물리화학 1 문제풀이 130403 김대형교수님
Chapter 1 Exercise (#1) A sample of 255 mg of neon occupies 3.00 dm 3 at 122K. Use the perfect gas law to calculate the pressure of the gas. Solution 1) The perfect gas law p = nnn V 2) T = 122 K, V = 3.00 dm 3 3) p = 1.22 11 2 mmm 0.00000 dd 3 aaa K 1 mmm 1 (111 K) 3.0 dd 3 = 4. 22 11 2 aaa
Chapter 1 Exercise (#2) Given that the density of air at 0.987 bar and 27 C is 1.146 kg m -3, calculate the mole fraction and partial pressure of nitrogen and oxygen assuming that (a) air consists only of these two gases, (b) air also contains 1.0 mole percent Ar. Solution (a) 1) For simplicity assume a container of volume 1 m 3. The total mass is m τ = n N2 M N2 + n O2 M O2 = 1111 g 2) Assuming that air is a perfect gas p τ V = n τ RR Then, n τ is the total amount of gas n τ = p τv RR = 0. 999 bbb 111 PP bbb 1 (1 m 3 ) 8. 3333 PP m 3 K 1 mmm 1 (333 K) = 33. 6 mmm 3) n N2 22. 0000 g mmm 1 + 33. 6 mmm n N2 33. 9999 g mmm 1 n N2 = 33. 16 mmm n O2 = 9. 41 mmm = 1111 g
Chapter 1 4) The mole fractions x N2 = 33. 16 mmm 33. 6 mmm = 0. 777 x O2 = 9. 41 mmm 33. 6 mmm = 0. 222 The partial pressures p N2 = p O2 = 0. 777 0. 999 bbb = 0. 777 bbb 0. 222 0. 999 bbb = 0. 222 bbb
Chapter 1 Solution (b) 1) For simplicity assume a container of volume 1 m 3. The total mass is m τ = n N2 M N2 + n O2 M O2 + n AA M AA = 1111 g 2) The total amount of gas n τ = n N2 + n O2 + n AA = 33. 6 mmm Since x AA = 0. 0000, nnn = 0. 333 mmm n N2 = 33. 94 mmm n O2 = 8. 22 mmm n AA = 0. 333 mmm
Chapter 1 3) The mole fractions x N2 = 33. 94 mmm 33. 6 mmm = 0. 782 x O2 = 8. 22 mmm 33. 6 mmm = 0. 208 x AA = 0. 333 mmm 33. 6 mmm = 0. 000 The partial pressures p N2 = 0. 782 0. 999 bbb = 0. 772 bbb p O2 = 0. 208 0. 999 bbb = 0. 205 bbb p AA = 0. 000 0. 999 bbb = 0. 0000 bbb
Chapter 1 Exercise (#3) A gas at 250 K and 15 atm has a molar volume 12 percent smaller than that calculated from the perfect gas law. Calculate (a) the compression factor under these conditions and (b) the molar volume of the gas. Which are dominating in the sample, the attractive or the repulsive forces? Solution (a) 1) Compression factor Z Z = pp m RR = V m V o m (VV: aaaaaa mmmmm vvvvvv, VVV: ppppppp ggg mmmmm vvvvvv) 2) V m = 0.88 V o m Z = 0.88888 V o m = 0. 88 (b) 1) VV = ZZZ p = 0.88 0.00000 dd3 aaa K 1 mmm 1 (222 K) 11 aaa = 1. 2 dd 3 mmm 1 2) Z < 1 Attractive forces dominate
Chapter 1 Exercise (#4) The critical constants of methane are p c = 45.6 atm, V c = 98.7 cm 3 mol -1, and T c = 190.6 K. Calculate the van der Waals parameters of the gas. Solution 1) b = VV 3, a = 22 b 2 pp = 33 2 c pp 2) Substituting the critical constants b = 99. 7 cc3 mmm 1 3 = 33. 9 cc 3 mmm 1 a = 3 99. 7 11 3 dd 3 mmm 1 2 44. 6 aaa = 1. 33 dd 6 aaa mmm 2
Chapter 1 Exercise (#5) A certain gas obeys the van der Waals equation with a = 0.50 m 6 Pa mol -2. Its volume is found to be 5.00 10-4 m 3 mol -1 at 273 K and 3.0 MPa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure? Solution 1) van der Waals equation 2) Substituting the data b = VV RR a (p + V 2 ) m b = 5. 00 11 4 m 3 mmm 1 8. 3333 J K 1 mmm 1 222 K 0. 55 m 3. 0 111 PP + 6 PP mmm 2 5. 00 11 4 m 3 mmm 1 2 = 0. 44 11 4 m 3 mmm 1
Chapter 1 3) Compression factor Z = pp m RR = 3. 0 111 PP (5. 00 11 4 m 3 mmm 1 ) 8. 3333 J K 1 mmm 1 (222 K) = 0. 66
Chapter 2 2.5(a) A sample of 4.50 g of methane occupies 12.7dm 3 at 310K. (a) calculate the work done when the gas expands isothermally against a constant external pressure of 200 Torr until its volume has increase by 3.3 dm 3. (b) Calculate the work that Would be done if the same expansion occurred reversibly. * 133.3 Pa = 1 Torr (a) p = p ee w = p ee ΔΔ p ee = (200 TTTT) (133.3PP T ooo) = 2.67 10 4 PP ΔΔ = 3.3dm 3 = 3.3 10 3 m 3 w = (2.67 10 4 PP) (3.3 10 3 m 3 ) = 88J
Chapter 2 2.5(a) A sample of 4.50 g of methane occupies 12.7dm 3 at 310K. (a) calculate the work done when the gas expands isothermally against a constant external pressure of 200 Torr until its volume has increase by 3.3 dm 3. (b) Calculate the work that Would be done if the same expansion occurred reversibly. * 133.3 Pa = 1 Torr (b) p = nn T V dd = ppp V 2 w = nnn 1 V dd = nnn ln V 2 V 1 V 1 4.50g n = = 0.2805 mmm 16.04g mml 1 RR = (8.3145J K 1 mml 1 ) (310K) = 2.577kk mml 1 w = (0.2805 mmm) (2.577kk mml 1 ) ln 16.0 dm3 12.7 dm 3 = 167J
2.13(a) When 3.0 mol O 2 is heated at a constant pressure of 3.25 atm, its temperature increases from 260 K to 285 K. Given that the molecular heat capacity of O 2 (g) at constant pressure is 29.4 J K -1 mol -1, calculate q, H, and U. q, H q p = C p ΔΔ = nc p ΔΔ = (3.0 mmm) (29.4J K 1 mml 1 ) (25K) = +2.2kk ΔΔ = q p = +2.2kk
2.13(a) When 3.0 mol O 2 is heated at a constant pressure of 3.25 atm, its temperature increases from 260 K to 285 K. Given that the molecular heat capacity of O 2 (g) at constant pressure is 29.4 J K -1 mol -1, calculate q, H, and U. U [Solution 1] [Solution 2] c p = c v + R c v = 21.1J K 1 mml 1 ΔΔ = c v c p ΔΔ ΔΔ = ΔΔ w = ΔΔ Δ(pp) = ΔΔ p ee ( nnt 2 p ee nnt 1 p ee = ΔΔ nnnn U = +1.6 kj
2.22(a) The standard enthalpy of decomposition of the yellow complex H 3 NSO 2 into NH 3 and SO 2 is +40 kj mol -1. Calculate the standard enthalpy of formation of H 3 NSO 2. Δ r H Φ (NH 3, g) = 46.11kk mml 1 Δ f H Φ (SO 2, g) = 296.83kk mml 1 NH 3 SO 2 (s) NH 3 (g) + SO 2 (g) Δ r H Φ = +40 kk mml 1 Δ f H Φ (NH 3 SO 2, s) = Δ r H Φ (NH 3, g) + Δ f H Φ (SO 2, g) Δ r H Φ = ( 46.11 296.83 40) kk mml 1
2.29(a) Set up a thermodynamic cycle for determining the enthalpy of hydration of Mg 2+ ions using the following data: enthalpy of sublimation of Mg(s), 167.2 kj mol -1 First and second ionization enthalpies of Mg(g); 7.646 ev and 15.035 ev; Dissociation enthalpy of Cl2(g), +241.6 kj mol -1 ; electron gain enthalpy of Cl(g), 3.78eV; enthalpy of solution of MgCl 2 (s), -150.5 kj mol -1 ; enthalpy of hydration of Cl - (g), 383.7 kj mol -1. Δ f H Φ (MMMM 2, s) = 641.32kk mml 1 1: sublimation of Mg(s), 167.2 kj mol -1, 2: first and second ionization enthalpies of Mg(g); 7.646 ev and 15.035 ev; = 737.7 kj mol -1, 1450.7 kj mol -1 3: dissociation enthalpy of Cl2(g), +241.6 kj mol -1 ; 4: electron gain enthalpy of Cl(g), 3.78eV = -364.7kJ mol -1, 5: enthalpy of solution of MgCl 2 (s), -150.5 kj mol -1 ; 6: enthalpy of hydration of Cl - (g), 383.7 kj mol -1 ; 7: formation of MgCl 2 (s), -641.32 kj mol -1.
3 4 2 2 6 1 7 5
( 150.5) ( 641.32) + (167.2) + (241.6) + 737.7 + 1450.7) + 2 ( 364.7) + 2 ( 383.7) + ΔH hyy (Mg 2+ ) = 0 1111kk mml 1
2.31(a) For a van der Waals gas, π T = a/v m2. Calculate U m for the isothermal expansion of nitrogen gas from an initial volume of 1.00 dm 3 to 24.8 dm 3 at 298K. What are the values of q and w? a = 1.352 dm 6 atm mol -2 b = 3.9 dm 3 mol -1 문제조건이부족해서추가합니다. 2.31(a) For a van der Waals gas, π T = a/v m2. Calculate U m for the reversible isothermal expansion of nitrogen gas from an initial volume of 1.00 dm 3 to 24.8 dm 3 at 298K. What are the values of q and w? a = 1.352 dm 6 atm mol -2 b = 3.9 dm 3 mol -1
2.31(a) For a van der Waals gas, π T = a/v m2. Calculate U m for the isothermal expansion of nitrogen gas from an initial volume of 1.00 dm 3 to 24.8 dm 3 at 298K. What are the values of q and w? a = 1.352 dm 6 atm mol -2 b = 3.9 dm 3 mol -1 U du m = V m,2 ΔU m = du m V m,1 U m T Vm dd + U m V m dd = 0 (IIIIIIIIII) du m = U m V m = V m,2 V m,1 T dv m = a V dv 2 m m a dv V m = a m T dv m 24.8dm 3 mml 1 1.00dm 3 mml 1 a V m dv m a = 24.8 dm 3 mml 1 + a 1.0 dm 3 = 0.9597a mmm dm 3 mml 1 = +131J mml 1
2.31(a) For a van der Waals gas, π T = a/v m2. Calculate U m for the isothermal expansion of nitrogen gas from an initial volume of 1.00 dm 3 to 24.8 dm 3 at 298K. What are the values of q and w? a = 1.352 dm 6 atm mol -2 b = 3.9 dm 3 mol -1 q, w w = ppv m w = p = V m,2 V m,1 RR V m b a V 2 m RR V m b a V 2 m dv m = V m,2 V m,1 RR V m b dv m V m,2 V m,1 w = ΔU m q a V m 2 dv m = q = V m,2 V m,1 V m,2 V m,1 RR V m b dv m RR V m b dv m + ΔU m
2.31(a) For a van der Waals gas, π T = a/v m2. Calculate U m for the isothermal expansion of nitrogen gas from an initial volume of 1.00 dm 3 to 24.8 dm 3 at 298K. What are the values of q and w? a = 1.352 dm 6 atm mol -2 b = 3.9 dm 3 mol -1 24.8dm 3 mml 1 RR q = V m b dv 24.8dm m = RRln V m b 3 mml 1 1.010dm 3 mml 1 1.010dm 3 mml 1 = (8.314J K 1 mml 1 ) (298K) ln = +8.05 10 3 J mml 1 w = ΔU m q = 7.92 10 3 J mml 1 24.8 3.9 10 2 1.00 3.9 10 2