problem18_87

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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18.87: a) From Eq. (18.21), 1 3 6 2 11 1 2 )) m 10 50 ( m) 10 0 . 5 ( 2 4 ( )) ( 2 4 ( - - - - × × = = λ π V N r π m. 10 5 . 4 11 × = b) s, m 703 mol) kg 10 008 . 1 ( K) K)(20 mol J 3(8.3145 3 = × - and the time between collisions is then s, 10 6.4 s) m (703 m) 10 5 . 4 ( 8 11 × = × about 20 yr. Collisions are not very important. c) Pa. 10 1.4 K) K)(20 J 10 381 . 1 )( m 10 50 ( ) ( 14 23 3 6 - - - × = × × = = kT V N p d) mR V N G π R πR V Nm G R GM v e ) ( ) 3 8 ( ) 3 4 )( ( 2 2 3 = = = kg) 10 67 . 1 )( m 10 50 )( kg m N 10 673 . 6 )( 3 8 ( 27 3 6 2 2 11 - - - × × × = π m) 10 46 . 9 10 ( 15 × × × s. m 650 = This is lower than s, v and the cloud would tend to evaporate. e) In equilibrium (clearly not thermal equilibrium), the pressures will be the same; from , NkT pV = nebula nebula ISM ISM
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