物理化學 (II) 第二次小考 (100%) 2007/12/14 姓名： R =8.3145 J/K‐mol; Cv(ideal gas) =3R/2; Cp(ideal gas) =5R/2; 1 atm =1.01325 bar 一、推導題 40% 1.
⎛ ∂T ⎞
The Joule coefficient, μJ, is defined as μ J = ⎜
⎟ . Show that μJ CV = p – αT/κ . ⎝ ∂V ⎠U
Ans: [3.30] μJ = (∂T/∂V)U and CV = (∂U/∂T)V so CV μJ = (∂U/∂T)V (∂T/∂V)U = ‐ (∂U/∂V)T [chain relation] = p – T (∂p/∂T)V (∂p/∂T)V = ‐(∂V/∂T)p (∂p/∂V)T = ‐ (∂V/∂T)p / (∂V/∂p)T = α /κT. Therefore μJ CV = p – αT/κT 2.
⎛ ∂H ⎞
⎛ ∂V ⎞
Derive the thermodynamic equation of state ⎜
⎟ =V −T ⎜
⎟ . Derive ⎝ ∂T ⎠P
⎝ ∂p ⎠T
⎛ ∂H ⎞
an expression of ⎜
⎟ for a perfect gas. ⎝ ∂p ⎠T
⎛ ∂H ⎞
⎛ ∂H ⎞ ⎛ ∂S ⎞ ⎛ ∂H ⎞
Ans: [3.26] ⎜
⎟ =⎜
⎟ +⎜
⎟ also H = H (p,S ), dH = T dS + V dp ⎟ ⎜
⎝ ∂p ⎠T ⎝ ∂S ⎠P ⎝ ∂p ⎠T ⎝ ∂p ⎠S
= ⎛⎜ ∂H ⎞⎟ dS + ⎛⎜ ∂H ⎞⎟ dp Thus ⎛⎜ ∂H ⎞⎟ = T and ⎝ ∂S ⎠P
⎝ ∂S ⎠P
⎝ ∂p ⎠S
⎛ ∂H ⎞
⎜
⎟ =V
⎝ ∂p ⎠S
Substitution yields, ⎛ ∂H ⎞
⎛ ∂S ⎞
⎛ ∂V ⎞
⎜
⎟ =T ⎜
⎟ +V = −T ⎜
⎟ +V ⎝ ∂T ⎠p
⎝ ∂p ⎠T
⎝ ∂p ⎠T
For ideal gas pV = nRT nR
⎛ ∂V ⎞
, hence ⎜
⎟ =
p
⎝ ∂T ⎠p
nRT
⎛ ∂H ⎞
+V = 0 ⎜
⎟ =−
∂
p
p
⎝
⎠S
3.
⎛ ∂ ⎛ Δr G
From the Gibbs–Helmholtz equation: ⎜
⎜
⎝ ∂T ⎝ T
⎛T ⎞
⎛ T ⎞
Δr G (T2 ) = Δr G (T1)⎜⎜ 2 ⎟⎟ + Δr H ⎜⎜1− 2 ⎟⎟ . ⎝ T1 ⎠
⎝ T1 ⎠
⎛ ∂ ⎛ Δr G
Ans: [3.16] ⎜
⎜
⎝ ∂T ⎝ T
− Δr H
⎞⎞
⎟⎟ =
T2
⎠ ⎠P
− Δr H
⎞⎞
, show that ⎟⎟ =
T2
⎠ ⎠P
dT
⎛ − Δr H ⎞
⎛ Δr G ⎞
⎟ dT = − Δr H ∫ 2 Integrate from T1 to T2. ⎟ = ∫⎜
2
T ⎠
T
⎠
⎝ T
∫ d ⎜⎝
Δr G (T2 )
T2
−
Δr G (T1)
T1
⎛ 1 1⎞
= Δr H ⎜⎜ − ⎟⎟ ⎝T2 T1 ⎠
⎛T ⎞
⎛ T ⎞
Δr G (T2 ) = Δr G (T1)⎜⎜ 2 ⎟⎟ + Δr H ⎜⎜1− 2 ⎟⎟ ⎝ T1 ⎠
⎝ T1 ⎠
4.
From Clapeyron equation
dp Δtrs S
=
, derive the Clausius–Clapeyron ΔtrsV
dT
Δvap H ⎛ 1 1 ⎞
⎜ −
⎟ χ. Specify all the R ⎝T T * ⎠
approximations and assumptions. equation p = p* exp(‐χ); χ =
Ans: Δ H
Δ S
dP
ΔP
=
= trs = vap
dT
ΔT
ΔtrsV T ΔvapV
At the liquid-vapor boundaries, ΔV m =V m (g ) -V m (l ) ≈V m (g ) ; ΔvapV ≈V (g ) ;V (g ) ≈ nRT
P
or V m (g ) ≈ RT P
Δvap H
Δ H
Δ H
dP
=
≈ vap
= vap 2 P ;
dT T (Vm (g )-Vm (l )) T Vm (g ) RT
dP
P
P2 / P °
∫
P1 / P °
= d ln
Δ H
Δ H
P
P
= vap 2 dT ; ln
= - vap + C
P°
RT
P ° RT
ΔvapH T dT
− ΔvapH ° ⎛ 1 1 ⎞
P
⎜⎜ − ⎟⎟ ;
=
; ln 2 =
2
∫
P
R TT
P1
R
⎝T2 T1 ⎠
dP
ln P2 = ln P1 −
2
1
⎡ − Δ H ° ⎛ 1 1 ⎞⎤
ΔvapH ° ⎛ 1 1 ⎞
⎜⎜ − ⎟⎟⎥ ⎜⎜ − ⎟⎟ ; P2 = P1 exp⎢ vap
R
R ⎝T2 T1 ⎠
⎝T2 T1 ⎠⎦
⎣
二、計算題 60% 1. What fraction of the enthalpy of vaporization of water ∆vapH (H2O) is spent on expanding the water vapor? ∆vapH = 40.656 kJ mol−1 Ans: [4.10a] ∆vapH = ∆vapU + ∆vap(pV); ∆vapH = 40.656 kJ mol−1 ∆vap(pV) = p ∆vap(V) = p(Vgas − Vliq) ≈ p Vgas = RT [per mole, perfect gas] ∆vap(pV) = (8.314 J K−1 mol−1) × (373.2 K) = 3102 kJ mol−1 Fraction = ∆vap(pV) / ∆vapH = (3102 kJ mol−1/ 40.656 kJ mol−1) = 0.0763 = 7.6 per cent 2. The heat capacity of chloroform (trichloromethane, CHCl3) in the range 240 K to 330 K is given by Cp,m/(J K–1 mol–1) = 91.47 + 7.5 × 10–2 (T/K). In a particular experiment, 1.00 mol CHCl3 is heated from 273 K to 300 K. Calculate the change in molar entropy of the sample. Ans: [3.2] Sm(Tf) ‐ Sm(Ti) = ʃ (Cp,m /T) dT = ʃ (a + bT)/T dT = a ln(T2/T1 )+ b(T2 − T1) ∆S = ʃ (91.47 + 7.5×10–2 (T/K)) /T ) dT = 91.47 ln(300/273) + 7.5 × 10–2 (300‐273) = 10.8 J K–1 for 1.00 mol. Therefore, for 1.00 mol, ∆S = +11 J K−1 3. 1.00 mol of perfect gas molecules at 27°C is expanded isothermally from an initial pressure of 3.00 atm to a final pressure of 1.00 atm in two ways: (a) reversibly, and (b) against a constant external pressure of 1.00 atm. Determine the values of q, w, ΔU, ΔH, ΔS, ΔSsurr, ΔStot for each path. Ans: [3.6] (Path a) reversible ΔU = ΔH = 0 (isothermal); w = ‐nRT ln(V f/V,i) = ‐nRT ln(pi/pf) = ‐(1.0 mol)(8.3145 J K‐1mol‐1)(300 K) ln (3.00 atm/1.00 atm) = ‐2.74x103 J = ‐2.74 kJ; q = ΔU ‐ w = +2.74 kJ; ΔS = qrev/T = +2.74x103 J/300 K = +9.13 J K‐1; ΔStot = 0; Ssurr = ΔStot ‐ ΔS =0‐9.13 J K‐1= ‐9.13 J K‐1 (Path b) constant external pressure w = ‐ p,ex(V f ‐ V,i) = ‐ p,ex nRT (pf ‐ p,i) = ‐nRT (p,ex /pf ‐ p,ex /p,i) = ‐(1.0 mol)(8.3145 J K‐1mol‐1)(300 K)(1.00 atm/1.00 atm – 1.00atm/3.00 atm) = ‐1.66x103 J = ‐1.66 kJ; ΔU = ΔH = 0 (isothermal); q =ΔU – w = 0 – (‐1.66 kJ) = +1.66 kJ; ΔS = qrev/T = +9.13 J K‐1 (state function); Ssurr = qsurr/T = ‐q/T = ‐1.66x103 J / 300 K = ‐5.53 J K‐1; ΔStot = ΔS + Ssurr = (9.13 – 5.53) J K‐1= ‐9.13 J K‐1 = +3.60 J K‐1; 4. When a certain liquid freezes at –3.65°C its density changes from 0.789 g cm–3 to 0.801 g cm–3. Its enthalpy of fusion ∆fusH = 8.68 kJ mol–1. Estimate the freezing point of the liquid at 100 MPa. [4.5b] Ans: ∆T = (∆fusV / ∆fusS )× ∆p = Tf (∆fusV / ∆fusH ) × ∆p = Tf (∆p M / ∆fusH ) ∆ (1/ρ) [Tf = −3.65 + 273.15 = 269.50 K] ∆T = (269.50 K) × (99.9 MPa)x M / 8.68 kJ mol−1 × [(1 / 0.789 g cm−3) – (1 / 0.801 g cm−3)] = (3.1017 × 106 KPa J−1 mol) × (M) × (+.01899 cm3/g) × (1x10‐6 m3 cm‐3) = (+5.889 × 10−2 K Pa m3 J−1 g−1 mol) M = (+5.889 × 10−2 Kg−1 mol) M ∆T = (46.07 g mol−1) × (+5.889 × 10−2 Kg−1 mol) = +2.71K Tf = 269.50 K + 2.71 K = 272 K 5.
Calculate ΔrG o (375 K) for the reaction 2 CO(g) + O2(g) → 2 CO2(g) from the values of ΔrG o (298K) and ΔrH o (298 K), and equation ⎛ T ⎞
⎛T ⎞
Δr G (T2 ) = Δr G (T1)⎜⎜ 2 ⎟⎟ + Δr H ⎜⎜1− 2 ⎟⎟ . Given that at 298 K, ⎝ T1 ⎠
⎝ T1 ⎠
∆fG o (CO2, g) = ‐394.36 kJ mol‐1, ∆fG o (CO, g) = ‐137.17 kJ mol‐1, ∆fH o (CO2, g) = ‐393.51 kJ mol‐1, ∆fH o (CO, g) = ‐110.53 kJ mol‐1. Ans: [3.16] ΔrG o (298 K) = 2×∆fG o (CO2, g) ‐ 2 × ∆fG o (CO, g) ‐ 1×∆fG o (O2, g) = 2x(‐394.36) ‐ 2x(‐137.17) ‐1x(0) = ‐514.38 kJ mol‐1 ΔrHo (298 K) = 2×∆fH o (CO2, g) ‐ 2 × ∆fH o (CO, g) ‐ 1×∆fH o (O2, g) = 2x(‐393.51) ‐ 2x(‐110.53) ‐1x(0) = ‐565.96 kJ mol‐1 ΔrG o (375 K) = (‐514.38 kJ mol‐1)(375 K/298 K) + (‐565.96 kJ mol‐1)x[1 ‐ (375 K/298 K)] = ‐501.05 kJ mol‐1 6.
Combine the barometric formula ( p = p0 e −M g h RT ) for the dependence of the pressure on altitude with the Clausius–Clapeyron equation, and predict how the boiling temperature of a liquid depends on the altitude and the ambient temperature. Take the mean ambient temperature as 20°C and predict the boiling temperature of water at 3000 m. Ans: The barometric formula: p = p0 exp(‐Mgh/RT) p = p* exp(‐χ); χ = (∆vapH/R) x[(1/T) – (1/T*)] Let T* = Tb the normal boiling point; then p* = 1 atm. Let T = Th, the boiling point at the altitude h. Take p0 = 1 atm. Boiling occurs when the vapor (p) is equal to the ambient pressure, that is, when p(T ) = p(h), and when this is so, T = Th. Therefore, since p0 = p*, p(T) = p(h) implies that exp(−Mgh/RT ) = exp{−(∆vapH/R) x [(1/Th) – (1/Tb)]} It follows that 1/Th = (1/Tb) + Mgh/(T∆vapH) where T is the ambient temperature and M the molar mass of the air. For water at 3000m, using M = 29 g mol‐1 1/Th = 1/373 K + (29×10‐3 kg mol‐1)(9.81 m s‐2)(3.000 × 103 m) /(293K) × (40.7×103 J mol‐1) = 1/373 K + 1/1.397× 104 K Hence, Th = 363K (90℃)