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Unit 6 (Chp 10):

Gases

John Bookstaver

St. Charles Community College

St. Peters, MO

2006, Prentice Hall, Inc.

Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

Characteristics of Gases

• Unlike liquids and solids, they…

expand to fill their containers.(indefinite volume)

are highly compressible.

have extremely low densities.

Atmospheric Pressure:weight of air per area

Pressure• Pressure is the amount

of force applied per area.

P =FA

22,000 lbs!!!(per sq. meter)

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Atmospheric Pressure

(weight of air)

empty space(a vacuum)

h760 mm

1 atm = 760 mmHg= 760 torr= 101.3 kPa

1 N1 m2

Units (at sea level)

Pressure

STP(standard T & P)273 K 1 atm

1) 657 mmHg to atm2) 830 torr to atm3) 0.59 atm to torr

657760

830760

0.59 x 760

Kinetic-Molecular Theory

KMT is a modelwhich explains the

Properties(P, V, T, n)

and

Behavior(motion, energy, speed, collisions)

of gases.

2) Gas pressure is caused by collisions with the container walls.

P = FA

P = FA

P = FA

1) Gas particles are in constantrandom motion.

5 Parts of Kinetic-Molecular Theory

Collisions are elastic(no KE lost).

(ideally)

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Ideally: Vgas = Vcontainer

5 Parts of Kinetic-Molecular Theory

4) Volume of particles is NEGLIGIBLE,compared to total volume of container.

Vgas = Vcontainer – Vparticles

In a 1.000 L container,

gas only expands into about 0.999 L of

volume

(negligible)1.000 L container

has 1.000 L of gas

3) Attractive forces (IMAFs) are NEGLIGIBLE.

IMAFs

5) Average KE of gas particles is……directly proportional to Kelvin Temp.

(K not oC)no negative temp’s, no negative energies, no negative volumes, etc.)

KEavg α T

5 Parts of Kinetic-Molecular Theory

Boyle’s Law (P & V)

1 atm 2 atm

10 L

5 L

(inversely proportional)

P ↑ , V ↓

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(directly proportional)

Charles’ Law (V & T)

how absolute zero was estimated

150 K

60 L30 L

300 K

T ↑ , V ↑

(directly proportional)

Lussac’s Law (P & T)

300 K

100 kPa 500 kPa

600 K

200 kPa

T ↑ , P ↑

Avogadro's Hypothesis• At the same ___ & ___, equal _________ of

gas must contain equal _________.

CO2

HeO2

1 mol He

2 mol Hen ↑ , V ↑

Avogadro's Law (V & n)

All at:P = 1 atmT = 25oCV = 1.0 L

moles (n)

add gas

volumesT P(particles)

(directly proportional)

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V 1/P (Boyle’s law)V T (Charles’s law)P T (Lussac’s law)V n (Avogadro’s law)

• So far we’ve seen that:

PVnT

= R

ideal gas constant: R = 0.08206 L∙atm/mol∙K

constant

PV = nRTIdeal Gas Law

(all gases same ratio)

NO Units of: mL , mmHg , kPa , grams , °C

given on exam

(changes in P,V,T,n)

P ↑ , V ↓ T ↑ , V ↑

T ↑ , P ↑

n ↑ , V ↑

(directly proportional)(inversely proportional)

PV = nRT

= Rconstant

given on exam

=P1V1

n1T1

P2V2

n2T2(initial) (final)

1. The pressure on a 411 mL sample of gas is decreased from 812 torr to 790 torr. How will volume of the gas be affected?

P1

P2

V2

V1

PV = nRT

= ?

(812)(0.411) = nRT

Ideal-Gas Changes PV = nRT

(790) (?) = nRT V increased

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2. A 10.0 L sample of a gas is collected at 25oC and then cooled to a new volume of 8.83 L while the pressure remains at 1.20 atm. How will the temperature in be affected?

V1

T1

V2 T2 = ?

Ideal-Gas Changes PV = nRT

PV = nRT

(1.20)(10.0)

(1.20)(8.83) = nR(?) T decreased

= nR(298)

=

HW p. 432 #1, 23, 26

Ideal-Gas Calc’s PV = nRT1. A 5.00 L He balloon has 1.20 atm at 0.00oC.

How many moles of He gas are in the balloon?

(1.20 atm)(5.00 L) = n (0.08206)(273 K)

(1.20)(5.00)(0.08206)(273)

= n

n = 0.268 mol He

PV = nRT R = 0.08206L∙atm∙mol–1∙K–1

How many atoms of He?

PT

V

n

Ideal-Gas Calc’s

Molar Mass

2. A sample of aluminum chloride gas weighing 0.0500 g at 350.oC and 760. mmHg of pressure occupies a volume of 19.2 mL.Calculate the Molar Mass of the gas.

133 g/mol

(760 mmHg)(0.0192 L) = n (62.36)(623 K)

n = 0.000376 mol

PV = nRT R = 62.36 L∙torrmol∙K

gramsmole

M = __0.0500 g_0.000376 mol

=AlCl3 =

n = mM

M = mn

so…

(given on exam)

mTPVM = ?

PV = nRT

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TPVV = ?@STP

Ideal-Gas Calc’s

3. A gas at 25.0oC and 1.20 atm occupies a volume of 4.28 L.What volume will it occupy at STP?

(1.20 atm)(4.28 L) = n (0.08206)(298 K)

n = 0.210 mol

PV = nRTR = 0.08206 L∙atm

mol∙K

PV = nRT

HW p. 438 #92, 29,

46, 35, 38

(1.00)V = (0.210)(0.08206)(273)

V = 4.70 L

• The volume of 1 mole of any gas at STPis:

Vm = _____

PVm = nRT

the ______ of ______of any gas at ____.

volume 1 mole

STP

(1.00 atm) Vm = (1 mol )(0.08206)(273 K)

Molar Volume:

22.4 L1 mol

but…ONLYat STP!!!

or22.4 L/mol22.4 L∙mol–1

Gas Stoich with Molar Volume1. Calculate the mass of NH4CI reacted with

Ca(OH)2 to produce 11.6 L of NH3(g) at STP.2 NH4Cl + Ca(OH)2 2 NH3 + CaCI2 + 2 H2O

2. Calculate the volume of CO2 gas produced when 9.85 g of BaCO3 is decomposed at STP.

BaCO3(s) BaO(s) + CO2(g)

27.7 g

1.12 L

11.6 L NH3 x 1 mol NH3 x22.4 L NH3

2 mol NH4Cl x2 mol NH3

53.49 g NH4Cl =1 mol NH4Cl

9.85 g BaCO3 x 1 mol BaCO3 x197.34 g BaCO3

1 mol CO2 x1 mol BaCO3

22.4 L CO2 =1 mol CO2

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3. What volume of O2 gas is produced from 490 g KClO3 at 298 K and 1.06 atm?

KClO3(s) KClO(s) + O2(g)

V = ____L O2

490 g KClO3 xg KClO3 mol KClO3

mol KClO3 x mol O2 mol O2122.55

=1

1 1

92.3 L O2

NOT at STP

Molar Mass of KClO3 is 122.55 g/mol

PV = nRTuse…

4.00

(1.06 atm) V = (4.00 mol )(0.08206)(298 K)(would be 89.6 L

if at STP)

Mole Fraction (XA)

PA = Ptotal x XA

XA =moles of A

total moles

• The mole fraction (XA) is like the % of total moles that is A, but without the % or x 100.

WS 6b#1-4

Ptotal = PA + PB + PC + …

Dalton’s Law of Partial Pressures

HWp. 436

#636466

Ptotal = PH2O + Pgas

• When one collects a gas over water, there is water vapor mixed in with the gas.

• To find only the pressure of the gas, one must subtract the water vapor pressure from the total pressure. Pgas = Ptotal – PH2O

equalize water levelsinside & outside =

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1. Calculate the mass of 0.641 L of H2 gas collected over water at 21.0oC with a total pressure of 750. torr.The vapor pressure of water at 21.0oC is 20.0 torr.

Ptotal = PH2O + Pgas PV = nRT

750. = 20.0 + PH2

PH2 = 730. torr

PH2 = 730/760 = 0.961 atm

(0.961)(0.641) = nH2 (0.08206)(294)

nH2 = 0.0255 mol

mH2 = 0.0514 g

↑ T , ↑ v↑ M , ↓ v

Study the models below. What can be said quantitatively about the molecular speed (v) of a gas in relation to its molar mass (M) and its temperature (T)?

Compare the molecular speed (v) of these gases:

1) at 25.0 oC (i) Helium (ii) Oxygen (O2)

2) at 50.0 oC (i) Helium (ii) Oxygen (O2)

3) Does the data support your conclusions from the models on the previous slide about effects of T and M on v ?

↑ T , ↑ v↑ M , ↓ v

1360 m/s 482 m/s

1420 m/s 502 m/s

KE = ½ mv2

(given on exam)

WHY?

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Distributions of Molecular Speed

average molecular speed (v)

KE = ½ mv2

(KMT)

(given on exam)

Therefore:

T & v are__________proportional

directly

Temp (K) & KEavg

are directly proportional T α KEavg

↑ T , ↑ v

Gases at the same Temp, have the same _____.

Speed vs. Molar MassKEavg

½ mv2 = ½ mv2

KE1 = ½ m1v12

Ar: M = 40 g/mol He: M = 4.0 g/mol

KE = ½ mv2

(at same T)

KEAr = KEHe

KE2 = ½ m2v22

↑ M , ↓ v ↓ M , ↑ v

M & v are__________proportional

inversely

Effusion

escape of gas particles

through a tiny hole

spread of gas particles throughout

a space

Diffusion↑T, ↑v

↑M,↓v

KE = ½ mv2

64 g/mol (SO2)16 g/mol (CH4)

_____ is __ times faster than _____.

CH4

SO2

2

KE = ½ mv2

½ mv2= ½ mv2

HW p.433 #8, 74, 76a

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(usually)

“Real” (non-Ideal) GasesIn the real world, the behavior of gases only conforms to the ideal-gas equation under “ideal” conditions.

Ideal: (High T) (Low P)

WHY?(weak IMAFs)

(negligible)

(high KE) (high Vtotal)

(ONLY under ideal conditions)

PV = nRT

(tiny Vparticles)(negligible)

Ideal Gas vs Non-Ideal Gas

) (V ) = nRT(P

“observed” P too lowb/c at low Temp

IMAFs not negligible,collisions less frequent

and of less force

“observed” V too highb/c at high P,

particle volumenot negligible

compared to total volume

(ideal P) (ideal V)

NON

T: ↓P: ↑

IDEAL

T: ↑P: ↓

more KEovercome IMAFsmore total volume

less KEsignificant IMAFsless total volume

IMAFs

Vgas = Vcontainer

n2aV2 − nb+

HW p.437 #81,82,83

– Vparticles

Slides #35-37 demonstrate calculation of gas changes

using the combined gas law.

This is no longer required on the new AP Exam

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1. The pressure on a 411 mL sample of gas is decreased from 812 torr to 790 torr. What will be the new volume of the gas?

= 422 mLP1

P2

V2

V1

P2V2

n2T2

P1V1

n1T1=

= ?

(812)(411) = (790) V2

(812)(411)(790)

= V2

Ideal-Gas Changes PV = nRT

PVnT

= R

2. A 10.0 L sample of a gas is collected at 25oC and then cooled to a new volume of 8.83 L while the pressure remains at 1.20 atm. What is the final temperature in oC?

V1

T1

V2 T2P2V2

n2T2

P1V1

n1T1=

= ?

(10.0)(298)

= (8.83)T2

T2 (10.0) = (8.83)(298)

(8.83)(298)(10.0)

=T2

Ideal-Gas Changes

= 263 K = –10 oC

PV = nRT

PVnT

= R

0.502 mol O2 x 2 mol O3 = 3 mol O2

3. A 13.1 L sample of 0.502 moles of O2 is held under conditions of 1.00 atm and 25.0oC. If all of the O2 is converted to Ozone (O3), what will be the volume of O3? = 8.73 L

n1

n2 = ?

V2

V1

O2 O33 2

0.335mol O3

n2

n1

P2V2

n2T2

P1V1

n1T1= (13.1)

(0.502)= V2 .

(0.335)

Ideal-Gas Changes PV = nRTHW p. 432 # 1, 23, 89, 26

PVnT

= R

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PV = nRT

0.500 mol O2 x 2 mol O3 = 3 mol O2

3. A 12.2 L sample of 0.500 moles of O2 is held under conditions of 1.00 atm and 25.0oC. If all of the O2 is converted to Ozone (O3), what will be the volume of O3?

n1

n2 = ?

V2

V1

O2 O33 20.333mol O3

n2n1

Ideal-Gas Changes PV = nRT

(1.00)(12.2) = (0.500) R (298)(1.00) (?) = (?) R (298)(1.00) V =

V = 8.19 L(0.333) R (298)

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