Chapter 5 - Gases

1)fluidity –ability to flow –mainly empty space –random arrangement 2)low density –part. very spread out 1000x further apart than solid or liquid –example – oxygen density –D–O 2(gas) = 1.43 g/L or g/mL –D–O 2(liquid) = 1140 g/L or 1.14 g/mL –D-O 2(solid) = 1429 g/L or g/mL

3)highly compressible –no definite volume –mainly empty space 4)fill any container(no def. shape) –no attractive forces, independent part. –expand to occupy all spaces of container

5)exert pressure –gas pressure – force created when gas particles collide with a surface press = force/area –force – measured in newtons »1 N = 1 kg▪m/s 2 pascal(Pa) = newton/m 2

–standard measures of pressure 101,325 Pa or kPa 760 mm Hg = in. Hg 760 torr 14.7 psi 1 atm –measured w/ barometer discovered by Torricelli air pressure, on an avg. day at sea level, will support 760 mm(29.92 in.) of Hg

aneroid barometer

kinetic molecular theory(KMT) –molecular model that describes behavior of gases –postulates of KMT 1)gas part. in constant random motion –no intermolecular forces 2)volume of a gas particle is negligible 3)collisions are elastic, no attractive or repulsive forces 4)avg. KE is directly related to Kelvin temp.

measurable properties of gases 1)pressure(P) kPa, mm Hg, atm 2)temp.(T) K, o C 3)volume(V) mL, L, cm 3 4)moles(n)

Boyle’s law V  P  or V  P  –volume of a gas is inversely related to the pressure

k = PVor V 1 P 1 = V 2 P 2

sample problem: A balloon has a volume of 5.75 L and a pressure of atm. If Aunt Edith sits on the balloon and decreases the volume to 1.75 L what is the pressure of the gas inside the balloon? k = PV k= atm x 5.75 L k = 5.61 L x atm P = k V P = 5.61 L x atm 1.75 L P = 3.21 atm P 1 V 1 = P 2 V 2 V 2 V 2 P 2 = P 1 V 1 V2V2 P 2 = atm x 5.75 L 1.75 L P 2 = 3.20 atm

Gay-Lussac’s Law T  P  or T  P  –the pressure of a gas is directly related to the temperature of the gas

k = Por P 1 P 2 TT 1 T 2 =

Sample problem – A balloon has a pressure of kPa at 25 o C, what is the pressure if the balloons temperature was at -15 o C? k = P T k = kPa 25 o C k = 5.4 kPa/ o C k x T = P x T T P = k x T P = 5.4 kPa x -15 o C o C P = -81 kPa P = kPa x 258 K K P = 117 kPa 298 K k = kPa/K

Sample problem – A spray paint can has a bursting pressure of 5.65 atm. If the can has a pressure of 1.30 atm at 25 o C, what temperature would the can burst? k = P T k = 1.30 atm 298 K k = atm/K k = P T T x k = P k k T = P k T = 5.65 atm atm/K T = 1.30 x 10 3 K = 1020 o C x T T x

When a gas is heated, the volume of the gas __________ ??INCREASES

Charles’ Law T  V  or T  V  – the volume of a gas is directly related to the temperature of the gas

k = VorV 1 V 2 TT 1 T 2 Charles also proved the change in volume of a gas was 1/273 the original volume for every o C change –what would happen to 1.0 L of gas if the temperature decreased by 273 o C???? =

Sample problem: If a weather balloon contains 45 L of hydrogen gas at 15 o C, what is the volume when it gets up in the atmosphere and the temperature is -45 o C? k = V T k = 45 L 288 K k = 0.16 L K k x T = V V = 0.16 L x 228 K K V = 36 L

Boyle’s Law - k = PV Gay-Lussac’s Law - k = P/T Charles’ Law - k = V/T Combined Gas Law - k = _____ or P 1 V 1 P 2 V 2 T 1 T 2 = PV T T

Sample problem – An aerosol can has propane gas as a propellant. What would the volume of propane be at STP if the volume of propane in the can is 135 mL at a temperature of 22 o C and pressure of 1.73 atm?

Sample problem – When I was 6, I accidentally let go of my helium balloon from the circus. The volume of the balloon when I let go was 2.49 liters at 18 o C and 1.22 atm of pressure. When the balloon was struck by a jet passing over at higher altitude, what was its volume if the temperature was 24 o C less and the pressure was 674 torr?

Avogadro’s law –equal volumes of gases, at the same T and P, contain the same # of particles (Avogadro’s principle) –V of a gas is directly related to the particles of gas(moles) n  V  or n  V  k = Vor V 1 V 2 nn 1 n 2 –basis for Avogadro’s #(6.023 x ) –molar volume – the volume of 1 mol of any STP L of any STP = 1 mol any gas =

Ideal gas law -describes the behavior of a gas that always and perfectly follows the gas laws -ideal gas properties -no IM forces -particles have no volume -no condensation -real gases follow gas laws most of the time -except at high P -except at low T -except any time intermolecular forces affect gas part.

Ideal Gas Law  PV = nRT P = pressure V = volume n = moles T = temperature R = L atm/mol K R = L kPa/mol K R = L torr/mol K R = PV nT

Sample problem – How many moles of carbon dioxide gas are there if 35.6 liters has T = 35 o C and 96.3 kPa?

Sample problem – How many liters of helium gas are there if 15.5 grams has T = 15 o C and kPa?

Dalton’s law of partial pressure – the total P of a mixture of gases is the sum of the partial P of each gas P total = P 1 + P 2 + P 3 + P n Sample problem

diffusion –the even spreading out of a substance from an area of high concentration to areas of low concentration effusion – passage of a gas thru a small opening or hole

Graham’s law of diffusion –the rate of diffusion of a gas is inversely related to the square root of the gases molar mass –compare rates of diffusion of different gases ט A M B ט B M A =

Sample problem – If helium gas and carbon dioxide gas effuse out of the same hole, how do their rates compare? rate He g/mol 3.32 rate CO g/mol 1.00 He effuses 3.32 times faster than CO 2 = =

Gas Stoichiometry gas volumes correspond to mole ratios 3 H 2 (g) + N 2 (g)  2 NH 3 (g) coefficients represent –3 moles hydrogen react with 1 mole nitrogen to form 2 moles ammonia –3 liters hydrogen react with 1 liter nitrogen to form 2 liters ammonia(for gases only) –67.23 liters STP react with liters STP to form liters STP

sample problem: How many liters of hydrogen gas will be produced at K and 96.0 kPa if 1.74 moles of sodium react with water? answer: 21.1 L H 2

sample problem: What volume of oxygen, collected at 25 o C and 101 kPa, can be prepared by decomposing 37.9 grams of potassium chlorate? answer: 11.4 L