Chapter18 - 18.1 Solve We can use the ideal-gas law in the...

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18.1. Solve: We can use the ideal-gas law in the form pV = Nk B T to determine the Loschmidt number ( N / V ): N V p k T = = × ( ) × ( ) ( ) = × B Pa J K 273 K m 1 013 10 1 38 10 2 69 10 5 23 25 3 . . .
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18.2. Solve: Nitrogen is a diatomic molecule, so r 1.0 × 10 10 m. We can use the ideal-gas law in the form pV = Nk B T and Equation 18.3 for the mean free path to obtain p : λ π π = ( ) = 1 4 2 4 2 2 2 N V r k T pr B = = × ( ) ( ) ( ) × ( ) p k T r B J k 293 K 1.0 m m 4 2 1 38 10 4 2 1 0 10 2 23 10 2 πλ π . . = 0.0228 Pa Assess: In Example 18.1 λ = 225 nm at STP for nitrogen. λ = 1.0 m must therefore require a very small pressure.
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18.3. Solve: (a) Air is primarily comprised of diatomic molecules, so r 1.0 × 10 10 m. Using the ideal-gas law in the form pV = Nk B T , we get N V p k T = = × × × × ( ) ( ) = × B 23 mm of Hg Pa 760 mm of Hg 1.38 10 J K 293 K m 1 0 10 1 013 10 3 30 10 10 5 12 3 . . . (b) The mean free path is λ π π = ( ) ( ) = × ( ) × ( ) = × 1 4 2 1 4 2 3 30 10 1 0 10 1 71 10 2 12 3 10 2 6 N V r . . . m m m Assess: The pressure p in the vacuum chamber is 1.33 × 10 8 Pa = 1.32 × 10 13 atm. A mean free path of 1.71 × 10 6 m is large but not unreasonable.
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18.4. Solve: (a) The mean free path of a molecule in a gas at temperature T 1 , volume V 1 , and pressure p 1 is λ 1 = 300 nm. We also know that λ π λ = ( ) 1 4 2 2 N V r V Although T 2 = 2 T 1 , constant volume ( V 2 = V 1 ) means that λ 2 = λ 1 = 300 nm. (b) For T 2 = 2 T 1 and p 2 = p 1 , the ideal gas equation gives p V Nk T p V Nk T p V Nk T 1 1 1 2 2 2 1 2 1 2 B B B = = ( ) = V V 2 1 2 Because λ V , λ 2 = 2 λ 1 = 2(300 nm) = 600 nm.
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18.5. Solve: Neon is a monatomic gas and has a radius r 5.0 × 10 11 m. Using the ideal-gas equation, N V p k T = = ( ) × ( ) × ( ) ( ) = × B 23 3 Pa 1.38 10 J / K 298 K m 150 1 013 10 3 695 10 5 27 . . Thus, the mean free path of a neon atom is λ π π = ( ) = × ( ) × ( ) = × 1 4 2 1 4 2 3 695 10 5 0 10 6 09 10 2 27 11 2 9 N V r . . . m m m 3 Since the atomic diameter of neon is 2 × (5.0 × 10 11 m) = 1.0 × 10 –10 m, λ = × × = 6 09 10 1 0 10 61 9 10 . . m m atomic diameters
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18.6. Solve: The number density of the Ping-Pong balls inside the box is N V = = 2000 1 0 . m 2000 m 3 3 With r = (3.0 cm)/2 = 1.5 cm, the mean free path of the balls is λ π = ( ) ( ) = = 1 4 2 2 N V r 0.125 m 12.5 cm
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18.7. Solve: (a) In tabular form we have Particle v x (m/s) v y (m/s) v x 2 (m/s) 2 v y 2 (m/s) 2 v 2 (m/s) 2 v (m/s) 1 20 30 400 900 1300 36.06 2 40 70 1600 4900 6500 80.62 3 80 10 6400 100 6500 80.62 4 60 20 3600 400 4000 63.25 5 0 50 0 2500 2500 50.00 6 40 20 1600 400 2000 44.72 Average 0 0 3800 59.20 The average velocity is r r r v i j avg = + 0 0 ˆ ˆ . (b) The average speed is v avg 59.2 m / s = . (c) The root-mean-square speed is v v rms avg 2 2 3800 m s 61.6 m / s = ( ) = = 2 / .
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18.8. Solve: (a) The average speed is v n n n avg m / s 220 m / s 11 20.0 m / s = = = = = 1 11 15 25 (b) The root-mean-square speed is v v rms avg = ( ) 2 = = = 1 11 2 15 25 1 2 n n n m / s = = 4510 11 1 2 20.2 m / s
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18.9. Solve: (a) The atomic mass number of argon is 40. This means the mass of an argon atom is m = × = × 40 10 6 64 10 27 26 u = 40(1.661 kg) kg .
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