Chapter 18

Chapter 18 - 18.1. Solve: We can use the ideal-gas law in...

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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 ): ( ) () 5 25 3 23 B 1.013 10 Pa 2.69 10 m 1.38 10 J K 273 K Np Vk T × == ×
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18.2. Solve: The volume of the nitrogen gas is 1.0 m 3 and its temperature is 20 ° C or 293 K. The number of gas molecules can be found as ( )( ) () ( ) 53 23 1 25 AA 1.013 10 Pa 1.0 m 6.02 10 mol 2.5 10 8.31 J mol K 293 K pV Nn N N RT × == = × = × According to Figure 18.2, 12% of the molecules have a speed between 700 and 800 m/s, 7% between 800 and 900 m/s, and 3% between 900 and 1000 m/s. Thus, the number of molecules in the cube with a speed between 700 m/s and 1000 m/s is (0.22)(2.51 × 10 25 ) = 5.5 × 10 24 .
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18.3. 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 : () B 22 1 42 kT NVr p r λ ππ == ( ) 23 B 2 2 10 1.38 10 J k 293 K 4 2 1.0 m 1.0 10 m p r πλ π × ⇒= = × = 0.023 Pa Assess: In Example 18.1 = 225 nm at STP for nitrogen. = 1.0 m must therefore require a very small pressure.
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18.4. 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 () 5 10 12 3 23 B 1.013 10 Pa 1.0 10 mm of Hg 760 mm of Hg 3.30 10 m 1.38 10 J K 293 K Np Vk T × ×× == × (b) The mean free path is ( ) 6 2 2 12 3 10 11 1.71 10 m 42 4 2 1.0 10 m NV r λ π −− = × 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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18.5. 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 2 1 42( /) V NVr λλ π = ⇒∝ Although T 2 = 2 T 1 , constant volume ( V 2 = V 1 ) means that 2 = 1 = 300 nm. (b) For T 2 = 2 T 1 and p 2 = p 1 , the ideal gas equation gives () 11 2 2 12 B1 B2 B 1 2 pV NkT Nk T == 21 2 VV ⇒= Because V , 2 = 2 1 = 2(300 nm) = 600 nm.
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18.6. Solve: Neon is a monatomic gas and has a radius r 5.0 × 10 11 m. Using the ideal-gas equation, () ( ) 5 27 3 23 B 150 1.013 10 Pa 3.695 10 m 1.38 10 J/K 298 K Np Vk T × == = × × Thus, the mean free path of a neon atom is ( ) 9 2 2 27 3 11 11 6.09 10 m 42 4 2 5.0 10 m NV r λ π −− = × ×× Since the atomic diameter of neon is 2 × (5.0 × 10 11 m) = 1.0 × 10 –10 m, 9 10 61 atomic diameters 1.0 10 m × ×
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18.7. Solve: The number density of the Ping-Pong balls inside the box is 3 3 2000 2000 m 1.0 m N V == With r = (3.0 cm)/2 = 1.5 cm, the mean free path of the balls is () 2 1 0.125 m 12.5 cm 42 NV r λ π =
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18.8. Solve: (a) The average speed is 25 avg 15 1 220 m/s m/s 20.0 m/s 11 11 n n vn = = ⎛⎞ == = ⎜⎟ ⎝⎠ (b) The root-mean-square speed is () 2 rms avg vv = 1 2 25 2 15 1 m/s 11 n n n = = = 1 2 4510 20.2 m/s 11
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18.9. Solve: (a) In tabular form we have Particle v x (m/s) v y (m/s) 2 x v (m/s) 2 2 y v (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 avg ˆˆ 00 vi j =+ GG G .
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Chapter 18 - 18.1. Solve: We can use the ideal-gas law in...

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