601_Physics ProblemsTechnical Physics

601_Physics ProblemsTechnical Physics - Chapter 21 e5.00 10...

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Chapter 21 603 P21.2 F = ×× = 500 10 2 468 10 300 100 14 0 23 26 .. . . ej e j bg kg m s s N and P F A == × = 14 0 10 17 6 4 . . N 8.00 m kPa 2 . P21.3 We first find the pressure exerted by the gas on the wall of the container. P NkT V NkT V RT V ⋅⋅ × 3 3 38314 293 800 10 913 10 3 5 AB 3 Nmmo lK K m Pa . . . a f Thus, the force on one of the walls of the cubical container is FP A × × = × 9 13 10 4 00 10 3 65 10 52 4 . Pa m N 2 e j . P21.4 Use Equation 21.2, P N V mv = F H G I K J 2 32 2 , so that K mv PV N Nn N N K PV N K av A A av A 3 av where atm Pa atm m mol molecules mol Jmo lecu le × 2 53 23 21 2 3 2 2 3 22 3 8 00 1 013 10 5 00 10 602 10 505 10 a f e j af . . . P21.5 P N V KE = 2 3 di Equation 21.2 N PV KE n N N × × × = 3 2 3 2 1 20 10 4 00 10 360 10 200 10 332 22 24 24 23 e j . . . . . molecules molecules molecules mol mol A P21.6 One mole of helium contains Avogadro’s number of molecules and has a mass of 4.00 g. Let us call m the mass of one atom, and we have Nm A gmo l = 400 . or m = × 664 10
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