Page:of 724 Automatic Zoom P 3000atm D 0.17in A S 4 D 2 A 0.023in 2 F PA g 32.174 ft sec 2 mass F g mass 1000.7lb m Ans. 1.7 P abs U g h P atm = U 13.535 gm cm 3 g 9.832 m s 2 h 56.38cm P atm 101.78kPa P abs U g h P atm P abs 176.808kPa Ans. 1.8 U 13.535 gm cm 3 g 32.243 ft s 2 h 25.62in P atm 29.86in_Hg P abs U g h P atm P abs 27.22psia Ans. Chapter 1 - Section A - Mathcad Solutions 1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this equation by setting t(F) = t(C). Guess solution: t 0 Given t 1.8t 32 = Find t( ) 40 Ans. 1.5 By definition: P F A = F mass g = Note: Pressures are in gauge pressure. P 3000bar D 4mm A S 4 D 2 A 12.566mm 2 F PA g 9.807 m s 2 mass F g mass 384.4kg Ans. 1.6 By definition: P F A = F mass g = 1 F Mars Kx F Mars 4 10 3 u mK g Mars F Mars mass g Mars 0.01 mK kg Ans. 1.12 Given: z P d d U g = and: U MP RT = Substituting: z P d d MP RT g = Separating variables and integrating: P sea P Denver P 1 P ́ μ μ ¶ d 0 z Denver z Mg RT § ̈ © · ¹ ́ μ μ ¶ d = After integrating: ln P Denver P sea § ̈ © · ¹ M g RT z Denver = Taking the exponential of both sides and rearranging: P Denver P sea e M g RT z Denver § ̈ © · ¹ = P sea 1atm M 29 gm mol g 9.8 m s 2 1.10 Assume the following: U 13.5 gm cm 3 g 9.8 m s 2 P 400bar h P U g h 302.3m Ans. 1.11 The force on a spring is described by: F = K s x where K s is the spring constant. First calculate K based on the earth measurement then g Mars based on spring measurement on Mars. On Earth: F mass g = Kx = mass 0.40kg g 9.81 m s 2 x 1.08cm F mass g F 3.924N K s F x K s 363.333 N m On Mars: x 0.40cm 2 Ans. w moon Mg moon w moon 18.767lbf Ans. 1.14 cost bulb 5.00dollars 1000hr 10 hr day cost elec 0.1dollars kW hr 10 hr day 70 W cost bulb 18.262 dollars yr cost elec 25.567 dollars yr cost total cost bulb cost elec cost total 43.829 dollars yr Ans. 1.15 D 1.25ft mass 250lb m g 32.169 ft s 2 R 82.06 cm 3 atm mol K T 10 273.15 ( )K z Denver 1 mi Mg RT z Denver 0.194 P Denver P sea e M g RT z Denver § ̈ © · ¹ P Denver 0.823atm Ans. P Denver 0.834bar Ans. 1.13 The same proportionality applies as in Pb. 1.11. g earth 32.186 ft s 2 g moon 5.32 ft s 2 ' l moon 18.76 ' l earth ' l moon g earth g moon ' l earth 113.498 M ' l earth lb m M 113.498lb m 3 and hence (since each B is negative ) δ ij > 0. If unlike interactions are stronger than like interactions, | B ij | > 1 2 | B ii + B jj | Hence δ ij < 0. For identical interactions of all molecular pairs, B ij = B ii = B jj , and δ ij = 0 The rationalizations of signs for H E of binary liquid mixtures presented in Sec. 16.7 apply approxi mately to the signs of δ 12 for binary gas mixtures. Thus, positive δ 12 is the norm for NP/NP, NA/NP, and AS/NP mixtures, whereas δ 12 is usually negative for NA/NA mixtures comprising solvating species. One expects δ 12 to be essentially zero for ideal solutions of real gases, e.g. , for binary gas mixtures of the isomeric xylenes. 16.8 The magnitude of Henry’s constant H i is reflected through Henry’s law in the solubility of solute i in a liquid solvent: The smaller H i , the larger the solubility [see Eq. (10.4)]. Hence, molecular f actors that influence solubility also influence H i . In the present case, the triple bond in acetylene and the doubl e bond in ethylene act as proton acceptors for hydrogen-bond form ation with the donor H in water, the triple bond being the stronger acceptor. No hydrogen bonds form between ethane and water. Because hydrogen-bond formation between unlike species promotes solubility through smaller values of G E and γ i than would otherwise obtain, the values of H i are in the observed order. 16.9 By Eq. (6.70), H αβ = T S αβ . For the same temperaature and pressure, less structure or or der means larger S . Consequently, S sl , S l v , and S s v are all positive, and so therefore are H sl , H l v , and H s v . 16.11 At the normal boiling point: H l v ≡ H v − H l = ( H v − H ig ) − ( H l − H ig ) = H R ,v − H R , l Therefore H R , l = H R ,v − H l v At 1(atm), H R ,v should be negligible relative to H l v . Then H R , l ≈− H l v . Because the normal boiling point is a representative T for typical liquid behavior, and because H R reflects intermolecular forces, H l v has the stated feature. H l v (H 2 O) is much larger than H l v (CH 4 ) because of the strong hydrogen bonding in liquid water. 16.12 By definition, write C l P = C ig P + C R , l P , where C R , l P is the residual heat capacity for the liquid phase. Also by definition, C R , l P = (∂ H R , l /∂ T ) P . By assumption (modest pressure levels) C ig P ≈ C v P . Thus, C l P ≈ C v P + ∂ H R , l ∂ T P For liquids, H R , l is highly negative, becoming less so as T increases, owing to diminution of intermolecular forces (see, e.g., Fig. 6.5 or Tables E.5 and E.6). Th us C R , l P is positive , and C l P > C v P . 16.13 The ideal-gas equation may be written: V t = n RT P = N N A · RT P ⇒ V t N = RT N A P The quantity V t / N is the average volume available to a particle, and the averag e length available is about: V t N 1 / 3 = RT N A P 1 / 3 V t N 1 / 3 = 83.14 cm 3 bar mol − 1 K − 1 × 300 K 6 . 023 × 10 23 mol − 1 × 1 bar × 10 6 cm 3 m − 3 1 / 3 = 34 . 6 × 10 − 10 m or 34.6 ̊ A For argon, this is about 10 diameters. See comments on p. 64 9 with respect to separations at which attractions become negligible. 723 Page:of 724 Automatic Zoom P 3000atm D 0.17in A S 4 D 2 A 0.023in 2 F PA g 32.174 ft sec 2 mass F g mass 1000.7lb m Ans. 1.7 P abs U g h P atm = U 13.535 gm cm 3 g 9.832 m s 2 h 56.38cm P atm 101.78kPa P abs U g h P atm P abs 176.808kPa Ans. 1.8 U 13.535 gm cm 3 g 32.243 ft s 2 h 25.62in P atm 29.86in_Hg P abs U g h P atm P abs 27.22psia Ans. Chapter 1 - Section A - Mathcad Solutions 1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this equation by setting t(F) = t(C). Guess solution: t 0 Given t 1.8t 32 = Find t( ) 40 Ans. 1.5 By definition: P F A = F mass g = Note: Pressures are in gauge pressure. P 3000bar D 4mm A S 4 D 2 A 12.566mm 2 F PA g 9.807 m s 2 mass F g mass 384.4kg Ans. 1.6 By definition: P F A = F mass g = 1 F Mars Kx F Mars 4 10 3 u mK g Mars F Mars mass g Mars 0.01 mK kg Ans. 1.12 Given: z P d d U g = and: U MP RT = Substituting: z P d d MP RT g = Separating variables and integrating: P sea P Denver P 1 P ́ μ μ ¶ d 0 z Denver z Mg RT § ̈ © · ¹ ́ μ μ ¶ d = After integrating: ln P Denver P sea § ̈ © · ¹ M g RT z Denver = Taking the exponential of both sides and rearranging: P Denver P sea e M g RT z Denver § ̈ © · ¹ = P sea 1atm M 29 gm mol g 9.8 m s 2 1.10 Assume the following: U 13.5 gm cm 3 g 9.8 m s 2 P 400bar h P U g h 302.3m Ans. 1.11 The force on a spring is described by: F = K s x where K s is the spring constant. First calculate K based on the earth measurement then g Mars based on spring measurement on Mars. On Earth: F mass g = Kx = mass 0.40kg g 9.81 m s 2 x 1.08cm F mass g F 3.924N K s F x K s 363.333 N m On Mars: x 0.40cm 2 Ans. w moon Mg moon w moon 18.767lbf Ans. 1.14 cost bulb 5.00dollars 1000hr 10 hr day cost elec 0.1dollars kW hr 10 hr day 70 W cost bulb 18.262 dollars yr cost elec 25.567 dollars yr cost total cost bulb cost elec cost total 43.829 dollars yr Ans. 1.15 D 1.25ft mass 250lb m g 32.169 ft s 2 R 82.06 cm 3 atm mol K T 10 273.15 ( )K z Denver 1 mi Mg RT z Denver 0.194 P Denver P sea e M g RT z Denver § ̈ © · ¹ P Denver 0.823atm Ans. P Denver 0.834bar Ans. 1.13 The same proportionality applies as in Pb. 1.11. g earth 32.186 ft s 2 g moon 5.32 ft s 2 ' l moon 18.76 ' l earth ' l moon g earth g moon ' l earth 113.498 M ' l earth lb m M 113.498lb m 3 and hence (since each B is negative ) δ ij > 0. If unlike interactions are stronger than like interactions, | B ij | > 1 2 | B ii + B jj | Hence δ ij < 0. For identical interactions of all molecular pairs, B ij = B ii = B jj , and δ ij = 0 The rationalizations of signs for H E of binary liquid mixtures presented in Sec. 16.7 apply approxi mately to the signs of δ 12 for binary gas mixtures. Thus, positive δ 12 is the norm for NP/NP, NA/NP, and AS/NP mixtures, whereas δ 12 is usually negative for NA/NA mixtures comprising solvating species. One expects δ 12 to be essentially zero for ideal solutions of real gases, e.g. , for binary gas mixtures of the isomeric xylenes. 16.8 The magnitude of Henry’s constant H i is reflected through Henry’s law in the solubility of solute i in a liquid solvent: The smaller H i , the larger the solubility [see Eq. (10.4)]. Hence, molecular f actors that influence solubility also influence H i . In the present case, the triple bond in acetylene and the doubl e bond in ethylene act as proton acceptors for hydrogen-bond form ation with the donor H in water, the triple bond being the stronger acceptor. No hydrogen bonds form between ethane and water. Because hydrogen-bond formation between unlike species promotes solubility through smaller values of G E and γ i than would otherwise obtain, the values of H i are in the observed order. 16.9 By Eq. (6.70), H αβ = T S αβ . For the same temperaature and pressure, less structure or or der means larger S . Consequently, S sl , S l v , and S s v are all positive, and so therefore are H sl , H l v , and H s v . 16.11 At the normal boiling point: H l v ≡ H v − H l = ( H v − H ig ) − ( H l − H ig ) = H R ,v − H R , l Therefore H R , l = H R ,v − H l v At 1(atm), H R ,v should be negligible relative to H l v . Then H R , l ≈− H l v . Because the normal boiling point is a representative T for typical liquid behavior, and because H R reflects intermolecular forces, H l v has the stated feature. H l v (H 2 O) is much larger than H l v (CH 4 ) because of the strong hydrogen bonding in liquid water. 16.12 By definition, write C l P = C ig P + C R , l P , where C R , l P is the residual heat capacity for the liquid phase. Also by definition, C R , l P = (∂ H R , l /∂ T ) P . By assumption (modest pressure levels) C ig P ≈ C v P . Thus, C l P ≈ C v P + ∂ H R , l ∂ T P For liquids, H R , l is highly negative, becoming less so as T increases, owing to diminution of intermolecular forces (see, e.g., Fig. 6.5 or Tables E.5 and E.6). Th us C R , l P is positive , and C l P > C v P . 16.13 The ideal-gas equation may be written: V t = n RT P = N N A · RT P ⇒ V t N = RT N A P The quantity V t / N is the average volume available to a particle, and the averag e length available is about: V t N 1 / 3 = RT N A P 1 / 3 V t N 1 / 3 = 83.14 cm 3 bar mol − 1 K − 1 × 300 K 6 . 023 × 10 23 mol − 1 × 1 bar × 10 6 cm 3 m − 3 1 / 3 = 34 . 6 × 10 − 10 m or 34.6 ̊ A For argon, this is about 10 diameters. See comments on p. 64 9 with respect to separations at which attractions become negligible. 723
