{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW13solutions

# HW13solutions - Physics 213 Homework 13 Solutions Spring...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

HW13solutions - Physics 213 Homework 13 Solutions Spring...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online