Chapter 5 Gases

Gas – neither definite shape or volume

Oxygen  Joseph Priestley ( ) ( )

 21% of atmosphere  Necessary for life (respiration   photosynthesis)

 Combines with Si, Al, Fe, Ca, Mg, etc. to form rocks and minerals

 Lewis structure  Colorless, odorless, tasteless gas  reactive

 Supports combustion

Ozone (O 3 )  Poisonous blue gas, pungent odor

Hydrogen

 Most abundant element  Rocket fuel (liquid)  Possible fuel of the future

 Colorless, odorless, tasteless gas  reactive

Nitrogen

 78% of atmosphere  Non reactive  Important to life on earth on earth  Nitrogen cycle

Carbon Dioxide  Colorless gas  Important to life

Tests for common gases

Characteristics of Gases 1.Expansion 2.Fluidity

(cont.) 3. Low density

(cont.) 4.Compressibility 5.Diffusion

Kinetic theory  Particles of matter are always in motion

Ideal gas  Imaginary gas, conforms perfectly to kinetic theory  Kinetic theory applies only to an ideal gas, but many gases are close to ‘Ideal’

Kinetic Theory of Gases 1.Gases consist of a lg. # of tiny particles, occupy a volume 1000x the volume of liquid or solid w/ the same # of particles 2.Constant motion 3.Elastic collisions of particles (no net loss of kinetic E)

(cont.) 4.No forces of attraction or repulsion 5.Avg. KE is proportional to Kelvin temp. of gas

Real Gas  Do not completely behave according to kinetic theory, e.g. exert attractive forces

4 measurable quantities of gases 1.Volume (V) 2.Pressure (P) 3.Temperature (T, always in K) 4.# of molecules (moles)

Temperature Scales  Absolute zero = o C (rounded to -273 o C) to -273 o C) = 0 K (not 0 o K) = 0 K (not 0 o K)  Therefore K = o C  0 o C = 273 K

Temp. conversions K = o C K = o C o C = ? K 25.0 o C = ? K = = 298 = 298K = 298K

Units of Pressure  Pressure at sea level & 0 o C =760 mmHg or 1 atm  Standard temperature and pressure (STP) pressure (STP) = 1 atm & 0 o C = 1 atm & 0 o C

Convert a pressure of atm to mmHg atm x 760 mmHg/ 1 atm = 631 mmHg

Write an equation for the relationship between P, V, and T (assume a constant # moles) P x V = a constant P x V = a constant P/ T = a constant P/ T = a constant V/ T = a constant V/ T = a constant PV/ T = a constant PV/ T = a constant P 1 V 1 / T 1 = P 2 V 2 / T 2

Boyle’s Law Boyle’s Law  Volume off a fixed mass of gas varies inversely with pressure at a constant temperature

Robert Boyle ( )  Born at Lismore Castle, Munster, Ireland, the 14th child of the Earl of Cork.  1662, delineated the quantitative relationship that the volume of a gas varies inversely with pressure.

Boyle’s Law  P 1 V 1 = k and P 2 V 2 = k  Therefore P 1 V 1 = P 2 V 2

A sample of oxygen gas occupies a vol. of 150 mL at a pressure of 720 mmHg. What would the volume be at 750 mmHg press.? P 1 V 1 = P 2 V 2 or V 2 = P 1 V 1 / P 2 P 1 V 1 = P 2 V 2 or V 2 = P 1 V 1 / P 2 = (720 mmHg)(150 mL)/ 750 mmHg = 144 mL = 140 mL oxygen gas

France, early 1800’s  Hot air balloons were extremely popular  Scientists were eager to improve the performance of their balloons. Two of the prominent French scientists were Jacques Charles and Joseph-Louis Gay- Lussac,

Charles’ Law  The volume of a fixed mass of gas varies directly with the Kelvin temperature at constant pressure Jacques Charles Jacques Charles

Charles’ Law  V 1 / T 1 = V 2 / T 2

A sample of Ne gas has a vol. of 752 mL at 25.0 deg. C. What is the vol. at 50.0 deg. C? T 1 = 25.0 deg. C = 298 K T 1 = 25.0 deg. C = 298 K T 2 = 50.0 deg C = 323 K T 2 = 50.0 deg C = 323 K V 1 = 752 mL V 1 = 752 mL V 1 / T 1 = V 2 / T 2 or V 2 = V 1 T 2 / T 1 V 1 / T 1 = V 2 / T 2 or V 2 = V 1 T 2 / T 1 = 752 mL x 323 K/ 298K = 752 mL x 323 K/ 298K = 815 mL = 815 mL

Gay-Lussac’s Law  The pressure of a fixed mass of gas varies directly with the Kelvin temp. at constant volume  P 1 / T 1 = P 2 / T 2

P 1 / T 1 = P 2 / T 2

A sample of N gas is at 3.00 atm of pressure at 25 deg C what would the pressure be at 52 deg C? P 1 = 3.00 atm T 1 = 25 deg C = 298 K T 2 = 52 deg C = 325 K P 2 = ? P 1 / T 1 = P 2 / T 2 or P 2 = P 1 T 2 / T 1

(cont.) = (3.00 atm)(325 K)/ 298K = 3.27 atm = 3.3 atm

Combined Gas Law  P 1 V 1 / T 1 = P 2 V 2 / T 2

A helium filled balloon has a vol. of 50.0 L at 25 deg C and 820. mmHg of pressure. What would the vol. be at 650 mmHg pressure and 10. deg C?  P 1 V 1 / T 1 = P 2 V 2 / T 2 or V 2 = P 1 V 1 T 2 / P 2 T 1  V 1 = 50.0 L  V 2 = ?  T 1 = 25 deg C = 298 K  T 2 = 10. deg C = 283 K  P 1 = 820. mmHg  P 2 = 650 mmHg

(cont) V 2 = (820. mmHg)(50.0 L)(283K) / V 2 = (820. mmHg)(50.0 L)(283K) / (650. mmHg )(298 K) =59.9 L =59.9 L = 60. L

Molar Volume of a Gas  One mole of gas (6.02 x molecules) has the same volume at STP (0 deg C, 1 atm) as any other gas

Standard Molar Volume of a Gas = 22.4 L / mol = 22.4 L / mol therefore 1 mol gas = 22.4 L (at STP)

What volume would mol of oxygen gas occupy at STP? mol x 22.4 L/ 1 mol mol x 22.4 L/ 1 mol = 1.52 L O 2

Ideal Gas Law PV = nRT PV = nRT n = number of moles n = number of moles R = ideal gas constant R = ideal gas constant = L. atm/ mol. K Volume must be in L Pressure must be in atm Temp must be in K

What is the P (in atm) exerted by a mol sample of nitrogen gas in a 10.0 L container at 298 K? PV = nRT or P= nRT/ V PV = nRT or P= nRT/ V = (0.500 mol)( L. atm/mol. K)(298K) 10.0 L = 1.22 atm

Gas Stoichiometry  volume-volume calculations e.g. L  L, use like mole ratios e.g. L  L, use like mole ratios

C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g ) How many L of oxygen are required for the complete combustion of L of propane? How many L of oxygen are required for the complete combustion of L of propane? 5 L O 2 : 1 L C 3 H 8 5 L O 2 : 1 L C 3 H L C 3 H 8 x 5 L O L C 3 H 8 x 5 L O 2 1 L C 3 H 8 1 L C 3 H 8 = 1.25 L O 2 = 1.25 L O 2

CaCO 3 (s)  CaO(s) + CO 2 (g) How many grams of calcium carbonate must be decomposed to produce 2.00 L of CO 2 at STP? How many grams of calcium carbonate must be decomposed to produce 2.00 L of CO 2 at STP? 2.00 L CO 2 x 1 mol CO 2 x 1 mol CaCO L CO 2 1 mol CO L CO 2 1 mol CO 2 x g CaCO 3 x g CaCO 3 1 mol CaCO 3 1 mol CaCO 3 = 8.94 g CaCO 3 = 8.94 g CaCO 3

Graham’s Law of effusion or diffusion

Graham’s Law of Effusion or Diffusion The rate of effusion or diffusion is inversely proportional to the square roots of their molar masses The rate of effusion or diffusion is inversely proportional to the square roots of their molar masses Rate A = M B Rate A = M B Rate B M A

Compare the rate of effusion of hydrogen and nitrogen gas rate H 2 = 28 = 14 = 3.7 rate H 2 = 28 = 14 = 3.7 rate N rate N therefore hydrogen effuses 3.7 times faster than nitrogen therefore hydrogen effuses 3.7 times faster than nitrogen