Recitation 06. n total = P total V/RT = (0.425 atm * 10.5 L) / ( L atm mol -1 K -1 * 338 K) = mol
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1 Recitatin 06 Mixture f Ideal Gases 1. Chapter 5: Exercise: 69 The partial pressure f CH 4 (g) is atm and that f O 2 (g) is atm in a mixture f the tw gases. a. What is the mle fractin f each gas in the mixture? b. If the mixture ccupies a vlume f 10.5 L at 65 0 C, calculate the ttal number f mles f gas in the mixture. c. Calculate the number f grams f each gas in the mixture. Slutin: (a) ml fractin f CH 4, χ CH4 = P CH4 / P ttal = atm / (0.175 atm atm) = χ O2 = χ CH4 = = (b) PV = nrt n ttal = P ttal V/RT = (0.425 atm * 10.5 L) / ( L atm ml -1 K -1 * 338 K) = ml (c) χ CH4 = n CH4 / n ttal n CH4 = χ CH4 * n ttal = * ml = 6.63 x10-2 ml CH 4 = 6.63 x10-2 ml CH 4 x g/ml CH 4 = 1.06 g CH 4 n O2 = x.161 ml = 9.47 x10-2 ml O 2 = 9.47 x10-2 ml O 2 x 32.0 g/ml O 2 = 3.03 g O 2 Real Gases 2. Chapter 5: Exercise: 89 Calculate the pressure exerted by ml N 2 in a L cntainer at C a. Using the ideal gas law. b. Using the van der Waals equatin. c. Cmpare the results Slutin:
2 (a) Ideal gas law: PV = nrt P = nrt/v = [0.500 ml * L atm ml -1 K -1 * ( )K] / L = atm (b) van der Waals equatin: [P + a(n/v) 2 ] * (V nb) = nrt Fr N 2, a = 1.39 atm L 2 /ml 2 and b = L/ml [P (0.500/1.000) 2 atm] * (1.000 L * L) = ml * L atm ml -1 K -1 * ( )K Or, (P atm) * ( L) = atm P = (12.24 L atm/ L) atm = = atm (c) The ideal gas law is high by 0.11 atm r (0.11/12.13)*100 = 0.91 % Thus, using ideal gas law we gt an ver estimated value. Intermlecular frces 3. Chapter 10: Exercise: 33 Ratinalize the difference in biling pints fr each f the fllwing pairs f substances: a. n-pentane, C nepentane, C b. HF 20 0 C HCl C c. HCl C LiCl C d. n-pentane C n-hexane 69 0 C Slutin: (a) Nepentane is mre cmpact than n-pentane. Therefre, there is less surface area cntact amng nepentane mlecules. This leads t weaker Lndn dispersin frces and a lwer biling pint. (b) HF is capable f the hydrgen bnding, but HCl is nt. (c) LiCl is inic and HCl is a mlecular slid with nly diple frces and Lndn dispersin frces. Inic frces are much strnger than the frces fr mlecular slids.
3 (d) N-Hexane is a larger mlecule, s it has strnger Lndn dispersin frce Phase diagrams 4. Chapter 10: Exercise: 91 Cnsider the phase diagram given belw. What phases are present at pints A thrugh H? Identify the triple pint, nrmal biling pint, nrmal freezing pint, and critical pint. Which phase is denser, slid r liquid? Slutin: A: Slid B: Liquid C: Gas/Vapur D: Slid + Vapur E. Slid + Liquid + Vapur F: Liquid + Vapur G: Liquid + Vapur H: Vapur The triple pint is E, critical pint is G Nrmal freezing pint: Temperature at which slid-liquid line is at 1 atm Nrmal biling pint: Temperature at which liquid-vapur line is at 1 atm
4 Since the slid-liquid line has a psitive slpe, the slid phase is denser than the liquid phase. 5. Heating curve f water is shwn belw: When ice is heated cntinuusly, the temperature increases with time. The heat evlved during this prcess is given as q (mass f the substance*heat capacity*change in temperature = m*c*t). The temperature f the system cntinues t increases until it reaches 0 C when ice starts melting t frm liquid water. During transfrmatin f ice t liquid water at 0 C, the temperature remains cnstant until all the ice is cnverted t liquid state. The heat evlved during the transitin at 0 C is called enthalpy f fusin H fusin (ttal heat evlved during the prcess is given as m*h fusin ). When all ice is cnverted t liquid water, again the temperature starts increasing till when liquid water starts cnverting t vapur at 100 C. Similar t heat f fusin, the temperature remains cnstant at 100 C until all the water in liquid state is cnverted t its vapur (steam). The heat evlved during this transitin is called enthalpy f vaprizatin, H fusin. When all water gets cnverted t steam, again the temperature increases.
5 6. Applicatin t heating curve: Chapter 10: Exercise 87 Hw much energy des it take t cnvert kh ice at 20 0C t steam at 250 0C? Specific heat capacities: ice, 2.03 J/g.0C; liquid, 4.2 J/g.0C, steam, 2.0 J/g.0C, Hvapur = 40.7 kj/ml, Hfus = 6.02 kj/ml. Slutin: q1 q2 q3 q4 q5 H2O (s) H2O (s) H2O (l) H2O (l) H2O (g) H2O (g) - 20 C 0 C 0 C 100 C 100 C 250 C 0.5 kg g ice = 500 g f ice = 500/18 mle f ice = mles f ice q1 = m *C* T = 500 g * 2.03 J/g. C * 20 C = 20 kj ( T = 0 C -20 C = 20 C)
6 q 2 = m *H fusin = ml * 6.02 kj/ml = 167 kj q 3 = m *C*T = 500 g * 4.2 J/g. C * 100 C = 210 kj (T = 100 C 0 C = 100 C) q 4 = m *H vapur = ml * 40.7 kj/ml = 1130 kj q 5 = m *C*T = 500 g * 2.0 J/g. C * 150 C = 150 kj (T = 250 C 100 C = 150 C) q = q 1 + q 2 + q 3 + q 4 + q 5 = 20 kj kj kj kj + 150kJ = 1677 kj
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