617_Physics ProblemsTechnical Physics

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

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Chapter 21 619 P21.49 Using PnkT VB = , Equation 21.30 becomes A = kT Pd B 2 2 π (1) (a) A = × ×× 138 10 293 2 1 013 10 3 10 10 936 10 23 51 0 2 8 . .. . JK K Pa m m ej af (b) Equation (1) shows that PP 11 22 AA = . Taking P A from (a) and with A 2 100 = . m, we find P 2 8 8 1 00 9 36 10 = × . . atm m m atm . (c) For A 3 10 310 10 . m, we have P 3 8 1 00 9 36 10 302 = × × = atm m 3.10 10 m atm 10 . Additional Problems P21.50 (a) n PV RT == × (. ) . . ) ) ( ) 1 013 10 4 20 3 00 2 50 8314 131 10 5 3 Pa m m m Jmo lK K . m o l Nn N N A ==× × 1 31 10 6 02 10 789 10 32 3 26 . mol . molecules mol molecules e j . (b) mn M × = 1 31 10 0 028 9 37 9 3 . mol . kg mol kg bg . (c) 1 2 3 2 3 2 1 3 81 0 2 9 3 6 0 71 0 0 3 2 1 mv B × = × −− J k K J molecule (d) For one molecule,
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