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

=???? ???-?=???

? ?

?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

V V V

-???

? ?

?-=2

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

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

?????+??-==--?

=20349–380+625=20.62 kJ

(2) Q V =?U=?H –?(pV)=?H –(p 2V 1–

p 1V 1)???

?-?-?=1

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

,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

102106.506p

V p p p ln nRT 332

=???===-

（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) p 1T 1=p 2T 2 K 5.136p T p T 2

12==

3222m 0028.04

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 =

nRT T /c nRT p nRT V 2=

TdT 2C

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

m 0246.0p RT V

体

积

变

化

：

32m 003026.0p RT V V ==

压力

Pa 821554V RT p 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

.V m

,p ==γ γγ-γγ

-=12

2111p T p T

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

11=??==-

γγ

211V p V p

kPa

7.11486.243.1p 3V

V p p 406

.121

12=??? ??=?

? ??=γ

K 9.224RT V p T 2

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

.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

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

H p

298

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

398

()??+?+?+?=dT cT

bT a H 2

298

398

298

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

复旦 物理化学 第一章 习题答案

复旦物理化学第一章习题答案