物理化学(复旦大学药学院) 第一章习题答案

物理化学(复旦大学药学院）第一章习题答案第一章 习题解答1. (1) Q =∆U -W =200-160=40 kJ (2) ∆U =Q +W =260-100=160 kJ2.()pR T T p nRp nRT p nRT V =-=-=∆1212 ∴W = -p ∆V = -R3. (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 160100101801092001801013251133ls.....M p -=⎪⎪⎭⎫ ⎝⎛⨯-⨯-=⎪⎪⎭⎫ ⎝⎛--=ρρ4. 在压力p 和房间容积V 恒定时，提高温度，部分空气溢出室外，因此室内气体量n 是温度的函数，TR pV n 1⋅=。

kJ 7721288298ln 32960100000ln 12m m 2121.R .**T T C R pV T dT C R pV dT nC Q ,p T Tm ,p T T,p p =====⎰⎰5. (1) 恒温可逆膨胀 J 429902501023*******2-=⨯-=-=..ln..VVln RT W(2) 真空膨胀 W = 0(3) 恒外压膨胀 W = -p外(V 2–V 3) =()122V V VRT--⎪⎪⎭⎫ ⎝⎛--=211V V RT ⎪⎭⎫ ⎝⎛-⨯⨯-=100250123733148....= -2327 J(4) 二次膨胀 W =W 1 + W 2 ⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫⎝⎛--=322111VV RT VV RTJ310310050105002501-=⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛--=..RT ..RT以上结果表明，功与具体过程有关，不是状态函数。

6. (1) 理想气体为系统，等压过程， Q =∆H >0 (2) 电热丝+理想气体为系统，等压过程，Q =0，∆U =W '>0，∆H =∆U +∆(pV )>07. ∆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 8.9. C p,m =29.07–0.836⨯103T +2.01⨯10-6T 2(1) Q p =∆H 10003003623m1001231108360210729d 21⎥⎦⎤⎢⎣⎡⨯⨯+⨯⨯-==--⎰T .T ..T C n T T ,p=20349–380+625=20.62 kJ(2) Q V =∆U =∆H –∆(pV )=∆H –(p 2V 2–p 1V 1) =∆H –nR (T 2–T 1)=20.62–R (1000-300)⨯10-3 =14.80 kJ10．（1）等温可逆膨胀 ∆U =∆H = 0Q = -W J 163115ln 102106506ln 33211121=⨯⨯⨯===-.p p V p p p ln nRT （2）等温恒外压膨胀 ∆U =∆H =0Q = -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 J11. （1）常压蒸发：Q =∆H = 40.7 kJ W = -p (V g –V l )≈ -pV g ≈ -RT = -8.314⨯373= -3.1 kJ ∆U =Q +W =37.6 kJ(2) 真空蒸发：∆H =40.7 kJ W =0 ∆U =Q =37.6 kJ 12.K 2273111.nRV p T ==(1) p 1T 1=p 2T 2 K 51362112.p T p T ==3222m 0028045136.R.p nRT V ===(2) ∆U =nC V ,m (T 2–T 1)=J 17022273513623-=-)..(R ∆H =nC p ,m (T 2–T 1)=J 28372273513625-=-)..(R (3) 以T 为积分变量求算： pT =C (常数) TCp =CnRT T /C nRT p nRT V 2===T T CnRdV d 2⋅=J 227022d 2d 12=--=-=⋅⋅-=-=⎰⎰⎰)T T (nR dT nR T CnRT T C V p W也可以用p 或V 为积分变量进行求算。

13．∆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 3111m 02460.p RT V == 体积变化：33332m 0030260.pRTV V === 压力Pa 821554222==V RT pW =W 1+W 2= -p 2(V 2–V 1)+0=-821554⨯(0.003026–0.0246)= 17724 J Q=∆U -W =1464–17724=–16260 J14. (1) ∆H =C p ,m (T 2–T 1)20922273252=-=).T (R T 2=373.8 K121122nRT nRT V p V p =2273837321052..p =⨯ p 2=0.684⨯105 Pa∆U = C V ,m (T 2–T 1)12552273837323=-=)..(R J W =∆U –Q =1255–1674=–419 J (2)2121状态＝恒容，中间状态恒温，状态−−−−−−→−−−−−→−W W状态函数变化同上 ∆U =1255 J ∆H =2092 JW =W 112VV ln nRT -==–8.314⨯273.2⋅ln2=–1574 JQ =∆U –W =1255+1574 = 2829 J15. 双原子分子 R C ,V 25m = 41.C C m.V m ,p ==γγγγγ--=122111p T p TK1224655032510122734141112112....p p T T ..=⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=--γγW =∆U =nC V ,m (T 2–T 1)()J 10202273122425-=-=..R16. （1）等温可逆膨胀 ∆U =∆H = 0Q = W J 1822412250ln 1527312-=-=-=.R .V V ln nRT （2）绝热可逆膨胀 C p .m =20.79 C V .m = C p .m -R =12.4766661476127920...C C m,V m ,p ===γTV γ-1=常数 K06160504122152731666112112...V V T T .=⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛=--γQ =0∆U =W =nC V ,m (T 2–T 1)=12.476⨯(160.06–273.15)= -1411 J∆H =nC p ,m (T 2–T 1)= 20.79⨯(160.06–273.15)= -2351 J17. (1) 4061828828mm .R..C C .V ,p =-==γmol 1755029843199303111.R..RT V p n =⨯==γγ2211V p V p =kPa 71148624319930340612112....V V p p .=⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=γK 92242122.RT V p T ==(2) ∆U =nC V ,m (T 2–T 1)=0.1755⨯(28.8–R )⨯(224.9–298)= -263 J ∆H =nC p ,m (T 2–T 1)=0.1755⨯28.8⨯(224.9–298))= -369 J18. 设物质量为1mol415253.R.R.C C m,V m ,p ===γ 始态温度：6240314810101025111...nR V p T =⨯⨯⨯== K 求终态温度：122111--=γγV T VT1553010624014112112..V V T T .=⎪⎭⎫⎝⎛=⎪⎪⎭⎫⎝⎛=--γK1779624001552512-=-=-=∆)..(R )T T (nC U m ,V J 2490624001552712-=-=-=∆)..(R )T T (nC H m ,p J Q =0 W =∆U = -1779 J若设物质量为n mol ，可如下计算：始态温度n .nR V p T 6240111==， 终态温度n.V V T T 015512112=⎪⎪⎭⎫ ⎝⎛=-γ 在计算∆U 、∆H 、W 时n 可消去，得相同的结果。

19. 证明 U =H –pV pppppT V p C T V p T H T U ⎪⎭⎫⎝⎛∂∂-=⎪⎭⎫ ⎝⎛∂∂-⎪⎭⎫ ⎝⎛∂∂=⎪⎭⎫ ⎝⎛∂∂20. 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 Tp ⎪⎪⎭⎫ ⎝⎛∂∂+⎪⎭⎫⎝⎛∂∂=V 不变，对T 求导 VT p VT 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 p21. 发酵反应 C 6H 12O 6(s) −→ 2C 2H 5OH(l) +2CO 2(g) ∆n =2∆U =∆H -∆nRT = -67.8 -2R ⨯298⨯10-3 = -72.76 kJ ⋅mol -122． n Q V +C ∆T =005817594210050=⨯+..Q .V Q V =–4807200 JC 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 -123．(1) 2H 2S(g)+SO 2(g) = 8H 2O(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) + O 2(g) = 2CO(g) ∆n = 1 Q V =–231.3 kJ∆r H m = Q V + ∆nRT = –231.3 +RT ⨯10–3 = –228.8 kJ(3) H 2(g)+Cl 2(g) = 2HCl (g) ∆n =0 Q V =–184 kJ ∆r H m = Q V =–184 kJ24. (1) ξ=4 mol (2) ξ=2 mol (3) ξ=8 mol25．2NaCl(s) + H2SO4(l) = Na2SO4(s) + 2HCl(g)∆f H O m(kJ⋅mol-1) –411 –811.3 –1383–92.3∆r H O m=∑(n∆f H O m)产物–∑(n∆f H O m)反应物= (–1383–2⨯92.3)–(–811.3–2⨯411) = 65.7 kJ⋅mol-1∆r U O m=∆r H︒m–∆nRT=65.7–2RT⨯10-3=60.7 kJ⋅mol-126．生成反应7C(s) + 3H2(g) + O2(g) = C6H5COOH(l)∆c H O m(kJ⋅ mol-1) –394 –286 –3230 ∆r H O m=∑(n∆c H O m)反应物–∑(n∆c H O m)产物= [7⨯(–394) + 3⨯(–286)] – (–3230)= –386 kJ⋅mol-127. 反应C(石墨) → C(金刚石)∆c H O m(kJ⋅mol-1) –393.5 –395.4∆r H O m=∆c H O m，石墨–∆c H O m，金刚石=–393.5–(–395.4)=1.9 kJ⋅mol-128. 3C2H2 (g) = C6H6 (l)∆f H O m (kJ⋅mol-1) 226.73 49.04∆c H O m (kJ⋅mol-1) –1299.6 –3267.5由生成热数据计算：∆r H O m=49.04–3⨯226.73=–631.15 kJ⋅mol-1∆r U O m=∆r H O m–∆nRT=–631.15+3⨯8.314⨯298.2⨯10-3= -623.71 kJ⋅mol-1 由燃烧热数据计算：∆r H O m=3⨯(–1299.6)–(–3267.5) =–631.3 kJ⋅mol-1∆r U O m=∆r H O m–∆nRT=–631.3+3⨯8.314⨯298.2⨯10-3= -623.86 kJ⋅mol-129. 反应KCl(s) → K+(aq, ∞) + Cl-(aq, ∞) ∆f H O m(kJ⋅mol-1) –435.87 ? –167.44 ∆r H O m=17.18 kJ⋅mol-1∆r H O m=∑(n∆f H O m)产物–∑(n∆f H O m)反应物17.18=[∆f H O m (K+,aq, ∞)–167.44]–(–435.87)∆f H O m (K+,aq, ∞)=–251.25 kJ⋅mol-130. 生成反应H2(g) + 0.5O2(g) = H2O(g)∆r H O m,298=–285.83 kJ⋅mol-1C p,m(J⋅K-1⋅mol-1) 28.824 29.355 75.291∆C p =75.291–(28.824+0.5⨯29.355)=31.79 J ⋅K -1 ⎰∆+∆=∆T C H H p d O m ,298r O m ,373r =–285.83+31.79⨯(373–298)⨯10-3 =–283.45 kJ ⋅mol -131. 反应N 2(g) + 3H 2(g) = 2NH 3(g) ∆r H O m,298=–92.88 kJ ⋅mol -1a 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 ⎰∆+∆=∆T C H H p d O m ,298r O m ,398r ()⎰∆+∆+∆+∆=T cT bT a H d 2O m ,298r39829832O m,298r 3121⎥⎦⎤⎢⎣⎡∆+∆+∆+∆=cT bT aT H 3982983723O m,298r )109117(31)10662(21)4162(⎥⎦⎤⎢⎣⎡⨯-+⨯+-+∆=--T .T .T .H=–92880+[–6241+2178–144]= –97086 J ⋅mol -132.-1⋅mol -1 ⋅mol -1 r m ，473∆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∆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 H O m,473=∆r H O m,291+∆H HI5296+69632+∆r H O m,473=49455+10592∆r H O m,473=–14.881 kJ ⋅mol-1。