18 1 Solve We can use the ideal gas law in the form pV NkBT to determine the Loschmidt number N V N V p k T B 5 1 013 10 Pa 1 38 10 J K 273 K 23 2 69 10 m 25 3 18 2 Solve The volume of the nitrogen gas is 1 0 m3 and its temperature is 20 C or 293 K The number of gas molecules can be found as N nN A N A pV RT 5 1 013 10 Pa 1 0 m 8 31 J mol K 293 K 3 6 02 10 mol 23 1 2 5 10 25 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 1025 5 5 1024 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 NkBT and Equation 18 3 for the mean free path to obtain p 1 38 10 4 2 J k 293 K 1 0 m 1 0 10 k T B 4 2 r 1 N V r k T B 4 2 0 023 Pa p 4 2 m pr 10 23 2 2 2 2 In Example 18 1 225 nm at STP for nitrogen 1 0 m must therefore require a very small Assess pressure 18 4 Solve law in the form pV NkBT we get a Air is primarily comprised of diatomic molecules so r 1 0 10 10 m Using the ideal gas N V p k T B 1 0 10 10 mm of Hg 5 1 013 10 Pa 760 mm of Hg J K 293 K 1 38 10 23 3 30 10 m 12 3 b The mean free path is 4 2 1 N V r 2 1 1 71 10 m 6 4 2 3 30 10 m 1 0 10 12 3 10 m 2 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 71 106 m is large but not unreasonable 18 5 Solve 300 nm We also know that a The mean free path of a molecule in a gas at temperature T1 volume V1 and pressure p1 is 1 4 2 1 N V r 2 V Although T2 2T1 constant volume V2 V1 means that 2 1 300 nm b For T2 2 T1 and p2 p1 the ideal gas equation gives p V 1 1 Nk T B 1 p V 2 2 Nk T B 2 p V 1 2 B T 12 Nk V 2 V 12 V 2 2 1 2 300 nm 600 nm Because 18 6 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 150 1 013 10 Pa 1 38 10 J K 298 K 23 5 3 695 10 m 27 3 Thus the mean free path of a neon atom is 1 N V r 2 4 2 1 6 09 10 m 9 4 2 3 695 10 m 5 0 10 27 3 11 m 2 Since the atomic diameter of neon is 2 5 0 10 11 m 1 0 10 10 m 9 6 09 10 m 1 0 10 m 10 61 atomic diameters 18 7 Solve The number density of the Ping Pong balls inside the box is With r 3 0 cm 2 1 5 cm the mean free path of the balls is N V 2000 3 1 0 m 2000 m 3 1 4 2 N V r 2 0 125 m 12 5 cm 18 8 Solve a The average speed is b The root mean square speed is v avg 1 11 n 25 n 15 n m s 220 m s 11 20 0 m s v rms v 2 avg 1 11 n 25 n 15 2 n 1 2 1 2 4510 11 m s 20 2 m s 18 9 Solve Particle a In tabular form we have vx m s vy m s 1 2 3 4 5 6 Average The average velocity is 20 40 80 60 0 40 0 30 70 10 20 50 20 0 cid 71 i 0 cid 71 v avg cid 71 0 j b The average speed is v avg 59 m s v2 m s 2 v m s 2 xv m s 2 400 1600 6400 3600 0 1600 2 yv m s 2 900 4900 100 400 2500 400 1300 6500 6500 4000 2500 2000 3800 36 06 80 62 80 62 63 25 50 00 44 72 59 20 c The root mean square speed is v rms v 2 avg 3800 m s 2 2 62 m s 18 10 Solve b The average speed is a The most probable speed is 4 0 m s v avg 2 2 m s 4 4 m s 3 6 m s 1 8 m s 2 4 3 1 4 6 m s c The root mean square speed is v rms 2 2 m s 2 4 4 m s 6 m s 2 1 8 m s 2 2 3 2 4 3 1 4 9 m s 18 11 Solve a The atomic mass number of argon is 40 This means the mass of an argon atom is m 40 u 40 1 661 10 27 kg 6 64 10 26 kg The pressure of the gas is p 1 3 N V mv 2 rms 1 3 2 00 10 m 25 3 6 64 10 26 kg 455 m s 2 9 16 104 Pa b The temperature of the gas in the container can be obtained from the ideal gas equation in the form pV NkBT T pV Nk B 9 16 10 Pa 3 25 2 00 10 m 1 38 10 4 23 J K 332 K 18 12 Model Pressure is due to random collisions of gas molecules with the walls Solve According to Equation 18 8 the collision rate with one wall is rate of collisions N coll t coll F net 2 mv x pA mv x 2 where Fnet pA is the force exerted on area A by the gas pressure However this equation assumed that all molecules are moving in the x direction with constant speed The rms speed vrms is for motion in three dimensions at varying speeds Consequently we need to replace vx not with vx avg which is zero but with v x v 2 x avg 2 v rms 3 v rms 3 With this change rate of collisions pA 3 mv 2 rms 3 2 101 300 …
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