Ch18Word - Chapter 18 CHAPTER 18 Kinetic Theory of Gases...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 18 Page 1 CHAPTER 18 – Kinetic Theory of Gases 1. ( a ) The average kinetic energy depends on the temperature: ! mv rms 2 = * kT = * (1.38 × 10 –23 J/K)(273 K) = 5.65 × 10 –21 J . ( b ) For the total translational kinetic energy we have K = N ( ! mv rms 2 ) = * nN A kT = * (2.0 mol)(6.02 × 10 23 molecules/mol)(1.38 × 10 –23 J/K)(293 K) = 7.3 × 10 3 J . 2. The average kinetic energy depends on the temperature: ! mv rms 2 = * kT , which gives v rms = (3 kT / m ) 1/2 = [3(1.38 × 10 –23 J/K)(6000 K)/(4 u)(1.66 × 10 –27 kg/u)] 1/2 = 6.1 × 10 3 m/s . 3. The average kinetic energy depends on the temperature: ! mv rms 2 = * kT . If we form the ratio for the two temperatures, we have ( v rms2 / v rms1 ) 2 = T 2 / T 1 = 373 K/273 K, which gives v rms2 / v rms1 = 1.17 . 4. The average kinetic energy depends on the temperature: ! mv rms 2 = * kT . We form the ratio at the two temperatures: ( v rms2 / v rms1 ) 2 = T 2 / T 1 ; (2) 2 = T 2 /293 K, which gives T 2 = 1172 K = 899°C . 5. ( a ) We find the mean speed from v rms = (? v )/ N = (6 + 2 + 4 + 6 + 0 + 4 + 1 + 8 + 5 + 3 + 7 + 8)/12 = 4.5 . ( b ) We find the rms speed from v rms = [(? v 2 )/ N ] 1/2 = [(6 2 + 2 2 + 4 2 + 6 2 + 0 2 + 4 2 + 1 2 + 8 2 + 5 2 + 3 2 + 7 2 + 8 2 )/12] 1/2 = 5.2 . Note that this is greater than the mean speed of 4.5. 6. The average kinetic energy depends on the temperature: ! mv rms 2 = * kT . We treat the small changes as differentials. We find the relationship between the changes by differentiating: mv rms d v rms = * k d T , or, after dividing by ! mv rms 2 = * kT , 2 d v rms / v rms = d T / T ; 2(0.010 v rms )/ v rms = d T /293.2 K, which gives d T = 5.9 K. Thus the new temperature is T + d T = 293.2 K + 5.9 K = 299.1 K = 25.9°C . 7. The average kinetic energy depends on the temperature: ! mv rms 2 = * kT . We form the ratio at the two temperatures, and use the ideal gas law: ( v rms2 / v rms1 ) 2 = T 2 / T 1 = P 2 V 2 / P 1 V 1 = P 2 / P 1 ; ( v rms2 / v rms1 ) 2 = 2, which gives v rms2 / v rms1 = v2 . 8. The average kinetic energy depends on the temperature: ! mv rms 2 = * kT , or kT / m = @ v rms 2 . With M the mass of the gas and m the mass of a molecule, we write the ideal gas law as PV = NkT = ( M / m ) kT , or P = ( M / V )( kT / m ) = ρ ( @ v rms 2 ) = @ v rms 2 , or v rms = (3 P / ) 1/2 .
Background image of page 1

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

View Full DocumentRight Arrow Icon
Chapter 18 Page 2 9. The average kinetic energy depends on the temperature: ! mv rms 2 = * kT . We form the ratio at the two temperatures, and use the ideal gas law: ( m 2 / m 1 )( v rms2 / v rms1 ) 2 = T 2 / T 1 = 1, so we have ( v rms2 / v rms1 ) 2 = ( m 1 / m 2 ), or v rms2 / v rms1 = ( m 1 / m 2 ) 1/2 . 10. We use the ideal gas law to find the temperature: PV = nRT ; (2.1 atm)(1.013 × 10 5 Pa/atm)(8.5 m 3 ) = (1300 mol)(8.315 J/mol · K) T , which gives T = 167 K. The average kinetic energy depends on the temperature: ! mv rms 2 = * kT , which gives v rms = (3 kT / m ) 1/2 = [3(1.38 × 10 –23 J/K)(167 K)/(28 u)(1.66 × 10 –27 kg/u)] 1/2 = 3.9 × 10 2 m/s . 11. ( a ) The average kinetic energy depends on the temperature: ! mv rms 2 = * kT , which gives v rms = (3 kT / m ) 1/2 = [3(1.38 × 10 –23 J/K)(273 K)/(32 u)(1.66 × 10 –27 kg/u)] 1/2 = 461 m/s . ( b ) The molecule, on the average, will have a component in one direction less than the average speed. If we take the rms speed as the average speed, from the analysis of the molecular motion, we know that ( v x 2 ) av = @ v rms 2 . Thus the time to go back and forth is t = 2 ¬ /( v x ) av ˜ 2 ¬ v3/ v rms .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

Page1 / 11

Ch18Word - Chapter 18 CHAPTER 18 Kinetic Theory of Gases...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online