ?U =?H = 0
Q = W = p 2 (V 2–V 1) = p 2V 2–p 2V 1= p 1V 1–p 2V 1= (p 1–p 2)V 1
=(506.6-101.3)?103?2?10–3 = 810 J 7. K 2.273nR
V p T
1
11
==
(1) p 1T 1=p 2T 2 K 5.136p T p T 2
1
12==
3222m 0028.04
R
5.136p nRT V ===
(2) ?U=nC V ,m (T 2–T 1)=J 1702)2.2735.136(R 23
-=- ?H=nC p,m (T 2–T 1)=J 2837)2.2735.136(R 2
5-=-
(3) 以T 为积分变量求算: pT=C(常数) T
C p =
C
nRT T /c nRT p nRT V 2=
==
TdT 2C
nR
dV ?=
J 2270)T T (nR 2dT nR 2dT C
nRT 2T C pdV W 12-=-==??==???
也可以用p 或V 为积分变量进行求算。
8. ?U=nC V ,m (T 2–T 1)=20.92?(370–300)=1464 J
?H=nC p,m (T 2–T 1)=(20.92+R)?(370–300)=2046 J 始态体积 31
1
1
m 0246.0p RT V
==
体
积
变
化
:
33
3
32m 003026.0p RT V V ==
=
压力
Pa 821554V RT p 2
2
2==
W=W 1+W 2=p 2(V 2–V 1)+0=821554?(0.003026–0.0246)=–17724 J
Q=?U+W=1464–17724=–16260 J
9. 双原子分子 R 25C m
,V = 4.1C
C
m
.V m
,p ==γ γγ-γγ
-=12
2111p T p T
K
1.224p 5.0p
2.273p p T T 4
.14
.11121
12=???
?
??=?
??
? ??=-γ
γ-
W=–?U=–nC V ,m (T 2–T 1)()J 10202.2731.224R 2
5=--=
10. (1) 406.1R 8.288
.28C
C m
.V m
,p =-==γ mol 1755.0R
298104.1p 3RT V p n 3
1
11=??==-
γγ
=
2
211V p V p
kPa
7.11486.243.1p 3V
V p p 406
.121
12=??? ??=?
??
? ??=γ
K 9.224RT V p T 2
1
22==
(2) ?U=nC V .m (T 2–T 1)=n (28.8–R)?(224.9–298)= –263 J ?H=nC p.m (T 2–T 1)=n ?28.8?(224.9–298)=
–369 J
11. 证明 U =H –pV p
p p p p T V p C T V p T H T U ??? ????-=??? ????-??? ????=??? ????
12. 证明 ?
??????????
????-??? ????-???
????=??? ????-??? ????=-V V p V p V p
T p V T H T H T U T H C C
(1)
H=f(T,p) dp
p H dT T H dH T p
???? ????+?
?
?
????=
V 不变,对T 求导 V T p V
T p p H T H T H ??? ?
??????? ????+???
????=?
?
?
???? 代
入(1)
???
?????-????
??????? ????-=??? ????+??? ???????? ????-=-V p H T p T p V T p p H C C T V V V T V p
13. n Q V +C ?T=0
05.817594.2Q 100
5
.0V
=?+ Q V =–4807200 J C 7H 16(l) + 11O 2(g) = 8H 2O(l) + 7CO 2(g) ?n =–4
?c H m = Q V + ?nRT =–4807200–4R ?298 = –4817100 J ?mol –1 =–4817.1 kJ ?mol –1
14.(1) 2H2S(g)+SO2(g) = 8H2O(l) + 3S(斜方) ?n =–3 Q V =–223.8 kJ
?r H m= Q V+ ?nRT = –223.8 + (–3)RT?10–3 = –231.2 kJ
(2) 2C(石墨) + O2(g) = 2CO2(g) ?n = 1 Q V =–231.3 kJ
?r H m= Q V+ ?nRT = –228.8 +RT?10–3 = –228.8 kJ
(3) 2H2(g)+Cl2(g) = HCl (g) ?n =0
Q V =–184 kJ
?r H m = Q V =–184 kJ
15.(1) ξ=4 mol (2) ξ=2 mol (3) ξ=8 mol
16.2NaCl(s) + H2SO4(l) = Na2SO4(s) + 2HCl(g)
?f H?m(kJ?mol–1) –411 –811.3 –1383–92.3
?r H?m=∑(ν?f H?m)产物–∑(ν?f H?m)反应物= (–1383–2?92.3)–(–811.3–2?411) = 65.7 kJ
?r U?m=?r H?m–?nRT=65.7–2RT?10–3=60.7
kJ
17. ?r H?m=∑(ν?c H?m)反应物–∑(ν?c H?m)产物= (–2?283–4?285.8)–(0–1370)=339.2 kJ
18.生成反应7C(s) + 3H2(g) + O2(g) = C6H5COOH(l)
?c H?m(kJ?mol–1) –394 –286 –3230 ?r H?m=∑(ν?c H?m)反应物–∑(ν?c H?m)产物= [7?(–394) + 3?(–286)] – (–3230)= –386 kJ
19. 反应C(石墨) → C(金刚石)
?c H?m(kJ?mol–1) –393.5 –395.4
?r H?m=?c H?m,石墨–?c H?m,金刚石=–393.5–(–395.4)=1.9 kJ
20.反应CH4(g)+2O2(g) = CO2(g) + 2H2O(l)
?f H?m(kJ?mol–1) –74.8 –393.5 –285.8
?r H?m=∑(ν?f H?m)产物–∑(ν?f H?m)反应物=[–393.5+2?(–285.5)]–(–74.8)=–890.3 kJ
21. 反应(COOH)2(s)+2CH3OH(l) = (COOCH3)2(l) + 2H2O(l)
?c H?m(kJ?mol–1) –251.5 –726.6 –1677.8 0
?r H?m=∑(ν?c H?m)反应物–∑(ν?c H?m)产物=[–251.5+2?(–726.6)]–(–1677.8)=–26.9 kJ
22.反应KCl(s) → K+(aq, ∞) + Cl–(aq, ∞) ?f H?m(kJ?mol–1) –435.87 ? –167.44
?r H?m=17.18 kJ
?r H?m=∑(ν?f H?m)产物–∑(ν?f H?m)反应物
17.18=[?f H?m(K+,aq, ∞)–167.44]–(–435.87)
?f H?m (K+,aq, ∞)=–251.25 kJ?mol–1
23. 生成反应H2(g) + 0.5O2(g) = H2O(g)?r H298=–285.8 kJ?mol–1
C p,m(J?K–1?mol–1) 28.83 29.16 75.31
?C p=75.31–(28.83+0.5?29.16)=31.9J?K–1
??
+
?
=
?dT
C
H
H p
298
r
373
r=–285.8+31.9?(373–298)?10–
3=–283.4 kJ?mol–1
24. 反应N2(g) + 3H2(g) = 2NH3(g) ?r H298=–92.888 kJ?mol–1
a b ?103 c ?107 N 2(g) 26.98 5.912 –3.376 H 2(g) 29.07 –0.837 20.12 NH 3(g) 25.89 33.00 –30.46 ?
–62.41
62.599
–117.904 ??+?=?dT C H H p
298
r
398
r
()??+?+?+?=dT cT
bT a H 2
298
r
398
298
3
2298
r cT 31bT 21aT H ??
?
????+?+?+?= 398
298
3723298
r T )109.117(31T )106.62(21T )41.62(H ?
??
????-+?+-+?=--
=–92880+[–6241+2178–144]=97086 J
25.
?H H2=nC p,m ?T=3.5R(473–291)=5296 J ?H HI =nC p,m ?T=2?3.5R(473–291)=10592 J
r 473
?H I2=?H1(s,291→386.7K) + ?H2(s→l) + ?H3(l,1386.7→457.5K) + ?H4(l→g)
+ ?H5(g,457.5→473K)
=55.64?(386.7–291)+16736+62.76?(457.5–386.7)+42677+3.5R?(473–457.5)
=69632 J
?H H2+?H I2+?r H473=?r H291+?H HI
5296+69632+?r H473=49455+10592
?r H473=–14881 J