1. Phys 250 Ch12 p1 Chapter 12: Gas Laws and Kinetic Theory Air Pressure at bottom of column of mercury: P =  gh, h≈76 cm pressure= atmospheric pressure, which support column of mercury Boyle’s Law: At constant temperature, the volume of a gas is inversely proportional to is pressure. => the product of pressure and volume for a sample of gas is fixed p 1 V 1 = p 2 V 2 (T = constant) Example: A cylinder with a height.20 m and cross sectional area pf 0.40 m 2 has a close fitting piston which compresses the cylinder volume to a height of.12 m. If the air started at atmospheric pressure and the temperature remains constant, what is the new pressure of the compressed air within the cylinder? vacuum air pressure

2. Phys 250 Ch12 p2 The Law of Charles and Gay-Lussac Thermal expansion at constant pressure  V = b V 0  T All gases have the same value for  ! starting at 0°C,  = 1/273 /°C Extrapolated: all gases have zero volume at  273°C (  460°F) All gases extrapolate to zero pressure in a constant volume gas thermometers at  273°C  Absolute Temperature Scales are natural scales, and Example: To what temperature would the air in a hot air balloon have to be heated so that its mass would be 0.980 times that of an equal volume of air at a temperature of 25C?

3. Phys 250 Ch12 p3 Ideal Gas Law Charles’s Law and Boyle’s Law can be combined to relate pressure, volume and absolute temperature for more general changes. n = number of moles R = Gas Constant R=8.314 J/mole·K The Ideal Gas Law is an example of an Equation of State (an equation which relates the variables describing the state of the system).

4. Phys 250 Ch12 p4 Microscopic Atoms, simplest and smallest subdivision Molecule, a combination of atoms bonded together Different atoms and/or molecules not bonded together Macroscopic Elements, chemically simplest materials Compound, can be chemically reduced Mixture, can be chemically or physically separated/simplified The type of atom determines the element, the type of molecule determines the compound, etc. Atoms and Molecules mass often expressed in atomic mass units 1 atomic mass unit = 1 u = 1.66x10-27 kg Isotopes of an element: extremely slight variation in chemical properties => slightly different types of atoms of same element (different masses for different atoms of same element). The weight of an atom in u is approximately the same as the molecular weight of the corresponding element in grams.

5. Phys 250 Ch12 p5 Example: A sample of a gas originally has a volume of 0.5 liters at room temperature (23ºC) and pressure (1.00 atm) is transferred to a container where it is cooled to 55C in a volume of.12 L. What is the new pressure of the gas? Example: What is the density of carbon dioxide gas at a temperature of 23ºC and atmospheric pressure?

6. Phys 250 Ch12 p6 Example: An automobile tire is filled to a gauge pressure of 240 kPa early in the morning when the temperature is 15ºC. After the car has been driven for the day, the temperature of the iar in the tires is 70ºC. Estimate the new gauge pressure.

7. Phys 250 Ch12 p7 Kinetic Theory of the Ideal Gas Container with volume V contains a large number N of identical molecules of mass m. Molecules act as point particles (size is small compared to intraparticle distances). Molecules are in constant motion and obey Newton’s Laws of motion. Molecules collide elastically with walls of container. Walls of container are perfectly rigid. Pressure from collisions: Each elastic collision exerts an impulse on the wall of the container. => Boyle’s Law: pressure is inversely proportional to volume vxvx vyvy v vxvx vyvy v

8. Phys 250 Ch12 p8 Each elastic collision exerts an impulse on the wall of the container For a molecule with v x to hit wall within a time dt, it must be within v x dt of the wall. The number of collisions is The total imulse on the wall is Impulse is also related to force Pressure is average force per area vxvx vyvy v vxvx vyvy v A v x dt

9. Phys 250 Ch12 p9 symetry!

10. Phys 250 Ch12 p10 Example: Estimate the rms speed of oxygen molecules at STP (standard Temperature and Pressure: 0ºC and 1 atm). Compare this with the speed of hydrogen molecules under the same conditions.

11. Phys 250 Ch12 p11 Internal Energy of an Ideal Gas average KE of one molecule: KE =3/2 kT for N molecules KE tot = N 3/2 kT but N = n N A, and R = N A k so U = 3/2 n N A k T = 3/2 nRT this is the Internal Energy of the Gas Example: A parade balloon contains 368 m 3 of helium at a pressure of 115 kPa. What is the internal energy of the helium in the balloon?