Ch. 10 GASES &KINETIC-MOLECULAR THEORY

Overview of Gases - physical states - behavior - density

Ideal GasPartial PressureMolar Mass of a Gas

Gas Laws - Charle’s - Boyle’s - Dalton’s - Graham’s

Kinetic Theory - postulates - relationships

Pressure - defined - labels - conversions

Effusion/Diffusion Real GasesVan Der Waals

Combined

Physical States

Chemical composition of gases ---- chemical behaviorAll gases similar in physical behavior

Temp (K) K = 273.15 + oC

Gas Behaviors

1. Vol of gas ∆es w/ P 2. Vol of gas ∆es w/ temp (highly compressible, temp, vol vol gas = vol container)3. Low viscosity 4. Low densities (free flow, little resistance) D = g / L

5. Miscible (gases mix any proportions; homogeneous)

Molecules farther apart in gas than liquid or solid DN2 @ 20oC; 1.25g/L -200oC;0.808g/mL -210oC;1.03g/mL

PRESSURE

Units: SI = N, Newton 1 Kg . m / s2

Pa: Pascal atm: atmosphere mmHg – torr bar

areaforceP

Conversions 1000 Pa =1 kPa 1 Pa = 1 N / m2

1 atm = 101.325 kPa1 atm = 760 mmHg = 760 torr1 mmHg(torr) = 133.322 Pa1 bar = 1*102 kPa = 1*105 Pa

Pressure:atm: atmospheresmm Hg or torr

Temperature: K Volume: L1 mol of gas = 22.4 L

LABELS

1 atm =760 mm Hg1 mm Hg = 1 torr

STP:

Standard Temp & Pressure273.15 K (0oC) -- 1 atm (760 mm Hg)

P1; V1; T1; n1 P2; V2; T2; n2

initial, starting, values

final, ending, values

Quantityn: # moles

4 Physical Variables P, V, T, n

3 LawsBoyle’s, Charle’s, Avogadro’s - changes of 1 variable to another - 2 variables held constant

IDEAL GAS - exhibits linear relation bet variables: V, P, T, n - not actual gas; N2, H2, O2, Noble gases

GAS BEHAVIORS & LAWS

EQUATIONS

Boyle’s: P vs V @ const T P1V1 = P2V2

Charle’s: T vs V @ const P V1T2 = V2T1

Avogadro’s: V vs n @ const P & T V1n2 = V2n1

Combined: No Const

Ideal Gas: PV = nRT

R: gas constant 8.314 J / mol-K 0.0821 atm-L / mol-K

Density of a GasD = m/v = MP / RTDalton’s: Ptotal = P1 + P2 +P3 +…….+ Pn

Boyle’s Law

V inversely proportional to PT, n: constant

V PP1V1 = P2V2

Charle’s Law

V directly proportional to TP, n: constant

T V2

2

1

1

T

V

T

V

n = 0.04P = 1 atm

n = 0.02P = 1 atm

n = 0.04P = 4 atm

Volume vs Temperature

The volume of a balloon is 3.50 L at 1.00 atm. What is the pressure if the balloon is compressed to 2.00 atm?

Boyle’s

P1V1 = P2V2P1 = P2 =V1 = V2 =

V decreased 1/3V decreased 2.5XV increased 3X

What is the effect on 1 mol of an Ideal Gas if1. P increased 3X @ const. T. . . . . . . . . . .2. T decreased 2.5X @ const. P. . . . . . . . . .3. n: added 2X more n @ const. P & T. . . . . . . .

1 atm X3.50 L 2.00 L

atm 1.75 L 00.2

L) atm)(3.50 (1

V

VP P

2

112

Sulfur hexafluoride gas occupies a vol of 5.10 L@ 198oC. The pressure is constant,what temp (K & oC) is needed to reduce the vol to 250. cL?

Charle’s

V1 = 5.10 L V2 = 2.50 L T1 = 273.15 + 198 T2 = X 471.15 K

L 10.5

K 15.471L 50.2 T2

2

2

1

1

T

V

T

V

231 K231 - 273.15 = -42.15 oC

When 2 variables (P, V, T), find effect on the third -- Combined Gas Law --

COMBINE GAS LAW

2

22

1

11

T

VP

T

VP

changes in T, P, & V nothing held constant

A container of 1.0 L is filled w/ a gas at STP. The vol is ed, pressure is increased to 6.0 atm while the temp is raised by 100oC. The new vol becomes ?????

0.23L

STP: T1 = 273.15 K P1= 1.0 atm V1 = 1.0 L T2 = 373.15 K P2 = 6.0 atm V2 = X

atm 0.6K 15.273

K 15.373L 0.1atm 1 V2

???? Standard Pressure 1 atm = 760 mmHg

Dalton’s: Ptotal = PO2 + PN2 +PCO2 + Pothers

What is the mole fraction of N2 present in the air?

for each: P1 = X1PT press gas = (mole fraction)*(total press)

X n

n :fraction mole

total

gas

Air contains oxygen, nitrogen, carbon dioxide, and traces of othergases at standard pressure. What is the partial pressure of PO2 @ 1 atm when the conditions are: PN2= 593.4 mmHg, PCO2= 0.30 mmHg, Pothers = 7.1 mmHg

PO2 = Ptotal - (PN2 +PCO2 + Pothers) = 760 - (593.4 + 0.30 + 7.1) = 159.2 mmHg

atm 0.209

Hg mm 760

atm 1 * Hg mm 159.2 atm

593.4 = X1(760) X1 = 0.781 or 78.1%

Ideal: - no intermolecular attractions - gas particles have no vol. - follows laws @ all condition* of T & P *deviate @ high P & low T - (not really exist)

Real: - can be liquified - sometimes solidified

REAL vs IDEAL GAS

Realize: P depends on total moles of gas

What is the pressure of a10.0 L container w/ mixture 2 gases (2.00 g H2; 8.00 g N2) @ 0oC.

mol0.286 1.00L 0.10

K 15.273KL/molatm 0.0821 P

mol 1.00 g 2.0

mol 1g 2.00

2H

n

mol 0.286 g 28.0

mol 1g 8.00

2N

n

Vol, temp, P?, mass

Ideal Gas: PV = nRT

2.88 atm.

Ideal Gas Law -- Limiting Reagent

An alkali metal reacts w/ a halogen to form metallic halides.What mass of lithium bromide forms when 5.25 L of Br gas@ 0.950 atm & 20oC reacts w/ 2.75 g lithium?

*Diff w/ other limiting rxns, using Ideal gas to find moles (n)

2 Li(s) + Br2(g) -----> 2 LiBr(s)P = .950 atm V = 5.25 L T1 = 293.15 K n = X

What are we looking at?? - use Ideal Gas Eq to find nBr2

- convert g Li to mols - determine limiting reagent

RT

PV n

2Br

K 15.293Kmol

Latm 0821.0

L 25.5atm 950.0

0.207 mol Br2

g 6.9

mol 1Li g 75.2 0.399 mol Li

If Br2 limiting??

Br mol 1

LiBr mol 2Br mol 207.0

22

If Li limiting??

Li mol 2

LiBr mol 2Li mol 399.0

0.414 mol LiBr

0.399 mol LiBr

Li limiting reagent, produces fewer mols LiBrLi limiting reagent, produces fewer mols LiBr

KINETIC - MOLECULAR THEORY

postulate 1: volpostulate 2: motionpostulate 3: collisions

1. Gases move randomly, no attractive forces, mix quickly together

2. Gases occupy less vol. than the vol. a gas occupies; more vol. between gases than gases themselves

3. K.E. of gas is proportional to K temp.; higher temp results in more K.E. results increase velocity

4. Total K.E. of particles is constant; more forceful collisions results in higher pressure

summarize

Explain how gases behave, not whyUnderstanding of P & T @ molecular level

Vol increase w/const T - T remains constant; means ave KE uned - also, urms, unedw/ incr in vol; - molecules longer dist bet collisons - fewer collisions per time - dec P (Boyle’s)

Observations of gas properties expressed in the diff gas laws are explained in Kinetic-molecular Theory

T increase w/const Vol - T incr; means incr ave KE - also, urms, increw/ vol const; - more collisons per time w/ wall - larger in momentum; harder hits - incr P (Charle’s)

3 factors influence P exerted by a gas: # of particles V of gas KE (T) of gas

Rates as applied to 2 diff gases ????

Diffusion - movement one gas through another - described by Graham’s

KE-molec Theory explains: @ a given T & P a gas w/ lower M effuses faster. Why?? - lower M, faster speed, effuses faster

Effusion - process gas escapes container into evacuated space - eff rate inversely propor to sq root molar mass

GRAHAM’S LAW of EFFUSION & DIFFUSION

M

1 α rateeff

time

1 rate

An unknown gas effuses in 14.2 min, an = vol of I2 effuses in 482 s.Mass, M, of unknown?

Also find M of unknown gas, compare rates with known gas

gas1

gas2

2

1

2gas

1gas

time1

time1

: Rate

Rate

M

M

Ratio of rates of 2 gases

793.7 g

03.8

2.14g 253.8

time

time

time1

time1

2

2

2

2

2

I

2

I

xI

2

x

I MM

M

2I

xmin 8.03

mins 60

s 482

2I

Ideal Gas Law: PV = nRT n = m/M & d = m/V

Finding - Density d = m/V - molar mass, M, of a gas

RTm

PVM

RT

Pd

V

m M

PV

mRT then

RT

PV

m n MM

P

dRT M

Mean Free Path & Collision Frequency

MFP: diameter of molec explain collisons -- ave dist molec travels bet collisions -- * larger molec involved in more collisions * more molec per V, more collisions

MFP

speed probable CF Collision Frequency

s

collisions

collisionm

sm

Expect properties of gases to deviate from ideal conditions

Gases at high T & low T deviate from Ideal-gas behavior - moderate P, PV/RT < 1, intermolecular attractions - high P, PV/RT > 1, molecular volume

Attractive forces weaker bet molec than covalent w/i molecImbalance of e- distribution causes attraction bet molecDist bet molec large, attraction 0, gas behaves IdealIncre P, molec closer, attraction becomes greater

Vol bet molec > vol of gas molec themselvesFree V = V containerIncre P results decre VMolec V more imprtnt than attraction

van dervan der WaalsWaals

van der Waals eqn: makes corrections for P & V

nRT nb - VV

an P

2

2

P & V: measured valuesn: moles of real gasa & b: van der Waals constants values for specific gas

A 48.0 L steel tank of oxygen gas contains 9.22 kg @ 25.0oC. How much P is exerted? Does gas show ideal behavior?

mol 288.1 g 32.0

mol 1g 9220 nO2

atm 147

L 48.0

K 15.298KmolatmL 08206.0mol 288.1

P ---- nRTPV

What is the more correct estimate of the P??

2

2

V

na -

nb - V

nRT P b = 0.0318 L-mol-1

a = 1.36 L2-atm-mol-2

P = 181.4 atm – 49.0 atm = 132 atm

ex1: What is the volume in liters of 64.0 g O2 gas at STP?

Given 64.0 g O2(g) at STP Need Volume in liters (L)STEP 1

Solution: convert mass of O2 to moles O2; molar volume of gas at STP used to calculate volume (L) of O2

STEP 2 Plan for solution

Conversion factorsSTEP 3

Set up problem cancel unitsSTEP 4

ex2: Butane, C4H10 , is used as a fuel for barbecues and as an aerosol propellant. If you have 108 mL butane at 715 mm Hg and 25ºC, what is the mass (g) of the butane?Solution

Organize the data, including R, in a table. 3 or more quantities (P,V,n, & T) are known, use ideal gas law to solve for unknown quantity. Pressure given in mm Hg, use R in mm Hg. Volume given in milliliters (mL) converted to liters (L). Temperature converted from to Kelvin.

STEP 1

STEP 2Rearrange solve for unknown. Divide both sides by RT, solve for moles, n:

Substitute values from table to calculate unknownSTEP 3

ex3: Magnesium reacts with HCl, a volume of 355 mL hydrogen gas is collectedover water at 26ºC.

If the pressure of the gases is 752 mm Hg, how many moles of H2(g) were collected? The vapor pressure of water at 26ºC is 25 mm Hg.

STEP 2 Rearrange to solve for unknown. Divide both by RT, solve for moles, n:

SolutionSTEP 1

Substitute values from table calculate unknown. First, determine the partial pressure H2 using Dalton’s law

STEP 3

ex4: Using the Ideal-Gas equation

Calcium carbonate, CaCO3(s), decomposes to give CaO(s) and CO2(g). A sample is decomposed, and CO2 collected in 250-mL flask. After complete, gas has pressure of 1.3 atm at 31 °C. How many moles CO2 gas?Solution

Analyze: Given volume (250 mL), pressure (1.3 atm), and temperature 31 °C of CO2 gas and calculate molesPlan: Given V, P, and T, can solve ideal-gas equation for unknown quantity, n.Solve: In analyzing & solving gas-law problems, helpful to tabulate information given in problems & convert values to units consistent with R(0.0821 L-atm/mol-K). Rearrange equation, solve for n

ex5: Calculating Gas Density

What is the density of carbon tetrachloride vapor at 714 torr and 125 °C?

Solution

Analyze: Calculate density gas given name, pressure, and temperature. From name, can write chemical formula and determine molar massPlan: Calculate the density. Before use equation, need to convert given quantities to appropriate units. Convert temperature to Kelvin & pressure to atmospheres. Calculate molar mass of CCl4

Solve: Temperature is 125 + 273 = 398 K. Pressure is (714 torr)(1 atm/760 torr) = 0.939 atm.Molar mass CCl4 is 12.0 + (4) (35.5) = 154.0 g/mol.Using quantities along Equation 10.10, have

ex6: Applying Dalton’s Law to the Partial Pressures

Gaseous mixture of 6.00 g O2 & 9.00 g CH4 placed in 15.0-L vessel at 0 °C. What is partial pressure of each gas, and what is total pressure in vessel?Solution

Analyze: Calculate pressure for 2 different gases in same volume, at same temperaturePlan: Each gas behaves independently, use ideal-gas eqn to calculate pressure each gas. Total pressure is sum of two partial pressuresSolve: First convert mass of each gas to moles:

Now calculate partial pressure of each gas:

Dalton’s law, total pressure is the sum:

ex7: Applying the Kinetic-Molecular Theory

Sample of O2 gas initially at STP is compressed to smaller volume at constant temperature. What effect does this change have on (a) the average kinetic energy of O2 molecules, (b) the average speed of O2 molecules, (c) the total number of collisions of O2 molecules with the container walls in a unit time, (d) the number of collisions of O2 molecules with a unit area of container wall per unit time?

Solution

Analyze: Apply concepts of kinetic-molecular theory to a gas compressed at constant temperaturePlan: Determine how each quantities in (a)–(d) affected by change in volume at constant temperatureSolve: (a) The average kinetic energy of the O2 molecules determined only by temperature. The average kinetic energy unchanged at constant temperature. (b) Average kinetic energy not change, average speed remains constant.(c) Total number of collisions with container walls per unit time must increase because molecules are moving within smaller volume but with same average speed as before. Must encounter wall more(d) Number of collisions with unit area of wall per unit time increases because total number of collisions with walls per unit time increases and area of the walls decreases

ex8: Calculating a Root-Mean-Square Speed

Calculate the rms speed, u, of an N2 molecule at 25 °C.

Solution

Analyze: Given identity of gas and temperature, quantities need to calculate rms speedPlan: Calculate rms speed use Equation 10.22Solve: Convert each quantity to SI units so all units compatible. Use R units of J/mol-K to cancel units

ex9: Applying Graham’s Law

Unknown gas composed of diatomic molecules effuses at rate only 0.355 times that of O2 at the same temperature. Calculate molar mass of unknown, and identify it.

Plan: Use Graham’s law, Equation 10.23, to determine molar mass of unknown gas. Let rx and represent rate of effusion and molar mass of unknown gas, written as follows:

Solve: From information given,

Solve for unknown molar mass, Need to find Diatomic of 254

g/molI2 = 2*126 = 252 g/mol

Solution

Analyze: Given rate of effusion of unknown gas relative to O2, ask to find molar mass and identity unknown. Need to connect rates of effusion to molar masses

ex10: Cyanogen composed of 46.2% C and 53.8% N by mass. At 25 °C and 751 torr, 1.05 g of cyanogen occupies 0.500 L. What is the molecular formula of cyanogen?

Molar mass of empirical formula, CN, is 12.0 + 14.0 = 26.0 g/mol. Divide molar mass of compound by its empirical formula (52.0 g/mol)/(26.0 g/mol) = 2.00. Molecule has twice as many atoms of each element as empirical formula, giving molecular formula C2N2.

Solution

Analyze: Need to determine molecular formula from data and data on properties of gaseous substanceHave 2 separate calculations(a) Plan: Use percentage composition of compound to calculate empirical formula, then determine molecular formula by comparing mass of empirical formula molar massSolve: Determine empirical formula, assume have 100-g sample & calculate moles each element in sample:

Ratio of the moles of the two elements is essentially 1:1, the empirical formula is CN.Determine the molar mass of the compound