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617_Physics ProblemsTechnical Physics

617_Physics ProblemsTechnical Physics - Chapter 21 P21.49 k...

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Chapter 21 619 P21.49 Using P n k T V B = , Equation 21.30 becomes A = k T Pd B 2 2 π (1) (a) A = × × × = × 1 38 10 293 2 1 013 10 3 10 10 9 36 10 23 5 10 2 8 . . . . J K K Pa m m e j a f e je j π (b) Equation (1) shows that P P 1 1 2 2 A A = . Taking P 1 1 A from (a) and with A 2 1 00 = . m, we find P 2 8 8 1 00 9 36 10 1 00 9 36 10 = × = × . . . . atm m m atm a f e j . (c) For A 3 10 3 10 10 = × . m, we have P 3 8 1 00 9 36 10 302 = × × = . . atm m 3.10 10 m atm 10 a f e j . Additional Problems P21.50 (a) n PV RT = = × × × = × ( . )( . . . ) ( . )( ) 1 013 10 4 20 3 00 2 50 8 314 293 1 31 10 5 3 Pa m m m J mol K K . mol N nN N A = = × × = × 1 31 10 6 02 10 7 89 10 3 23 26 . mol . molecules mol molecules e je j . (b) m nM = = × = 1 31 10 0 028 9 37 9 3 . mol . kg mol kg e j b g . (c) 1 2 3 2 3 2 1 38 10 293 6 07 10 0 2 23 21 m v
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