1. Unit 6 (Chp 10): Gases Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 6 (Chp 10): Gases
John Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. Characteristics of Gases
Unlike liquids and solids, they…
expand to fill their containers.
(indefinite volume)
are highly compressible.
have extremely low densities.

3. Pressure F P = A Atmospheric Pressure: weight of air per area
Pressure is the amount of force applied per area.
22,000 lbs!!!
(per sq. meter)
P = F A
Atmospheric Pressure:
weight of air per area

4. Pressure Units (at sea level) STP (standard T & P) 273 K 1 atm
empty space
(a vacuum)
Units (at sea level)
1 atm = 760 mmHg
760 torr
101.3 kPa h
760 mm
1 N
1 m2
STP
(standard T & P)
273 K 1 atm
Atmospheric Pressure
(weight of air)
657
760
1) 657 mmHg to atm
2) 830 torr to atm
3) 0.59 atm to torr
830
760
0.59 x 760

5. Kinetic-Molecular Theory
KMT is a model which explains the
Properties
(P, V, T, n)
and
Behavior
(motion, energy, speed, collisions)
of gases.

6. 5 Parts of Kinetic-Molecular Theory
Gas particles are in constant random motion.
2) Gas pressure is caused by collisions with the container walls.
P = F A
Collisions are elastic
(no KE lost).
(ideally)
P = F A
P = F A

7. 5 Parts of Kinetic-Molecular Theory
3) Attractive forces (IMAFs) are NEGLIGIBLE.
IMAFs
4) Volume of particles is NEGLIGIBLE,
compared to total volume of container.
Ideally: Vgas = Vcontainer
Vgas = Vcontainer – Vparticles
In a L container,
gas only expands into about
0.999 L of volume
1.000 L container
has L of gas
(negligible)

8. 5 Parts of Kinetic-Molecular Theory
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
video clip – Describing the invisible properties of gas (TEDEd) (3 min)

9. (inversely proportional)
Boyle’s Law (P & V)
P ↑ , V ↓
(inversely proportional)
10 L
5 L
1 atm
2 atm

10. (directly proportional)
Charles’ Law (V & T)
T ↑ , V ↑
(directly proportional)
60 L
30 L
how absolute zero was estimated
300 K
150 K

11. (directly proportional)
Lussac’s Law (P & T)
T ↑ , P ↑
(directly proportional)
500 kPa
100 kPa
200 kPa
300 K
600 K

12. (directly proportional)
Avogadro's Hypothesis
At the same ___ & ___, equal _________ of gas must contain equal _________. T P
volumes
moles (n)
(particles)
All at:
P = 1 atm
T = 25oC
V = 1.0 L O2 He
CO2
Avogadro's Law (V & n)
n ↑ , V ↑
2 mol He
1 mol He
add gas
(directly proportional)

13. PV = nRT Ideal Gas Law PV = R nT R = 0.08206 L∙atm/mol∙K
So far we’ve seen that:
V  1/P (Boyle’s law)
V  T (Charles’s law)
P  T (Lussac’s law)
V  n (Avogadro’s law)
constant PV nT
= R
(all gases same ratio)
R = L∙atm/mol∙K
ideal gas constant:
given on exam
PV = nRT
Ideal Gas Law
NO Units of: mL , mmHg , kPa , grams , °C

14. PV = nRT P1V1 n1T1 P2V2 n2T2 = R = P ↑ , V ↓ T ↑ , V ↑ T ↑ , P ↑
given on exam
P1V1
n1T1
P2V2
n2T2
= R =
(changes in P,V,T,n)
constant
(initial)
(final)
(inversely proportional)
(directly proportional)
P ↑ , V ↓
T ↑ , V ↑
T ↑ , P ↑
n ↑ , V ↑

15. PV = nRT Ideal-Gas Changes PV = nRT V1
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? P2 V2
= ? P1
PV = nRT
(812)(0.411) =
nRT
(790) (?) =
nRT
V increased

16. PV = nRT Ideal-Gas Changes PV = nRT V1
A 10.0 L sample of a gas is collected at oC and then cooled to a new volume of L while the pressure remains at 1.20 atm. How will the temperature in be affected? T1 V2 T2
= ?
PV = nRT
(1.20)(10.0) =
nR(298)
(1.20)(8.83) =
nR(?)
T decreased =

17. PV = nRT Ideal-Gas Changes PV = nRT O2  O3 3 2 n1 V1
A 12.2 L sample of 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? V2
PV = nRT
HW p #1, 23, 26
(1.00)(12.2) =
(0.500) R (298)
(1.00) (?) =
(?) R (298)
(1.00) V =
(0.335) R (298)
O2  O3 3 2 V =
8.19 L
0.500 mol O2 x 2 mol O3 =
3 mol O2
0.333
mol O3 n2 n1
n2 = ?

18. PV = nRT Ideal-Gas Calc’s PV = nRT V P T
A 5.00 L He balloon has 1.20 atm at 0.00oC.
How many moles of He gas are in the balloon? n
PV = nRT
R =
L∙atm∙mol–1∙K–1
(1.20 atm)(5.00 L) =
n ( )(273 K)
(1.20)(5.00)
( )(273) = n
How many atoms of He?
n =
0.268 mol He

19. PV = nRT Ideal-Gas Calc’s PV = nRT m T P V M = ?
A sample of aluminum chloride gas weighing g at 350.oC and 760 mmHg of pressure occupies a volume of 19.2 mL.
Calculate the Molar Mass of the gas.
PV = nRT
R = L∙atm
mol∙K
(1.00 atm)( L) =
n ( )(623 K) m M
(given on exam)
n =
mol
n =
__ g_
mol
M =
so… m n
Molar Mass
grams
mole
M = =
AlCl3 =
133 g/mol

20. PV = nRT Ideal-Gas Calc’s PV = nRT HW p. 438 #92, 29, 46, 35, 38 T P V
@STP
A gas at 25.0oC and 1.20 atm occupies a volume of 4.28 L.
What volume will it occupy at STP?
R = L∙atm
mol∙K
PV = nRT
(1.20 atm)(4.28 L) =
n ( )(298 K)
n =
0.210 mol
(1.00)V =
(0.210)( )(273)
V =
4.70 L

21. PVm = nRT Molar Volume: Vm = _____ 22.4 L 1 mol the ______ of ______
of any gas at ____.
volume
1 mole
STP
PVm = nRT
(1.00 atm) Vm =
(1 mol )( )(273 K)
The volume of 1 mole of any gas at STP will be:
Vm = _____
22.4 L
1 mol
but…
ONLY at STP!!!

22. Gas Stoich with Molar Volume
The reactions below occurred at STP.
Calculate the mass of NH4CI reacted with Ca(OH)2 to produce 11.6 L of NH3(g) .
2 NH4Cl + Ca(OH)2  2 NH3 + CaCI2 + 2 H2O
Calculate the volume of CO2 gas produced when 9.85 g of BaCO3 is decomposed.
BaCO3(s)  BaO(s) + CO2(g)
11.6 L NH3 x
1 mol NH3 x
22.4 L NH3
2 mol NH4Cl x
2 mol NH3
53.49 g NH4Cl =
1 mol NH4Cl
27.7 g
1.12 L
9.85 g BaCO3 x
1 mol BaCO3 x
g BaCO3
1 mol CO2 x
1 mol BaCO3
22.4 L CO2 =
1 mol CO2

23. KClO3(s)  KClO(s) + O2(g)
NOT at STP
PV = nRT
use…
What volume of O2 gas is produced from 490 g KClO3 at 298 K and 1.06 atm?
KClO3(s)  KClO(s) + O2(g) 1
mol KClO3 1
mol O2
490 g KClO3 x x =
4.00
mol O2
122.55
g KClO3 1
mol KClO3
(1.06 atm) V =
(4.00 mol )( )(298 K)
(would be 89.6 L if at STP)
= ____L O2
92.3 L O2
Molar Mass of KClO3 is g/mol

24. Dalton’s Law of Partial Pressures
Ptotal = PA + PB + PC + …
Mole Fraction (XA) HW
p. 436
#63 64 66
moles of A
XA =
WS 6b
#1-4
total moles
PA = Ptotal x XA
The mole fraction (XA) is like the % of total moles that is A, but without the % or x 100.

25. = Ptotal = PH2O + Pgas Pgas = Ptotal – PH2O
equalize water level inside & outside
Ptot Patm =
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

26. PV = nRT Ptotal = PH2O + Pgas
Calculate the mass of 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.
750. = PH2
nH2 = mol
PH2 = torr
mH2 = g
PH2 = 730/760 = atm
(0.961)(0.641) = nH2 ( )(294)

27. 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)?
↑ M , ↓ v
↑ T , ↑ v

28. KE = ½ mv2 WHY? ↑ M , ↓ v ↑ T , ↑ v (given on exam)
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)
Does the data support your conclusions from the models on the previous slide about effects of T and M on v ?
1360 m/s
482 m/s
1420 m/s
502 m/s
WHY?
↑ M , ↓ v
↑ T , ↑ v

29. Distributions of Molecular Speed
Temp (K) & KEavg
are directly proportional
T α KEavg
(KMT)
average molecular speed (v)
KE = ½ mv2
(given on exam)
Therefore:
T & v are __________
proportional
directly
↑ T , ↑ v

30. ½ mv2 = ½ mv2 ↑ M , ↓ v ↓ M , ↑ v Speed vs. Molar Mass KE1 = ½ m1v12
Gases at the same Temp, have the same _____.
KEavg
KE = ½ mv2
Ar: M = 40 g/mol
He: M = 4.0 g/mol
KE1 = ½ m1v12
KE2 = ½ m2v22
KEAr = KEHe
(at same T)
½ mv2 = ½ mv2
M & v are __________
proportional
inversely
↑ M , ↓ v
↓ M , ↑ v

31. KE = ½ mv2 Effusion Diffusion ↑T, ↑v ↑M,↓v KE = ½ mv2 ½ mv2 = ½ mv2
escape of gas particles through a tiny hole
spread of gas particles throughout a space
↑M,↓v
½ mv2 = ½ mv2
64 g/mol (SO2)
16 g/mol (CH4)
CH4
_____ is __ times faster than _____. 2
SO2
HW p.437 #8, 74, 76a

32. “Real” (non-Ideal) Gases
In the real world, the behavior of gases only conforms to the ideal-gas equation under “ideal” conditions.
(usually) PV
= nRT
(ONLY under ideal conditions)
Ideal: (High T) (Low P)
(high KE)
(high Vtotal)
WHY?
(weak IMAFs)
(tiny Vparticles)
(negligible)
(negligible)

33. Ideal Gas vs Non-Ideal Gas
T: ↑
P: ↓
NON
T: ↓
P: ↑
more KE/speed
weaker IMAFs
more avg. dist.
less KE/speed
stronger IMAFs
less avg. dist.
HW p.437 #81,82,83
IMAFs
(ideal P)
(ideal V)
n2a V2
) (V ) = nRT (P +
− nb
“observed” V too high b/c
size of particles
not negligible
compared to total volume
“observed” P too low b/c attractive forces not negligible,
collisions less frequent and of less force
Vgas = Vcontainer
– Vparticles

34. Slides #34-37 demonstrate calculation of gas changes using the combined gas law. His is no longer required on the new AP Exam

35. PV = nRT Ideal-Gas Changes P1V1 n1T1 P2V2 n2T2 = PV = R nT V1
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? P2 P1 V2
= ?
P1V1
n1T1
P2V2
n2T2
= 422 mL =
(812)(411) =
(790) V2 PV nT
(812)(411)
(790) = V2
= R

36. PV = nRT Ideal-Gas Changes P1V1 n1T1 P2V2 n2T2 = PV = R nT V1
A 10.0 L sample of a gas is collected at oC and then cooled to a new volume of L while the pressure remains at 1.20 atm. What is the final temperature in oC? T1 V2
P1V1
n1T1
P2V2
n2T2 T2
= 263 K
= ?
= –10 oC =
(8.83)(298)
(10.0) T2 =
(10.0)
(298)
(8.83) T2 = PV nT
T2 (10.0) =
(8.83)(298)
= R

37. PV = nRT Ideal-Gas Changes P1V1 n1T1 P2V2 n2T2 = 3 O2  O3 2 PV nT = R
HW p. 432 # 1, 23, 89, 26 V1 n1
A 13.1 L sample of 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? V2
= 8.73 L
P1V1
n1T1
P2V2
n2T2
n2 = ?
(13.1)
(0.502) V
(0.335) = = 3
O2  O3 2 n2 PV nT
0.502 mol O2 x 2 mol O3 =
3 mol O2
0.335
mol O3
= R n1