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Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 13 Temperature & Ideal Gases. PV = NkT N = PV/(kT) N = (13.5 * 1.013 x 10 5 N/m 2 * .00195 m 3 ) ( 1.38 x 10 -23 J/K * 293 K) N = 6.60 x 10 23. CAPA 7 & 8.

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Physics 2053C – Fall 2001

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  1. Physics 2053C – Fall 2001 Chapter 13 Temperature & Ideal Gases Dr. Larry Dennis, FSU Department of Physics

  2. PV = NkT N = PV/(kT) N = (13.5 * 1.013 x 105 N/m2 * .00195 m3 ) ( 1.38 x 10-23 J/K * 293 K) N = 6.60 x 1023 CAPA 7 & 8 A scuba tank has a volume of 3900 cm3. For very deep dives, the tank is filled with 50% (by volume) pure oxygen and 50% pure helium. 7. How many oxygen molecules are there in the tank if it is filled at 20°C to a gauge pressure of12.5 atm?

  3. CAPA 7 & 8 A scuba tank has a volume of 3900 cm3. For very deep dives, the tank is filled with 50% (by volume) pure oxygen and 50% pure helium. 8. How many helium molecules are there in the tank if it is filled at 20°C to a gauge pressure of12.5 atm? PV = NkT The same number as there are oxygen molecules. N = 6.60 x 1023

  4. Kinetic Theory of Gasses • Gases contain a large number of molecules moving in random directions with a variety of speeds. • Molecules are very far apart and don’t exert forces on one another except when they collide. • Molecules obey Newton’s Laws. • Collisions are perfectly elastic.

  5. Kinetic Theory of Gasses • The kinetic energy of the gas is directly related to it’s temperature. • KE = ½ m(v2)ave = 3/2 kT • Only depends on temperature. • Vrms = (V2)ave ( root mean square velocity ) • Vrms =  (3kT)/m

  6. Vrms =  (3kT)/m T = 100 K f KEtotal = 3/2 NkT KEone = 3/2 kT T = 200 K T = 300 K v RMS Velocity Gas molecules have a distribution of speeds. The range of speeds depends on the temperature.

  7. CAPA 9 & 10 A scuba tank has a volume of 3900 cm3. For very deep dives, the tank is filled with 50% (by volume) pure oxygen and 50% pure helium. 9. What is the ratio of the average kinetic energies of the two types of molecules? KE = 3/2 kT (for each molecule) Since the gases are at the same temperatures they have the same kinetic energies. Ratio = 1.0

  8. Vrms = (3KT/m) Vrms(He)/Vrms(O2) = ( m(O2)/m(He) ) Vrms(He)/Vrms(O2) = ( 2*16/(4) ) Vrms(He)/Vrms(O2) = 8 = 2.83 CAPA 9 & 10 A scuba tank has a volume of 3900 cm3. For very deep dives, the tank is filled with 50% (by volume) pure oxygen and 50% pure helium. 10. What is the ratio of the rms speeds of the two types of molecules?

  9. Quiz #8 • Temperature Scales • Fahrenheit  Centigrade  Kelvin • Ideal Gas Law • PV = nRT or PV = NkT • Gauge Pressure • Pay careful attention to units.

  10. Quiz #8 • Kinetic Theory of Gasses • Motion of molecules, Temperature • KE = 3/2 kT (average kinetic energy of a molecule). • KE = 3/2 nRT (kinetic energy of a mol of molecules. • Avogadro’s Number • (6.02 x 1023 molecules/mol) • Vrms = (3kT/m)½ average speed of a molecule.

  11. Quiz #8 • Questions, Chap. 13: 21 • Problems, Chap. 13: 5, 28, 35, 47, 88

  12. Problem 28 • What are the following temperatures on the Kelvin scale? (K = C + 273.15 ) • 86 °C •  K = 86 + 273 = 359 K • 78 °F  C = 5/9(F-32) •  C = 5/9*(78-32) =25.5  K = 299 K • -100 °C •  K = -100 + 273  K = 173 °C • 5500 °C •  K = 5500 + 273  K = 5773 °C

  13. Find T from PV = nRT T = PV/(nR) T = 4.2 atm * 7600 L 0.0821 (L-atm)*1800 mol/(mol K) T = 216 K Vrms =  (3kT)/m m(N2) = 4.7 x 10-26 kg Problem 88 What is the rms speed of nitrogen molecules contained in a 7.6 m3 volume at 4.2 atm if the total amount of nitrogen is 1800 mols.

  14. Vrms =  (3kT)/m m(N2) = 4.7 x 10-26 kg Vrms = [(3*1.38 x 10-23 J/K * 216 K)/4.7 x 10-26 kg ]½ Vrms = 436 m/s Problem 88 What is the rms speed of nitrogen molecules contained in a 7.6 m3 volume at 4.2 atm if the total amount of nitrogen is 1800 mols. T = 216 K

  15. Next Time • Quiz #8. • Begin Chapter 14 – Heat. • Please see me with any questions or comments. See you on Wednesday.

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