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- 1. Chapter 5 The Gas Laws pp
- 10. A Boyle’s Law Relationship
- 11. V P (at constant T)
- 12. V 1/P (at constant T) Slope = k
- 13. PV P (at constant T) CO 2 O 2 22.41 L atm Ne
- 17. Tr 51A Charles’ Law Relationship
- 18. V (L) T (ºC) He H 2 O CH 4 H 2 -273.15ºC
- 22. Figure 5.10 p. 195 Balloons Holding 1.0 L of Gas at 25º C and 1 atm . Each contains 0.041 mol of gas, or 2.5 x 10 22 molecules, even though their masses are different ( equal numbers, different masses ).
- 35. Figure 5.11 A Mole of Any Gas Occupies a Volume of Approximately 22.4 L at STP How many particles are in each box? 6.022 x 10 23 particles
- 52. Figure 5.12 p. 206 Partial Pressure of Each Gas in a Mixture
- 62. Figure 5.13 The Production of Oxygen by Thermal Decomposition of KCIO 3
- 73. Figure 5.18 Path of One Particle in a Gas. Any given particle will continuously change its course as a result of collisions with other particles, as well as with the walls of its container.
- 75. number of particles Molecular Velocity 273 K
- 76. number of particles Molecular Velocity 273 K 1273 K
- 77. number of particles Molecular Velocity 273 K 1273 K 2273 K
- 80. Figure 5.21 p. 219The Effusion of a Gas into an Evacuated Chamber Rate of effusion is inversely proportional to the square root of the mass of the gas molecules.
- 84. Figure 5.23 HCI( g ) and NH 3 ( g ) Meet in a Tube
- 92. Figure 5.24 p. 222 Plots of PV/nRT Versus P for Several Gases (200 K) Behavior is close to ideal ONLY at low pressures (less than 1 atm)
- 93. Figure 5.25 Plots of PV/nRT Versus P for Nitrogen Gas at Three Temperatures Deviations are smaller at higher temperatures.
Notas do Editor
- Z5e 188 Section 5.1 Pressure
- Z5e 189 Fig. 5.2
- Z5e 189 Fig. 5.3
- Z5e 189 Units of Pressure 96.5 kPa 724 torr 0.952 atm
- Z5e 190 Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro
- Hrw 314 ref. Fig 10-10
- Z5e 192 Fig. 5.6 Plot of PV vs P at P below 1 atm Ideal gas expected to have constant value of PV Boyle’s law is good approximation at relatively low pressures
- (20.5 L)(742 torr) = (9.80 atm x 760 torr/atm)(x); so x = 2.04 L (30.6)(740) = (x)(750); x = 30.2 torr = 402 kPa
- (20.5 L)(742 torr) = (9.80 atm x 760 torr/atm)(x); so x = 2.04 L (30.6)(740) = (x)(750); x = 30.2 torr = 402 kPa
- Z5e 193 Charles’ Law
- Hrw 316 and rf hrw 318 fig. 10-12.
- Z5e 194 Fig. 5.8 Plot V vs. T Gas samples contain different numbers of moles
- (247)/(22 + 273) = (x)(273 + 98); x = 311 ml
- (40.5)/(26.4 + 273) = (81.0)/x; x = 593 K = 320. Celsius
- Z5e 195 Avogadro’s Law
- Z5e 195 Fig. 5.10
- V is constant, so (3.8)/(295) = (x)/373; x = 4.8 atm (3.8)(175)/(273 + 22) = (743/760)(x)/(273 + 22); x = 680 ml
- V is constant, so (3.8)/(295) = (x)/373; x = 4.8 atm (3.8)(175)/(273 + 22) = (743/760)(x)/(273 + 22); x = 680 ml
- Z5e 196 Section 5.3 The Ideal Gas Law Also 62.396 l torr/k mol = 8.3145 J/k mol
- PV = nRT T = (1.85)(47.3)/(.0821 * 1.62) = 658 K n = PV/RT = 6.51 mol; n * M = 651 * 83.8 = 546 g Use combined law; x = 3.99 L (can convert torr to atm, but they cancel out so don’t need to for this problem).
- PV = nRT T = (1.85)(47.3)/(.0821 * 1.62) = 658 K n = PV/RT = 6.51 mol; n * M = 651 * 83.8 = 546 g Use combined law; x = 3.99 L (can convert torr to atm, but they cancel out so don’t need to for this problem).
- PV = nRT T = (1.85)(47.3)/(.0821 * 1.62) = 658 K n = PV/RT = 6.51 mol; n * M = 651 * 83.8 = 546 g Use combined law; x = 3.99 L (can convert torr to atm, but they cancel out so don’t need to for this problem).
- Z5e Section 5.4 Gas Stoichiometry
- Z5e 202
- Balance 1st! Stoich MW HgO 216.59; Hg 200.59; O2 32; ans. 0.212 L Use ideal gas law; ans. 0.537 L
- Balance 1st! Stoich MW HgO 216.59; Hg 200.59; O2 32; ans. 0.212 L Use ideal gas law; ans. 0.537 L
- MW NaHCO 3 84 Not at STP so use PV=nRT; ans. 21.5 g Use stoich to get moles HCl, then mole ratio --> moles CO 2 , then moles/L = concentration. Or, use Chas law to correct to 24.2 L, then stoich to get 1.08 mol HCl = 1.08 mole CO 2 ; so concentration is 0.415 M
- MW NaHCO 3 84 Not at STP so use PV=nRT; ans. 21.5 g Use stoich to get moles HCl, then mole ratio --> moles CO 2 , then moles/L = concentration. Or, use Chas law to correct to 24.2 L, then stoich to get 1.08 mol HCl = 1.08 mole CO 2 ; so concentration is 0.415 M
- T & P constant so cancel out. L proportional to moles, so 10 L watch Kg! Ans. Rounded to 16 500 (from calc ans of 16 470)
- T & P constant so cancel out. L proportional to moles, so 10 L watch Kg! Ans. Rounded to 16 500 (from calc ans of 16 470)
- LR = O 2 , ans. 313 g ans. 52 L correct to STP = 783 L; ans. 895 g.
- Z5e 205
- Hrw 345
- Hrw additional sample problem 11-5
- Hrw additional sample problem 11-5
- Hrw additional sample problem 11-6, modified so that density is used, where density is rounded down to 3.36 g/L from 3.365 g/L.
- D = MP/RT = 0.68 g/L n = 0.0083; M = 97; EF = 49 so molecular formula = C 2 H 2 Cl 2
- D = MP/RT = 0.68 g/L n = 0.0083; M = 97; EF = 49 so molecular formula = C 2 H 2 Cl 2
- D = MP/RT = 0.68 g/L n = 0.0083; M = 97; EF = 49 so molecular formula = C 2 H 2 Cl 2
- Z5e 205 Section 5.5 Dalton’s Law of Partial Pressures
- Z5e 206 Fig 5.12
- Z5e 205 Section 5.5 Dalton’s Law of Partial Pressures
- Z5e 207
- 592/752 = 0.787 0.787 x 5.25 = 4.13 atm Note: can use mole fraction regardless of units (don’t have to convert). Also, Air pressure is 752 (so this is the total pressure in the denominator)
- 592/752 = 0.787 0.787 x 5.25 = 4.13 atm Note: can use mole fraction regardless of units (don’t have to convert). Also, Air pressure is 752 (so this is the total pressure in the denominator)
- P1V1 = P2V2; total V = 9 L CH4 (4)(2.70) = (9)(x); x = 1.2 atm N2 ans. 0.763 O2 = 0.294 Sum = 2.127 = ans.
- Z5e 209
- Z5e 209 Fig. 5.13
- Make sure reaction is balanced MW of ammonium nitrate = 80. G Use spider to get moles of N20 = .0325, then use PV = nRT, but first have to: Convert 21 torr to 2.8 kPa 94.0 - 2.8 = 91.2 kPa N2O convert P to atm d/t R ( or use 62.4)!!! Convert C to K Ans. 6.6 L
- Make sure reaction is balanced MW of ammonium nitrate = 80. G Use spider to get moles of N20 = .0325, then use PV = nRT, but first have to: Convert 21 torr to 2.8 kPa 94.0 - 2.8 = 91.2 kPa N2O convert P to atm d/t R ( or use 62.4)!!! Convert C to K Ans. 6.6 L
- Z5e 211 Section 5.6 The Kinetic Molecular Theory of Gases
- Z5e 216, 217
- M rms = sq root 3RT/u Must express R using joules and M = Kg/mol (not g, d/t R) R = 8.3145 J/K*Mol 411 m/s 1928 m/s 324 m/s
- Z5e 218 Fig. 5.18: Any given particle will continuously change its course as a result of collisions with other particles, as well as with the walls of its container.
- Z5e Section 5.7 Effusion and Diffusion
- Z5e 219 Fig 5.21: Rate of effusion is inversely proportional to the square root of the mass of the gas molecules.
- Z5e 220
- Z5e 221 Diffusion
- a. ans. 40.96 b. rate NH3/rate HCl = 2.47/2.47 = 1/1 = sq rt (.0025*17/x*36.5) = .00177 mol HCl. Alternate method: (0.251/2.47 divided by (x)/2.47) = (sq rt 36.5)/(sq rt 17) = 0.00177 mol HCl. c. Rate N2is faster than rate x by 55/38 times = 1.45 times faster. Then, 1.45/1 = Sq Rt M x divided by sq rt N 2 and answer is 59 g/mol (sig figs)
- Z5e Section 5.8 Real Gases
- Z5e 224
- Z5e 222: Behavior is close to ideal ONLY at low pressures (less than 1 atm)
- Z5e 222
- a. P = nRT/V = 12.2 atm b. P = nRT/V - nb - a (n/v) 2 = 11.0 atm
- Z5e 225