2.  A “model” is an approximation that attempts to explain observable behavior, allows for future predictions in experimentation.  Based on behavior of individual gas particles: (atoms: Ne,He ; molecules H2,Cl2,CO2)

3.  4 Postulates 1. The volume of individual particles is assumed to be negligible (zero). 2. The Particles are at great distances from each other – between them, empty space. 3. The particles are assumed to exert no forces on each other, nor do they attract or repel each other.

4. 4. The average KE of a “collection” of gas particles is assumed to be directly proportional to Kelvin.  ***This model works only for ideal gases***  Real Gases do have volume, they take up space, and they do exert forces on each other b/t them (melting/boiling)  We use Ideal gases + their behaviors in Chem. 1

5. 1. The Kelvin Scale: - An index of gas motion - NOT a measurement of heat 2. T, temperature is directly proportional to KE, kinetic energy, energy in motion. - heat a gas sample T ^, KE ^, motion ^

6. - “absolute zero” – 0K – all motion stops - a gas sample at 100K has half the KE of a gas sample at 200K - K = Degrees Celsius + 273 - 273K is standard temperature

7. 1. Individual gas particles exert a force on the side of their containers (balloon). 2. Atmospheric Pressure: Gravity pulling particles 3. Atmospheric pressure can change with altitude, greater elevation, less pressure, less gas molecules

8.  Kinetic Theory works for IDEAL gas behavior – theoretical for almost all situations of T + P  Remember! T – in Kelvin, measure of KE of gas particles

9.  Remember… Kinetic theory asserts 2 assumptions about gas particles. 1. Gases have no volume, REAL gases do, they are matter therefore they take up some space, though very little. 2. No attractive forces exist b/t gas molecules, but REAL gases do have them, otherwise how could they take condense back to the liquid state.

10. 1 ATM = 760 mmHg = 760 torr = 101.3 KPa  ATM= atmospheres  mmHg= millimeters of mercury  Torr= Torricelli's  KPa= kilo Pascal's Ex: convert 9.23 atmospheres of pressure to KPa 9.23 atm x 101.3 KPa = 9.35 x 10 2 KPa 1 atm Ex: 99.2 KPa to mmHg 99.2 KPa x 760 mmHg = 744 mmHg 101.3 KPa

11.  Dalton’s Law of Partial Pressures - P Total = P 1 + P 2 + P 3 + …. + P 8 at a constant T and V The total pressure exerted by a mixture of gases is the sum of the partial pressure of each gas.

12.  Air is a mix of gas. The partial pressures are: PN 2 = 593.4mmHg, PCO 2 = 0.3mmHg, Pothers= 7.1mmHg, oxygen is also a component. Calculate partial pressure of oxygen at a barometric pressure of 1 atm.

13.  Boyle’s Law for Pressure – P 1 V 1 =P 2 V 2 Constant Temperature, an inverse relationship b/t P + V, as P increases V decreases and vice versa.

14.  P1V1 = P2V2 Example) A 153 cm 3 sample of N 2 gas originally at a P of 82.34 KPa will occupy what volume at standard pressure?

15.  Charles Law for Temperature – Volume Changes V 1 = V 2 T 1 T 2 Constant P, a direct relationship b/t V + T, as V increases, T must also increase and vice versa. T in Kelvin (can’t have a negative volume or motion)

16. V 1 = V 2 T 1 T 2 Example) A balloon has a volume of 98 cm 3 on a 32 C day. If the temperature the following day is 48C, what is the volume?

17.  Guy-Lussac’s Law for Pressure – Temperature Changes P 1 = P 2 T 1 T 2 Constant V, a direct relationship b/t P + T, as one increases/decreases so does the other

18.  Combined Gas Law: P 1 V 1 = P 2 V 2 T 1 T 2 Ideal Gas Law Ideal Gas Law: PV=nRT or gRT/FM

19.  P 1 V 1 = P 2 V 2 T 1 T 2 Ex) A sample of diborane gas, B 2 H 6, a substance that bursts into flame when exposed to air, has a P of 345 torr at a T of -15C and a V of 3.48 L. If conditions are change such that the T is 36C and the P is 0.616 atm, what will be the V?

20.  PV=nRT n=# of moles How many moles of Argon gas can be found in a cylindrical light tube with a volume of 3.7 L and a under a pressure of 162 KPa. The T in the tube is 350 K.

21.  PV= g (R)(T) FM Ex) How many grams of carbon dioxide are in your lungs at a T of 37C and under a pressure of 768 mmHg. Your lung capacity is 4.8 L.

22.  PFM = g = d RT V Ex) Calculate the density of NO2 at 300K if its under a P of 6 atm in a 5.0 L container.