2. Nov. 05, 2001 Dr. Larry Dennis, FSU Department of Physics1 Physics 2053C – Fall 2001 Chapter 13 Temperature & Ideal Gases

3. 2 CAPA 7 & 8 A scuba tank has a volume of 3900 cm 3. 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 of 12.5 atm? 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

4. 3 CAPA 7 & 8 A scuba tank has a volume of 3900 cm 3. 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 of 12.5 atm? PV = NkT The same number as there are oxygen molecules. N = 6.60 x 10 23

5. 4 Kinetic Theory of Gasses 1. Gases contain a large number of molecules moving in random directions with a variety of speeds. 2. Molecules are very far apart and don’t exert forces on one another except when they collide. 3. Molecules obey Newton’s Laws. 4. Collisions are perfectly elastic.

6. 5 Kinetic Theory of Gasses The kinetic energy of the gas is directly related to it’s temperature. KE = ½ m(v 2 ) ave = 3/2 kT Only depends on temperature. V rms =  (V 2 ) ave ( root mean square velocity ) V rms =  (3kT)/m

7. 6 RMS Velocity T = 100 K T = 200 K T = 300 K v f V rms =  (3kT)/m KE total = 3/2 NkT KE one = 3/2 kT Gas molecules have a distribution of speeds. The range of speeds depends on the temperature.

8. 7 CAPA 9 & 10 A scuba tank has a volume of 3900 cm 3. 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

9. 8 CAPA 9 & 10 A scuba tank has a volume of 3900 cm 3. 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? V rms =  (3KT/m) V rms (He)/V rms (O 2 ) =  ( m(O 2 )/m(He) ) V rms (He)/V rms (O 2 ) =  ( 2*16/(4) ) V rms (He)/V rms (O 2 ) =  8 = 2.83

10. 9 Quiz #8 Temperature Scales Fahrenheit  Centigrade  Kelvin Ideal Gas Law PV = nRT or PV = NkT Gauge Pressure Pay careful attention to units.

11. 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 10 23 molecules/mol) V rms = (3kT/m) ½ average speed of a molecule.

12. 11 Quiz #8 Questions, Chap. 13: 21 Problems, Chap. 13: 5, 28, 35, 47, 88

13. 12 Problem 28 What are the following temperatures on the Kelvin scale? (K = C + 273.15 ) 1. 86 °C  K = 86 + 273 = 359 K 2. 78 °F  C = 5/9(F-32)  C = 5/9*(78-32) =25.5  K = 299 K 3. -100 °C  K = -100 + 273  K = 173 °C 4. 5500 °C  K = 5500 + 273  K = 5773 °C

14. 13 Problem 88 What is the rms speed of nitrogen molecules contained in a 7.6 m 3 volume at 4.2 atm if the total amount of nitrogen is 1800 mols. V rms =  (3kT)/m m(N 2 ) = 4.7 x 10 -26 kg 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

15. 14 Problem 88 What is the rms speed of nitrogen molecules contained in a 7.6 m 3 volume at 4.2 atm if the total amount of nitrogen is 1800 mols. V rms =  (3kT)/m m(N 2 ) = 4.7 x 10 -26 kg V rms = [(3*1.38 x 10 -23 J/K * 216 K)/4.7 x 10 -26 kg ] ½ V rms = 436 m/s T = 216 K

16. 15 Next Time Quiz #8. Begin Chapter 14 – Heat. Please see me with any questions or comments. See you on Wednesday.