p42_019 - . 43 10 28 m 3 ) / (2 . 7 10 25 m 3 ) = 3 . 1 10...

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19. (a) The ideal gas law in the form of Eq. 20-9 leads to p = NkT/V = nkT . Thus, we solve for the molecules per cubic meter: n = p kT = (1 . 0 atm)(1 . 0 × 10 5 Pa / atm) (1 . 38 × 10 23 J / K)(273 K) =2 . 7 × 10 25 m 3 . (b) Combining Eqs. 42-2, 42-3 and 42-4 leads to the conduction electrons per cubic meter in copper: n = 8 . 96 × 10 3 kg / m 3 (63 . 54)(1 . 67 × 10 27 kg) =8 . 43 × 10 28 m 3 . (c) The ratio is (8
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Unformatted text preview: . 43 10 28 m 3 ) / (2 . 7 10 25 m 3 ) = 3 . 1 10 3 . (d) We use d avg = n 1 / 3 . For case (a), d avg = (2 . 7 10 25 m 3 ) 1 / 3 which equals 3 . 3 nm. For case (b), d avg = (8 . 43 10 28 m 3 ) 1 / 3 = 0 . 23 nm ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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