# Gases I. Physical Properties.

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Gases I. Physical Properties

Real vs. Ideal Gases: Ideal gas = an imaginary gas that conforms perfectly to all the assumptions of the kinetic-molecular theory We will assume that the gases used for the gas law problems are ideal gases.

Real Gases vs. Ideal Gases:
Real gas = a gas that does not behave completely according to the assumptions of the kinetic-molecular theory.  All real gases deviate to some degree from ideal gas behavior. However, most real gases behave nearly ideally when their particles are sufficiently far apart and have sufficiently high kinetic energy.

Causes of non-ideal behavior:
The kinetic-molecular theory is more likely to hold true for gases whose particles have NO attraction for each other.

Causes of non-ideal behavior:
1. High pressure (low volume): Space taken up by gas particles becomes significant  Intermolecular forces are more significant between gas particles that are closer together

Causes of non-ideal behavior:
2. Low temperature: Gas particles move slower so intermolecular forces become more important

KMT (Kinetic Molecular Theory)
Kinetic Molecular Theory (KMT) = the idea that particles of matter are always in motion and that this motion has consequences. theory developed in the late 19th century to account for the behavior of the atoms and molecules that make up matter

KMT (Kinetic Molecular Theory)
based on the idea that particles in all forms of matter are always in motion and that this motion has consequences SOLID LIQUID GAS

KMT (Kinetic Molecular Theory)
can be used to explain the properties of solids, liquids, and gases in terms of the energy of particles and the forces that act between them

B. Kinetic Molecular Theory – Ideal gas
KMT describing particles in an IDEAL gas: have no volume. have elastic collisions. are in constant, random motion. don’t attract or repel each other. have an avg. KE directly related to Kelvin temperature.

C. Kinetic Molecular Theory - Real Gases
KMT describing particles in a REAL gas: have their own volume attract each other Gas behavior is most ideal… at low pressures at high temperatures in nonpolar atoms/molecules

D. Characteristics of Gases- using KMT
Gases expand to fill any container. KMT - gas particles move rapidly in all directions without significant attraction or repulsion between particles

D. Characteristics of Gases- using KMT
Gases are fluids (like liquids). KMT - No significant attraction or repulsion between gas particles; glide past each other

D. Characteristics of Gases- using KMT
Gases have very low densities. KMT - particles are so much farther apart in the gas state sodium in the solid state: sodium in the liquid state: sodium in the gas state:

D. Characteristics of Gases- using KMT
Gases can be compressed. KMT - gas particles are far apart from one anther with room to be “squished” together

D. Characteristics of Gases- using KMT
Gases undergo diffusion & effusion. KMT – gas particles move in continuous, rapid, random motion Effusion

E. Temperature Always use Kelvin temperature when working with gases. ºF ºC K -459 32 212 -273 100 273 373 K = ºC + 273

F. Pressure Which shoes create the most pressure?

F. Pressure Why do gases exert pressure?
Gas particles exert a pressure on any surface with which they collide! More collisions = increase in pressure!

F. Pressure Barometer = measures atmospheric pressure
The height of the Hg in the tube depends on the pressure The pressure of the atmosphere is proportional to the height of the Hg column, so the height of the Hg can be used to measure atmospheric pressure! Mercury Barometer

F. Pressure Pressure UNITS 101.3 kPa (kilopascal) 1 atm 760 mm Hg
760 torr

Standard Temperature & Pressure
G. STP STP Standard Temperature & Pressure 0°C (exact) 1 atm (exact) 273 K kPa 760 mm Hg (exact) 760 torr (exact)

Gases II. The Gas Laws P V T

GAS LAW PROBLEMS- MUST USE KELVIN
K = ºC

A. Boyle’s Law P V PV = k

A. Boyle’s Law The pressure and volume of a gas are INVERSELY related at constant mass & temp P1V1 = P2V2 P V

A. Boyle’s Law Real life application: When you breathe, your diaphragm moves downward, increasing the volume of the lungs. This causes the pressure inside the lungs to be less than the outside pressure so air rushes in.

A. Boyle’s Law Ex: Halving the volume leads to twice the rate of collisions and a doubling of the pressure.

A. Boyle’s Law As the volume increases, the pressure decreases!

B. Charles’ Law V T

B. Charles’ Law The volume and temperature (K) of a gas are DIRECTLY related at constant mass & pressure V1_ = V2__ T T2 V T

B. Charles’ Law Real life application: Bread dough rises because yeast produces carbon dioxide. When placed in the oven, the heat causes the gas to expand, and the bread rises even further.

B. Charles’ Law As temperature decreases, the volume of the gas decreases Liquid nitrogen’s temp. is about 63K or -210 ºC or -346 ºF!

B. Charles’ Law As temperature decreases, the volume of the gas decreases

NUMBER OF PARTICLES & PRESSURE ARE CONSTANT!!!
B. Charles’ Law NUMBER OF PARTICLES & PRESSURE ARE CONSTANT!!!

C. Guy-Lussac’s Law P T

C. Guy-Lussac’s Law _P1_ = P2__ T1 T2
The pressure and temperature (K) of a gas are DIRECTLY RELATED at constant mass & volume P T _P1_ = P2__ T T2

C. Guy-Lussac’s Law When the temp. of a gas increases (KE increases) and gas particles move faster and hit container walls more frequently and collisions are more forceful

C. Guy-Lussac’s Law Real life application: The air pressure inside a tire increases on a hot summer day.

V P PV T = k P1V1 T1 = P2V2 T2 Boyles PV D. Combined Gas Law
Charles’ T P Guy Lussac’sT PV T Boyles PV = k P1V1 T1 = P2V2 T2

E. EXAMPLE Problems You need your calculators!

Gas Law Problems CHARLES’ LAW V1_ = V2__ T V WORK: GIVEN:
1.) A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW V1_ = V2__ T T2 GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K WORK: T V V2 = V1T2 T1 V2= (473 cm3)(367 K) (309 K) V2 = 562 cm3 C. Johannesson

Gas Law Problems BOYLE’S LAW P1V1 = P2V2 P V WORK: GIVEN: V2 = P1V1
2.) A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW P1V1 = P2V2 P GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa V WORK: V2 = P1V1 P2 V2 = (150.kPa)(100.mL) 200.kPa V2 = 75.0 mL C. Johannesson

Gas Law Problems P T V V1 = 7.84 cm3 V2 = P1V1T2 P1 = 71.8 kPa P2T1
3.) A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW P1V1 = __P2V T T2 GIVEN: V1 = 7.84 cm3 P1 = 71.8 kPa T1 = 25°C = 298 K V2 = ? P2 = kPa T2 = 273 K P T V WORK: V2 = P1V1T2 P2T1 V2 = (71.8 kPa)(7.84 cm3)(273 K) (101.3 kPa) (298 K) V2 = 5.09 cm3 C. Johannesson

Gas Law Problems GUY LUSSAC’S LAW P1_ = P2__ T P GIVEN: WORK:
4.) A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GUY LUSSAC’S LAW P1_ = P2__ T T2 GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? WORK: T2 = P2T1 P1 P T T2 = (560. torr)(296K) 765 torr T2 = 217 K C. Johannesson

III. Standard Molar Volume, Gas Densities & Molar Mass
Gases III. Standard Molar Volume, Gas Densities & Molar Mass

A. Standard Molar Volume
The volume occupied by one mole of a gas at STP L/mol or about 22.4 L/mol In other words, one mole of any ideal gas at STP will occupy 22.4 L

B. Example Problems-standard molar volume
1.) A chemical reaction is expected to produce mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP? 0.068 mol O L O2 1 mol O2 = 1.52 L O2

B. Example Problems-standard molar volume
1.) A chemical reaction is expected to produce mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP? 0.068 mol O L O2 1 mol O2 = 1.52 L O2

B. Example Problems-standard molar volume
2.) A chemical reaction produced 98.0 mL of SO2 at STP. What mass (in grams) of the gas was produced? g SO2 98.0 mL SO L SO mol O2 = 1000 mL SO2 22.4 L SO2 1 mol SO2 0.280 g SO2

C. Density The ratio of an object’s mass to its volume. D = M V
The volume (and density) of a gas will change when pressure and temperature change. Densities are usually given in g/L at STP

C. Density For an ideal gas: Density (at STP) = molar mass
standard molar volume

D. Example Problems- density
1.) What is the density of CO2 in g/L at STP? g CO mol CO2 1 mol CO L CO2 = 1.96 g/L CO2

D. Example Problems- density
2.) What is the molar mass of a gas whose density at STP is 2.08 g/L. 2.08 g L 1 L 1 mol = 46.6 g/mol

IV. More Gas Laws: Ideal Gas Law & Avogadro's Law
Gases IV. More Gas Laws: Ideal Gas Law & Avogadro's Law

A. Avogadro’s Law Equal volumes of gases contain equal numbers of moles at constant temp & pressure V n

UNIVERSAL GAS CONSTANT
B. Ideal Gas Law Boyle’s, Charles, and Avogadro’s laws are contained within the ideal gas law! Merge the Combined Gas Law with Avogadro’s Law: V n PV T PV nT = R = k UNIVERSAL GAS CONSTANT R= Latm/molK R=8.315 dm3kPa/molK You don’t need to memorize these values!

PV=nRT B. Ideal Gas Law P= pressure V = volume n = moles
T = temperature (in K) R = the ideal gas constant GAS CONSTANT R= Latm/molK R=8.315 dm3kPa/molK You don’t need to memorize these values!

C. Ideal Gas Law Problems
1.) Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L. PV = nRT GIVEN: P = ? atm n = mol T = 16°C = 289 K V = 3.25 L R = Latm/molK WORK: P = nRT V P =(0.412 mol)(0.0821Latm/molK) (289K) 3.25L P = 3.01 atm C. Johannesson

C. Ideal Gas Law Problems
2.) Find the volume of 85 g of O2 at 25°C and kPa. PV = nRT GIVEN: V = ? n = 85 g T = 25°C = 298 K P = kPa R = dm3kPa/molK WORK: 85 g O2 1 mol O2 = 2.7 mol g O2 O2 = 2.7 mol V = nRT P V =(2.7 mol)(8.315 dm3kPa/molK) (298K) 104.5 kPa V = 64 dm3 C. Johannesson

C. Ideal Gas Law Problems
3.) At 28C and 98.7 kPa, 1.00 L of an unidentified gas has a mass of 5.16 g. Calculate the number of moles of gas present and the molar mass of the gas. PV = nRT GIVEN: V = 1.00 L n = ? T = 28°C = 301 K P = 98.7 kPa R = LkPa/molK WORK: n = PV RT (98.7 kPa) (1.00L) (8.314 LkPa/molK) (301 K) n = mol = m = g mol = 131 g/mol C. Johannesson

IV. Two More Laws: Dalton’s Law & Graham’s Law
Gases IV. Two More Laws: Dalton’s Law & Graham’s Law

Ptotal = P1 + P2 + P3 ... A. Dalton’s Law
The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. Ptotal = P1 + P2 + P3 ...

A. Dalton’s Law Ptotal = P1 + P2 + P3 ...

A. Dalton’s Law When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 AND water vapor.

A. Dalton’s Law 1.) Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor. GIVEN: PH2 = ? Ptotal = 94.4 kPa PH2O = 2.72 kPa WORK: Ptotal = PH2 + PH2O 94.4 kPa = PH kPa PH2 = 91.7 kPa Look up water-vapor pressure on gas formula sheet for 22.5°C. Sig Figs: Round to least number of decimal places. C. Johannesson

A. Dalton’s Law GIVEN: Pgas = ? Ptotal = 742.0 torr PH2O = 42.2 torr
2.) A gas is collected over water at a temp of 35.0°C when the barometric pressure is torr. What is the partial pressure of the dry gas? The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor. GIVEN: Pgas = ? Ptotal = torr PH2O = 42.2 torr WORK: Ptotal = Pgas + PH2O 742.0 torr = PH torr Pgas = torr Look up water-vapor pressure on gas formula sheet for 35.0°C. Sig Figs: Round to least number of decimal places. C. Johannesson

B. Graham’s Law Diffusion =
Spreading of gas molecules throughout a container until evenly distributed. Effusion = Passing of gas molecules through a tiny opening in a container

KE = ½mv2 B. Graham’s Law Speed of diffusion/effusion
Kinetic energy is determined by the temperature of the gas. At the same temp & KE, heavier molecules move more slowly. Larger m  smaller v KE = ½mv2

KE = ½mv2 KE = kinetic energy m= molar mass v = velocity
B. Graham’s Law KE = kinetic energy m= molar mass v = velocity KE = ½mv2

B. Graham’s Law Graham’s Law = Rate of effusion of a gas is inversely related to the square root of its molar mass. The equation shows the ratio of Gas A’s speed to Gas B’s speed. C. Johannesson

C. Graham’s Law Problems
1.) Determine the relative rate of effusion for krypton and bromine. The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”. Kr effuses times faster than Br2.

C. Graham’s Law Problems
2.) A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? Put the gas with the unknown speed as “Gas A”. C. Johannesson

C. Graham’s Law Problems
3.) An unknown gas diffuses 4.0 times faster than O2. Find its molar mass. The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0. Square both sides to get rid of the square root sign. C. Johannesson

VI. Gas Stoichiometry at Non-STP Conditions
Gases VI. Gas Stoichiometry at Non-STP Conditions

A. Gas Stoichiometry Moles  Liters of a Gas: STP - use 22.4 L/mol
Non-STP - use ideal gas law Non-STP Given liters of gas? start with ideal gas law Looking for liters of gas? start with stoichiometry conv.

B. Volume  Mass Stoich. Thinking: Mole Ratio If gas A is at STP:
Gas volume A  moles A  moles B  mass B If gas A is at STP: ? L “A” 1 mol “A” ? mol “B” molar mass(g) “B” 22.4 L “A” ? mol “A” mol “B” Mole Ratio

B. Volume  Mass Stoich. If gas “A” is NOT at STP:
Use PV = nRT to find moles of “A” ? mol “A” ? mol “B” molar mass (g) “B” ? mol “A” 1 mol “B”

Given liters: Start with Ideal Gas Law and calculate moles of O2.
B. Volume  Mass 1.) How many grams of Al2O3 are formed from L of O2 at 97.3 kPa & 21°C? 4 Al O2  2 Al2O3 15.0 L non-STP ? g GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = dm3kPa/molK WORK: n = PV RT n= (97.3 kPa) (15.0 L) (8.315dm3kPa/molK) (294K) n = mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT  C. Johannesson

Use stoich to convert moles of O2 to grams Al2O3.
B. Volume  Mass 1.) How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al O2  2 Al2O3 15.0L non-STP ? g Use stoich to convert moles of O2 to grams Al2O3. 0.597 mol O2 2 mol Al2O3 3 mol O2 g Al2O3 1 mol Al2O3 = 40.6 g Al2O3

B. Volume  Mass ? g 12.6L 12.6 L SO2 1 mol SO2 22.4L SO2 1 mol S8
2.) What mass of sulfur is required to produce 12.6 L of sulfur dioxide at STP according to the equation: S8 (s) O2 (g)  8 SO2 (g) ? g 12.6L 12.6 L SO2 1 mol SO2 22.4L SO2 1 mol S8 8 mol SO2 g S8 1 mol S8 = 18.0 g S8

C. Mass  Volume Thinking: If gas “B” is at STP:
Mass A  moles A  moles B  gas volume B If gas “B” is at STP: ? g “A” 1 mol “A” ? mol “B” L “B” molar mass (g) “A” ? mol “A” mol “B”

C. Mass  Volume If gas “B” is NOT at STP:
? g “A” mol “A” ? mol “B” molar mass (g) “A” ? mol “A” Then use PV = nRT to find volume of “B”

C. Mass  Volume 2 Na + Cl2  2 NaCl ? L 10.4 g = 5. 07 L Cl2
1.) What volume of chlorine gas at STP is needed to react completely with 10.4 g of sodium to form NaCl? 2 Na + Cl2  2 NaCl 10.4 g ? L 22.4 L Cl2 10.4 g Na 1 mol Cl2 1 mol Na g Na 2 mol Na 1 mol Cl2 = L Cl2

C. Mass  Volume CaCO3  CaO + CO2 5.25 g ? L non-STP 5.25 g CaCO3
2.) What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC? CaCO3  CaO CO2 5.25 g ? L non-STP Looking for liters: Start with stoich and calculate moles of CO2. 5.25 g CaCO3 1 mol CaCO3 100.09g 1 mol CO2 CaCO3 = 1.26 mol CO2 Plug this into the Ideal Gas Law to find liters.

C. Mass  Volume 2.) What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 1.26 mol T = 25°C = 298 K R = dm3kPa/molK WORK: V = nRT P V = (1.26mol)(8.315dm3kPa/molK) (298K) (103 kPa) V = 30.3 dm3 CO2 C. Johannesson