Chemistry 431 Problem Set 10 Fall 2018 Solutions

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1 Chemistry 43 Problem Set 0 Fall 208 Solutions. A lead bullet is fired at a wooden plank. At what speed must it be traveling to melt upon impact if its initial temperature is 00. C and the heating of the plank is neglected? he melting point of lead is 327. C, the specific heat of solid lead is.254 J g K, and the heat of fusion of lead is 5.8 kj g. 0 2 mv2 = m0.254 J g K 3 g 0 327K 00K + m J g 3 g kg kg v 2 = 0 3 2[ ] m 3 v = 3228 m s 2. Seven grams of ice at -0.0 C are dropped into 80.0 g of water at 30. C and the system is allowed to reach equilibrium adiabatically. Calculate the final temperature of the system. he specific heat of water is 4.8 J g K, the specific heat of ice is 2.09 J g K, and the heat of fusion of ice is J g. q + q 2 = 0 7 g2.09 J K g 0 K + 7 g334.7 J g +7 g 273 K4.8 J g K + 80 g4.8 J g K 303 K = 0 = K = 20.7 C 3. One mole of liquid water is originally contained in a capsule that is enclosed in an evacuated tank of fixed volume that is thermostated to 00. C. he capsule is broken and the liquid water is allowed to evaporate into the previously evacuated tank. After equilibrium is established at 00. C, the pressure of the water vapor in the tank is found to be 0.5 atmospheres. Assuming the water vapor to be an ideal gas and neglecting the volume of the liquid water compared to that of the vapor, calculate U, H, S, A, G, q and w for the process. he enthalpy of vaporization of water at 00. C and. atmospheres pressure is 44. kj mol. w = 0 H = J

2 Figure : A representation of the flask discussed in problem 4 U = H P V = H P 2 V 2 P V H P 2 V 2 = H n = J mol8.344 J mol K 373 K = J q = U = J S = H + n P i P f = J 373 K + mol8.344 J mol K 0.5 = 24 J K G = H S = J 373 K24 J K = 2252 J A = U S = J 373 K24 J K = 5353 J Notice that A < 0 because the process is spontaneous at constant and V. G is unrestricted in sign for such a process. 4. In the container pictured in Figure, the left hand side initially consists of water in equilibrium with its vapor and the right hand side is evacuated. he two sides, each of volume 20. liters, are connected with a closed stop-cock and the entire system is placed in a heat bath thermostated to 25. C. he vapor pressure of water at 25. C is 24. mm and the enthalpy of vaporization of water is 44. kj mol. If the stopcock is opened, calculate U, H, S, A, G, q and w when equilibrium is reached. You may assume some liquid water remains on the left hand side at the end of the process, and you may neglect the volume of the liquid compared to the vapor. You may also 2

3 assume the vapor obeys the ideal gas law. n = P V = atm20 L L atm mol K 298 K = mol H = mol44000 J mol = 36 J S = H = 36 J = 3.8 J K 298 K G = H S = 0 U = H P V = H n = 36 J mol8.344 J mol K 298 K = 072 J A = U S = 072 J 298 K3.8 J K = 64 J w = 0 q = U = 072 J As with the previous problem, the process is spontaneous and constant and V. A < 0 and G can be anything. 5. A 0.0 g sample of liquid water at 373 K is placed in a capsule, and the capsule containing the water is put in a large, evacuated tank of fixed volume. he capsule is broken at a fixed temperature of 373 K, and the pressure of the water vapor in the previously evacuated tank is found to be bar. During the evaporation of the liquid water heat is absorbed from the surroundings, and a measurement of the heat finds q = 22.7 kj. Assuming the enthalpy of vaporization of water to be independent of temperature, use the data to calculate the vapor pressure of water at 50.0 C. 0.0 g U = J n = 8.0 g mol = mol mu = J = J mol mol m H = m U+ P V = m U+ = J mol J mol K 373 K = J mol P 2 = vap,mh P 2 P.0 bar J mol = J mol K P = 0.2 bar 6. he vapor pressure of solid ammonia is given by P = and that of liquid ammonia is given by P = where the pressure is expressed in millimeters of mercury K 323 K

4 a What is the temperature of ammonia at the triple point? = = 95.2K b What are the latent enthalpies of sublimation and vaporization of ammonia at the triple point? P = H + constant sub,m H = 3754K sub,m H = 3754K8.344J mol K = 3.4kJ mol vap,m H3063K8.344J mol K = 25.47kJ mol c What is the latent heat of fusion of ammonia at the triple point? solid liquid vapor solid so that fusion,m H = sub,m H vap,m H = kJ mol = 5.67kJ mol 7. he vapor pressure of solid SO 2 is atm at 77.0K and atm at 95.8K. he vapor pressure of liquid SO 2 is atm at 209.6K and.32 0 atm at 225.3K. Calculate the temperature of sulfur dioxide at the triple point. P 2 = sub,mh P = sub,mh 2 sub,m H P 2 = vap,mh P = vap,mh vap,m H 95.8 K 77.0 K = K K K = 33. K P solid = K 4 + K solid

5 = K solid K solid = 7.35 P liquid = 33. K + K liquid.32 0 = K liquid K liquid = 2.67 At the triple point K = 33. K = 200.K 8. he density of liquid bismuth is 0. kg liter, and the density of solid bismuth is 9.67 kg liter. he normal melting point of bismuth is 270. C and the enthalpy of fusion of bismuth is kj kg. Calculate the melting point of bismuth under 00. atmospheres pressure. P 2 P = fus,mh fus,m V 2 2 = fus,mv fus,m H P 2 P = 0. L kg L kg 00atm atm 5278J kg 6 L atm 8.344J = J L atm = = 543K exp = 542.6K 9. he temperature at the triple point of xenon is 6.40K and the normal boiling point of xenon is 65.K. Given that the vapor pressure of liquid xenon is 0.92 atmospheres at 63.0K, calculate the vapor pressure of xenon at the triple point. Assume the enthalpy of vaporization of xenon is temperature independent. P 2 = vap,mh P = vap,mh vap,m H P 63.0 K 65. K = 74.9 K atm = 74.9 K 65. K 6.40 K P = atm 5

6 0. he enthalpy of vaporization of water is 44. kj mol, and the enthalpy of fusion of ice is 6. kj mol. If the vapor pressure of ice at 0. C is atmospheres, what is the vapor pressure of ice at -20. C? sub,m H = vap,m H + fus,m H P atm = kJ mol = 50kJ mol 50000J mol = 8.344J mol K P = atm 273K 253K. he triple point of methanech 4 occurs at t =90.70 K and P t =0.70 bar. he critical point of methane occurs at c =90.6 K and P c =46.0 bar. Given the enthalpy of sublimation of methane is sub,m H=9.700 kj mol, calculate the melting temperature of methane when the total pressure is.000 bar. he densities of liquid and solid methane are respectively ρ l = g cm 3 and ρ s = g cm 3. You can assume the densities of the liquid and solid are pressure and temperature independent, and you can assume the enthalpies of fusion, sublimation and vaporization are also temperature and pressure independent. P 2 = vap,mh P = vap,m H J mol K K vap,m H = J mol 90.6 K fus,m H = sub,m H vap,m H = 00. J mol P 2 = P + fus,mh fus,m V J mol.000 bar = 0.70 bar + cm g cm g = K L bar J L 0 3 cm K 6.00 g mol 2. he triple point of argon is t = K and P t = bar, and the critical point of argon is c = K and P c = bar. At = 8.44 K, the vapor pressure of solid argon is measured to be P = bar. Assuming the enthalpy changes for all phase transition in argon are temperature independent, and given the densities of solid and liquid argon are respectively ρ solid =.66 g cm 3 and ρ liquid =.400 g cm 3, calculate the standard freezing point of argon. liquid-vapor : P 2 = vap,mh P 2 6

7 = vap,m H J mol K K vap,m H = 6680 J mol K solid-vapor : = sub,m H J mol K 8.44 K sub,m H = 7700 J mol K bar bar = fus,m H = sub,m H vap,m H = 020 J mol solid-liquid : P 2 P = fus,mh fus,m V L bar 020 J mol J 39.9 g mol cm g cm3.66 g = K 3. a Using an Euler-Maxwell relation, show that γ P σ, = V σ P, dg = Sd + V dp + γdσ γ V = P σ b For a spherical bubble of radius, r, show that σ, V σ P, = r 2.,P L 0 3 cm K V = 4 3 πr3 V σ dv = 4πr 2 dr σ = 4πr 2 dσ = 8πrdr,P = 4πr2 dr 8πrdr = r 2 7

8 c Use parts a and b to derive the Laplace equation P in P out = 2γ r. γ P σ, = r 2 dγ = r 2 dp or dp = 2dγ r P in P out = 2γ r 4. What is the ratio of the vapor pressure of mercury above the meniscus of a capillary 0 7 meters in radius to the vapor pressure of mercury above a flat surface at 20 C? he surface tension of mercury is N m, and the density of mercury is 3.6 gm cm 3. Assume a zero contact angle mercury has a convex meniscus. P v P v0 = V m,l = 2cm3 /3.6g g mol m/0 2 cm N/m J mol K 293 K0 7 m P v P v0 =.058 2γ r = Use the data of problem number 4 to predict the capillary depression of mercury in a tube. mm diameter at 20. C. h = 2γ ρgr h = 3.6 g kg ρ = cm g cm = kg m kg s kg m m s m m = 0.04 m =.4 cm 6. A sample of benzene of mass 00. grams is dispresed as droplets of radius.0 µm. he surface tension of benzene is N m and its density is 0.88 gm cm 3. What is the change in the Helmholtz free energy for the dispersal? What is the minimum 8

9 work necessary to bring about the dispersal? da = Sd P dv + γdσ A = γ σ constant and V Let N be the number of droplets. he volume of the droplets is N 4 3 πr3 = m ρ where m is the mass and ρ is the density of the fluid. hen he surface area of the droplets is N = 3m 4πr 3 ρ hen = 3m rρ = σ = σ = N4πr 2 = 3m 4πr 3 ρ 4πr2 ignoring the surface area of the inital sample w = A = γ σ kg s 2 0. kg 0 6 m kg/0 3 m 3 = 9.5 J 7. By how much is the vapor pressure of benzene changed when it is dispersed in the form of small droplets of radius a 0 µm; b 0.0 µm; both at 25. C? { } 2γVm,l P mist = P bulk exp r V m,l = 78.2 g 0.88 g cm = 89 3 cm3 mol 2γV m,l = N m m 3 mol J mol = m 9

10 For r =. 0 5 m For r =. 0 7 m P mist P bulk P mist P bulk = exp = exp { } m = m { } m = m 8. he vapor pressure of of liquid methanol CH 3 OH is 0.52 bar at 50.0 C, the standard boiling point of methanol is 64.3 C, the surface tension of methanol at 298.K is N m, and the density of methanol at 298.K is g cm 3. Assuming the enthalpy of vaporization of methanol is temperature independent, calculate the vapor pressure of methanol at 298.K when the liquid is dispersed into droplets of radii meters. P 2 = vap,mh P = vap,mh vap,m H = 4967 K K 323 K = 4967 K P K 298 K P 0 = 0.43 bar { } 2γVm,l P = P 0 exp r cm N m 3 32 g m g mol 0 = 0.43 bar exp 2 cm. 0 8 m8.344 J mol K 298.K = 0.54 bar 3 9. Given that for liquid water, the vapor pressure at 298.K is bar, the enthalpy of vaporization is J mol, the surface tension is N m 2 and the density is.00 g cm 3, calculate the temperature at which water droplets of radii m have a vapor pressure of bar. Assume that the enthalpy of vaporization, the density and the surface tension of water are all temperature independent. P flat P flat 298 = vap,mh 298 K 0

11 { vap,m H P flat = P flat 298 exp 298. K } { } 2γVm,l P drop = P flat exp r { vap,m H = P flat 298 exp 298. K + 2γV } m,l r { J mol bar = bar exp J mol K 298. K.00 cm N m 2 3 m g g cm mol m8.344 J mol K = 356. K 20. At 298 K the surface tension and density of ethanol C 2 H 5 OH are respectively γ = J m 2 and ρ = g cm 3. When 0.0 g of liquid ethanol are dispersed into small spherical droplets, the change in the Helmholtz free energy is measured to be A = J and the vapor pressure of the dispersed droplets is found to be bar. Calculate the vapor pressure of the bulk sample of liquid ethanol at 298K when the sample is not dispersed into droplets. A = N4πr 2 γ N = A 4πr 2 γ Mρ = N 4 3 πr3 = 4 A 3 πr3 4πr 2 γ cm 3 m 3 r = 3Mγ 30.0 g J m 2 ρ A = g 0 2 cm = m J [ ] 2γVm,l P = P flat exp r cm J m g m g mol 0 bar = P flat exp 2 cm m8.344 J mol K 298 K P flat = bar

12 2..00 moles of liquid water at 25.0 C are dispersed into spherical droplets, and it is found the vapor pressure of the droplets is exactly 4 times the vapor pressure of water above a flat surface. Given the density of water is.00 g cm 3 and the surface tension of water is N m, calculate the radii of the water droplets and the change in the Helmholtz free energy associated with the dispersal. { } 2γVm,l P = P 0 exp r = 2γV m,l P = 2γV m,l r = P 0 r P/P 0 cm N m 3 8 g m 3 g mol 0 2 cm J mol K = m 298 K 4 A = Nγ σ = Nγ4πr 2 N 4 3 πr3 = m ρ N = 3m 4πr 3 ρ = 38 g 4π m 3 g 0 2 cm 3 = cm 3 m A = N m 4π m 2 = 59 J 2

Problem Set 10 Solutions

Chemistry 360 Dr Jean M Standard Problem Set 10 Solutions 1 Sketch (roughly to scale) a phase diagram for molecular oxygen given the following information: the triple point occurs at 543 K and 114 torr;