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=02. (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 33ls=⨯⎪⎭⎫ ⎝⎛⨯-⨯=⎪⎪⎭⎫ ⎝⎛ρ-ρ=3. (1)恒温可逆膨胀J 4299025.01.0ln 2.373314.8V V lnRT W 12=⨯==(2)真空膨胀 W = 0(3)恒外压膨胀 W = p外(V 2–V 3) =()122V V VRT-⎪⎪⎭⎫ ⎝⎛-=21V V 1RT ⎪⎭⎫ ⎝⎛-⨯⨯=1.0025.012.373314.8= 2327 J(4)二次膨胀 W=W 1 + W 2 ⎪⎪⎭⎫⎝⎛-+⎪⎪⎭⎫ ⎝⎛-=3221V V 1RT V V 1RTJ31031.005.01RT 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 J5. C p,m =29.07–0.836⨯103T+2.01⨯10–6T 2(1) Q p =∆H 10003003623T T m ,p T 1001.231T 10836.02107.29dT C n 21⎥⎦⎤⎢⎣⎡⨯⨯+⨯⨯-==--⎰=20349–380+625=20.62 kJ(2) Q V =∆U=∆H –∆(pV)=∆H –(p 2V 1–p 1V 1)⎪⎪⎭⎫⎝⎛-⋅-∆=1122nRT V VnRT 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 46.29300100020621T Q C ⋅⋅=-=∆=6．（1）等温可逆膨胀 ∆U =∆H = 0Q =W J 163115ln102106.506pplnV p p p ln nRT 33211121=⨯⨯⨯===-（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 J 7. K 2.273nRV p T111==(1) p 1T 1=p 2T 2 K 5.136p T p T 2112==3222m 0028.04R5.136p 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 25-=-(3) 以T 为积分变量求算: pT=C(常数) TC p =CnRT T /c nRT p nRT V 2===TdT 2CnRdV ⋅=J 2270)T T (nR 2dT nR 2dT CnRT 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 3111m 0246.0p RT V==体积变化：33332m 003026.0p RT V V ===压力Pa 821554V RT p 222==W=W 1+W 2=p 2(V 2–V 1)+0=821554⨯(0.003026–0.0246)=–17724 JQ=∆U+W=1464–17724=–16260 J9. 双原子分子 R 25C m,V = 4.1CCm.V m,p ==γ γγ-γγ-=122111p T p TK1.224p 5.0p2.273p p T T 4.14.1112112=⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛=-γγ-W=–∆U=–nC V ,m (T 2–T 1)()J 10202.2731.224R 25=--=10. (1) 406.1R 8.288.28CC m.V m,p =-==γ mol 1755.0R298104.1p 3RT V p n 3111=⨯⨯==-γγ=2211V p V pkPa7.11486.243.1p 3VV p p 406.12112=⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=γK 9.224RT V p T 2122==(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 J11. 证明 U =H –pV pp p p p T V p C T V p T H T U ⎪⎭⎫ ⎝⎛∂∂-=⎪⎭⎫ ⎝⎛∂∂-⎪⎭⎫ ⎝⎛∂∂=⎪⎭⎫ ⎝⎛∂∂12. 证明 ⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫⎝⎛∂∂-⎪⎭⎫ ⎝⎛∂∂-⎪⎭⎫⎝⎛∂∂=⎪⎭⎫ ⎝⎛∂∂-⎪⎭⎫ ⎝⎛∂∂=-V V p V p V pT p V T H T H T U T H C C（1）H=f(T,p) dpp H dT T H dH T p⎪⎪⎭⎫ ⎝⎛∂∂+⎪⎭⎫⎝⎛∂∂=V 不变，对T 求导 V T 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 p13． n Q V +C ∆T=005.817594.2Q 1005.0V=⨯+ 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 –114．(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 =0Q V =–184 kJ∆r H m = Q V =–184 kJ15.(1) ξ=4 mol (2) ξ=2 mol (3) ξ=8 mol16．2NaCl(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=60.7kJ17. ∆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 kJ18．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 kJ19. 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 kJ20.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 kJ21. 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 kJ22.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–123. 生成反应H2(g) + 0.5O2(g) = H2O(g)∆r H298=–285.8 kJ⋅mol–1C 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⎰∆+∆=∆dTCHH p298r373r=–285.8+31.9⨯(373–298)⨯10–3=–283.4 kJ⋅mol–124. 反应N2(g) + 3H2(g) = 2NH3(g) ∆r H298=–92.888 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 ∆–62.4162.599–117.904 ⎰∆+∆=∆dT C H H p298r398r()⎰∆+∆+∆+∆=dT cTbT a H 2298r39829832298r cT 31bT 21aT H ⎥⎦⎤⎢⎣⎡∆+∆+∆+∆= 3982983723298r T )109.117(31T )106.62(21T )41.62(H ⎥⎦⎤⎢⎣⎡⨯-+⨯+-+∆=--=–92880+[–6241+2178–144]=97086 J25.∆H H2=nC p,m ∆T=3.5R(473–291)=5296 J ∆H HI =nC p,m ∆T=2⨯3.5R(473–291)=10592 Jr 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 HI5296+69632+∆r H473=49455+10592∆r H473=–14881 J。