ch9 - 9 9.1 9.2 9.3 Gases Their Properties and Behavior...

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9 Gases: Their Properties and Behavior 9.1 1.00 atm = 14.7 psi 1.00 mm Hg x atm 1 psi 14.7 x Hg mm 760 atm 1 = 1.93 x 10 2 psi 9.2 1.00 atmosphere pressure can support a column of Hg 0.760 m high. Because the density of H 2 O is 1.00 g/mL and that of Hg is 13.6 g/mL, 1.00 atmosphere pressure can support a column of H 2 O 13.6 times higher than that of Hg. The column of H 2 O supported by 1.00 atmosphere will be (0.760 m)(13.6) = 10.3 m. 9.3 The pressure in the flask is less than 0.975 atm because the liquid level is higher on the side connected to the flask. The 24.7 cm of Hg is the difference between the two pressures. Pressure difference = 24.7 cm Hg x 1 76 0 atm . cm Hg = 0.325 atm Pressure in flask = 0.975 atm 0.325 atm = 0.650 atm 9.4 The pressure in the flask is greater than 750 mm Hg because the liquid level is lower on the side connected to the flask. Pressure difference = 25 cm Hg x Hg cm 1 Hg mm 10 = 250 mm Hg Pressure in flask = 750 mm Hg + 250 mm Hg = 1000 mm Hg 9.5 (a) Assume an initial volume of 1.00 L. First consider the volume change resulting from a change in the number of moles with the pressure and temperature constant. n V = n V f f i i ; V f = mol 0.3 mol) L)(0.225 (1.00 = n n V i f i = 0.75 L Now consider the volume change from 0.75 L as a result of a change in temperature with the number of moles and the pressure constant. T V = T V f f i i ; V f = K 300 K) L)(400 (0.75 = T T V i f i = 1.0 L There is no net change in the volume as a result of the decrease in the number of moles of gas and a temperature increase. (b) Assume an initial volume of 1.00 L. First consider the volume change resulting from a change in the number of moles with 195
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the pressure and temperature constant. n V = n V f f i i ; V f = mol 0.3 mol) L)(0.225 (1.00 = n n V i f i = 0.75 L Now consider the volume change from 0.75 L as a result of a change in temperature with the number of moles and the pressure constant. T V = T V f f i i ; V f = K 300 K) L)(200 (0.75 = T T V i f i = 0.5 L The volume would be cut in half as a result of the decrease in the number of moles of gas and a temperature decrease. 9.6 n = K) (273.15 mol K atm L 06 0.082 L) 10 x atm)(1.000 (1.000 = RT PV 5 = 4.461 x 10 3 mol CH 4 CH 4 , 16.04 amu; mass CH 4 = (4.461 x 10 3 mol) mol 1 g 16.04 = 7.155 x 10 4 g CH 4 9.7 C 3 H 8 , 44.10 amu; V = 350 mL = 0.350 L; T = 20 o C = 293 K n = 3.2 g x H C g 44.10 H C mol 1 8 3 8 3 = 0.073 mol C 3 H 8 P = V nRT = L 0.350 K) (293 mol K atm L 06 0.082 mol) (0.073 = 5.0 atm 9.8 P = 1.51 x 10 4 kPa x kPa 101.325 atm 1 = 149 atm; T = 25.0 o C = 298 K n = K) (298 mol K atm L 06 0.082 L) atm)(43.8 (149 = RT PV = 267 mol He 9.9 The volume and number of moles of gas remain constant. T P = T P = V nR f f i i ; T f = atm 2.15 K) atm)(273 (2.37 = P T P i i f = 301 K = 28 o C 9.10 (a) The temperature has increased by about 10% (from 300 K to 325 K) while the amount and the pressure are unchanged. Thus, the volume should increase by about 10%. 196
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(b) The temperature has increased by a factor of 1.5 (from 300 K to 450 K) and the pressure has increased by a factor of 3 (from 0.9 atm to 2.7 atm) while the amount is unchanged. Thus, the volume should decrease by half (1.5/3 = 0.5).
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