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

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 * x 10 5 N/m 2 * m 3 ) ( 1.38 x J/K * 293 K) N = 6.60 x 10 23

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

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.

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

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.

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

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

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 molecules/mol) V rms = (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 ) °C  K = = 359 K °F  C = 5/9(F-32)  C = 5/9*(78-32) =25.5  K = 299 K °C  K =  K = 173 °C °C  K =  K = 5773 °C

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 kg Find T from PV = nRT T = PV/(nR) T = 4.2 atm * 7600 L (L-atm)*1800 mol/(mol K) T = 216 K

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 kg V rms = [(3*1.38 x J/K * 216 K)/4.7 x kg ] ½ V rms = 436 m/s T = 216 K

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