II. Gas Pressure A. Pressure is force per unit area (f/a) 1. result of particle collisions 2. measured by a barometer 3. influenced by temperature, gas volume, and the number of gas particles a. as the number of particle collisions increases the pressure increases
Pressure at Sea Level 14.7 psi = 1.0 atm = 760 mm of Hg = 750 Torr = 101.3 kPa = 1,013 mbars
Kinetic Theory A. Assumptions 1. gas particles do not attract each other 2. gas particles are very small 3. particles are very far apart 4. constant, random motion 5. elastic collisions 6. kinetic energy varies with temperature
B. Properties of Gases 1. low density (grams/liter) 2. can expand and can be compressed 3. can diffuse and effuse a. rate related to molar mass b. diffusion is the movement of particles from an area of greater concentration to an area of lesser concentration c. effusion is the movement of gas particles through a small opening
II. The Gas Laws A.Boyle’s Law (P1V1 = P2V2 )inverse relationship 1.As the volume of a gas increases the pressure decreases (temperature remains constant) 2.Example A sample of gas in a balloon is compressed from 7.00 L to 3.50 L. The pressure at 7.00L is 125 KPa. What will the pressure be at 2.50L if the temperature remains constant? P1 = 125 KPa P2 = X V1 = 7.00L V2 = 3.50L (125)(7.00) = (X) (3.50) X = 250.KPa
As volume increases the pressure decreases when temperature remains constant
Boyle’s Law • Pressure is related to 1/Volume Slope (k) = relationship between P and 1/V P = k(1/V)
B. Charles’ LawV1 = V2must use kelvin T1 T2 temperature • As the temperature of a gas increases the volume increases (direct relationship) 2. ExampleA gas sample at 20.0 C occupies a volume of 3.00 L. If the temperature is raised to 50.0 C, what will the volume be if the pressure remains constant? V1 = 3.00L V2 = X T1 = 293K T2 = 323K 3.00 = X 293X = (3)(323) X = (3)(323) 293 323 293 X = 3.31 L
C. Gay Lussac’s LawP1 = P2 T1 T2 1. as the temperature increases the pressure increases when the volume remains constant 2. Example The pressure of a gas in a tank is 4.00 atm at 200.0C. If the temperature rises rises to 800.0C, what will be the pressure of the gas in the tank? P1 = 4.00 atm P2 = X T1 = 473K T2 = 553K 4.00 = X 473X = (4)(553) X = (4)(553) • 553 473 X = 4.68 atm
D. Combined Gas LawP1 V1 = P2 V2 T1 T2 1. Combines Boyle’s, Charle’s and Gay Lussac’s 2. Example A gas at 70.0KPa and 10.0C fills a flexible container with an initial volume of 4.00L If the temperature is raised to 60C and the pressure is raised to 80.0 KPa, what is the new volume? P1 = 70.0 KPa P2 = 80.0 KPa V1 = 4.00 V2 = X T1 = 283K T2 = 333K (70.0)(4.00) = (80.0)(X) 283 333 X = (33.3)(70.0)(4.00) (2.83)(80.0) X = 41.2L
E. Dalton’s Law of Partial Pressures Ptotal = P1 + P2 + P3 + .....Pn The total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture 1. Example Find the total pressure for a mixture that contains four gases with partial pressures of 5.00 kPa, 4.56 kPa, 3.02 kPa and 1.20kPa.
2. Suppose two gases in a container have a total pressure of 1.20 atm. What is the pressure of gas B if the partial pressure of gas A is 0.75 atm? 3. What is the partial pressure of hydrogen gas in a mixture of hydrogen and helium if the total pressure is 600.0mmHg and the partial pressure of helium is 439 mmHg?
III. Avogadro’s Principle A. Equal volumes of gases at the same temperature and pressure have the same number of particles B. Molar Volume (22.4 L at STP) 1. volume of one mole of gas particles at STP(standard temperature and pressure) 0C and 1.00 atm (760mm Hg) * 1 mole of any gas at STP = 22.4 L 2. conversion factors 1 mol22.4 L 22.4 L 1 mol
Equal volumes of gases at the same temperature and pressure contain the same number of particles
C. Sample Problems 1. Calculate the volume occupied by .250 mol of oxygen gas at STP. 2. Calculate the number of moles of methane gas in a 11.2 L flask at STP.
3. Calculate the volume of 88.0 g of CO2 at STP. 4. How many grams of He are found in a 5.60L balloon at STP?
5. Calculate the density of H2 at STP. D = molar mass molar volume 6. Calculate the molar mass of a gas that has a density of 3.2 g/L.
IV. Ideal vs Real Gases A. Ideal compared to Real Gases 1. ideal gas a) particles do not have volume b) there are no intermolecular attractions c) all particle collisions are elastic d) obey all kinetic theory assumptions
2. real gases behave like ideal gases except when a) pressure is very high b) temperatures are low c) molecules are very large d) spaces between particles is small (small volume)
B. Ideal Gas Law - PV = nRT 1. pressure ( atm,mm Hg, KPa) 2. volume (liters) 3. temperature (kelvin) 4. number of moles (n) 5. R = constant (L) (pressure unit*) (mol) (K)
unit for pressure determines which constant must be used in the Ideal Gas Law PV = nRT a) R = 62.4 (pressure is mm Hg) b) R = .0821 (pressure is atm) c) R = 8.314 (pressure is KPa)
C. Application Problems (PV = nRT) 1. How many moles of O2 are in a 2.00L container at 2.00 atm pressure and 200K? 2. Calculate the volume occupied by 2.00 mol of N2 at 300K and .800 atm pressure.
3. What is the pressure in mm Hg of .200 moles of gas in a 5.00 L container at 27C? 4. Calculate the number of grams of oxygen in a 4.00 L sample of gas at 1.00 atm and 27 C.
V. Gas Stoichiometry A. Coefficients and Gas Volume 1. Gay Lussac’s Law of Combining Volumes a) gases in chemical reactions react with each other in whole number ratios at a constant temperature and pressure CH4 + 2O2 -----> CO2 + 2H2O 1 volume 2 volumes 1 volume 2 volumes 1 mole 2 moles 1 mole 2 moles 22.4L 2 x 22.4L 22.4L 2 x 22.4L