06_Petrucci10e_SSM - CHAPTER 6 GASES PRACTICE EXAMPLES 1A...

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122 CHAPTER 6 GASES PRACTICE EXAMPLES 1A The pressure measured by each liquid must be the same. They are related through P = g h d Thus, we have the following g h DEG d DEG = g h Hg d Hg . The g ’s cancel; we substitute known values: 9.25 m DEG ×1.118 g/cm 3 DEG = h Hg × 13.6 g/cm 3 Hg 3 Hg 3 1.118g/cm 9.25m 0.760mHg, = 0.760m Hg =760.mmHg 13.6g/cm hP  1B The solution is found through the expression relating density and height: TEG TEG Hg Hg hd We substitute known values and solve for triethylene glycol’s density: 9.14 m TEG × d TEG = 757 mmHg ×13.6 g/cm 3 Hg . Using unit conversions, we get d TEG = 33 0.757 m 13.6 g/cm 1.13 g/cm 9.14 m  2A We know that P gas = P bar + Δ P with P bar = 748.2 mmHg. We are told that Δ P = 7.8 mmHg. Thus, P gas = 748.2 mmHg + 7.8 mmHg = 756.0 mmHg. 2B The difference in pressure between the two levels must be the same, just expressed in different units. Hence, this problem is almost a repetition of Practice Example 6-1. h Hg =748.2 mmHg – 739.6 mmHg=8.6 mmHg. Again we have g h g d g = g h Hg d Hg . This becomes h g × 1.26 g/cm 3 glycerol = 8.6 mmHg ×13.6 g/cm 3 Hg h g 3 mmHg gcmH g gcmg l y c e r o l mm glycerol 86 136 126 93 3 . ./ 3A A = r 2 (here r = ½(2.60 cm × 1 m 100 cm ) = 0.0130 m)    0.0130 m) 2 = 5.31 × 10 -4 m 2 F = m × g = (1.000 kg)(9.81 m s -2 ) = 9.81 kg m s -2 = 9.81 N P = A F = -4 2 9.81 N 5.31 10 m = 18475 N m -2 or 1.85 × 10 4 Pa P (torr) = 1.85 × 10 4 Pa × = 139 torr 3B Final pressure = 100 mb. 101, 325 Pa 100 mb 1013.25 mb = 1.000 × 10 4 Pa The area of the cylinder is unchanged from that in Example 6-3, (1.32 × 10 -3 m 2 ). P = F A = 1.000 × 10 4 Pa = -3 2 F 1.32×10 m
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Chapter 6: Gases 123 Solving for F, we find F = 13.2 (Pa)m 2 = 13.2 (N m -2 )m 2 = 13.2 N F = m × g = 13.2 kg m s -2 = -2 m 9.81 m s total mass = mass of cylinder + mass added weight = m = g F = -2 -2 13.2 kg m s 9.81 m s = 1.35 kg An additional 350 grams must be added to the top of the 1.000 kg (1000 g) red cylinder to increase the pressure to 100 mb. It is not necessary to add a mass with the same cross sectional area. The pressure will only be exerted over the area that is the base of the cylinder on the surface beneath it. 4A The ideal gas equation is solved for volume. Conversions are made within the equation.  3 3 3 3 1 mol NH 0.08206L atm 20.2g NH 25 273 K 17.03g NH mol K 24.4 L NH 1atm 752 mmHg 760 mmHg nRT V P      4B The amount of Cl g 2 b g is 0.193 mol Cl 2 and the pressure is 0.980 atm (0.993 barr × (1 atm/1.01325 barr) = 0.980 atm). This information is substituted into the ideal gas equation after it has been solved for temperature. 11 0.980atm 7.50 L 464K 0.193mol 0.08206Latmmol K PV T nR  5A The ideal gas equation is solved for amount and the quantities are substituted. 10.5atm 5.00L 2.11molHe 0.08206L atm 30.0 273.15 K mol K PV n RT  5B n = PV RT = -7 3 3 -1 -1 1 atm 1000L 6.67 10 Pa 3.45 m 101325 Pa 1m (0.08206 L atm K mol ) (25 273.15)K = 9.28 × 10 -10 moles of N 2 molecules of N 2 = 9.28 × 10 -10 mol N 2 × 23 2 2 6.022 10 molecules of N 1 mole N molecules of N 2 = 5.59 × 10 14 molecules N 2 6A The general gas equation is solved for volume, after the constant amount in moles is cancelled.
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06_Petrucci10e_SSM - CHAPTER 6 GASES PRACTICE EXAMPLES 1A...

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