HW13solutions

HW13solutions - Physics 213 Homework 13 Solutions Spring...

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Unformatted text preview: Physics 213 Homework 13 Solutions Spring 2008 18.30. IDENTIFY: K av = 3 2 kT (for one particle, or N=1) v rms = 3 RT M = 3 kT m . SET UP: M Ne = 20.180 g/mol , M Kr = 83.80 g/mol and M Rn = 222 g/mol . EXECUTE:(a) K av = 3 2 kT depends only on the temperature so it is the same for each species of atom in the mixture. (b) v rms,Ne v rms,Kr = M Kr M Ne = 83.80 g/mol 20.18 g/mol = 2.04 . v rms,Ne v rms,Rn = M Rn M Ne = 222 g/mol 20.18 g/mol = 3.32 . v rms,Kr v rms,Rn = M Rn M Kr = 222 g/mol 83.80 g/mol = 1.63. EVALUATE: The average kinetic energies are the same. The gas atoms with smaller mass have larger v rms . 18.37. IDENTIFY and SET UP: Apply the analysis of Section 18.3. EXECUTE: (a) K . E . = 1 2 m ( v 2 ) av = 3 2 kT = 3 2 (1.38 ¡ 10 ¢ 23 J/molecule £ K)(300 K) = 6.21 ¡ 10 ¢ 21 J (b) We need the mass m of one molecule: m = M N A = 32.0 ¡ 10 ¢ 3 kg/mol 6.022 ¡ 10 23 molecules/mol = 5.314 ¡ 10 ¢ 26 kg/molecule Then 1 2 m ( v 2 ) av = 6.21 ¡ 10 ¢ 21 J = K.E. (from part (a)) gives ( v 2 ) av = 2(6.21 ¡ 10 ¢ 21 J) m = 2(6.21 ¡ 10 ¢ 21 J) 5.314 ¡ 10 ¢ 26 kg = 2.34 ¡ 10 5 m 2 / s 2 (c) v rms = ( v 2 ) rms = 2.34 ¡ 10 5 m 2 / s 2 = 484 m/s . (Alternatively, you could use v rms =(3kT/m) 1/2 instead.) (d) p = mv rms = (5.314 ¡ 10 ¢ 26 kg)(484 m/s) = 2.57 ¡ 10 ¢ 23 kg £ m/s (e) Time between collisions with one wall is t = 0.20 m v rms = 0.20 m 484 m/s = 4.13 ¡ 10 ¢ 4 s In a collision ¡ v changes direction, so ¡ p = 2 mv rms = 2(2.57 ¢ 10 £ 23 kg ¤ m/s) = 5.14 ¢ 10 £ 23 kg ¤ m/s F = dp dt so F av = ¡ p ¡ t = 5.14 ¢ 10 £ 23 kg ¤ m/s 4.13 ¢ 10 £ 4 s = 1.24 ¢ 10 £ 19 N (f) pressure = F / A = 1.24 ¡ 10 ¢ 19 N/(0.10 m) 2 = 1.24 ¡ 10 ¢ 17 Pa (due to one atom) (g) pressure = 1 atm = 1.013 ¡ 10 5 Pa Number of atoms needed is...
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This note was uploaded on 08/27/2008 for the course PHYS 2213 taught by Professor Perelstein,m during the Spring '07 term at Cornell.

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HW13solutions - Physics 213 Homework 13 Solutions Spring...

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