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
22,000 lbs!!!
Pressure is the amount of force applied per area.
(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
STP
(standard T & P)
273 K 1 atm
1 N
1 m2
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 that 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) Collisions with the walls of the container cause gas pressure.
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 gas particles is negligible.
(compared to total volume of container)
Ideally: Vgas = Vcontainer
Vgas = Vcontainer – Vparticles
In a L container,
gas fills 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)
video clip

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 P1V1 n1T1 P2V2 n2T2 = R = = P ↑ , V ↓
given on exam
NOT
given on exam
P1V1
n1T1
P2V2
n2T2
P1V1
n1T1
P2V2
n2T2
= R = =
constant
(initial)
(final)
(changes in P,V,T,n)
(inversely proportional)
(directly proportional)
P ↑ , V ↓
T ↑ , V ↑
T ↑ , P ↑
n ↑ , V ↑

15. 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

16. 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

17. 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

18. How many molecules of He?
Ideal-Gas Equation
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 molecules of He?
n =
0.268 mol He

19. PV = nRT Ideal-Gas Equation PV = nRT HW p. 438 #92, 29, 46, 38, 35 m T
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. 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!!!

21. Gas Stoich with Molar Volume
Assume 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 completely 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

22. 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)
= ____L O2
92.3 L O2
Molar Mass of KClO3 is g/mol

23. 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 a % of total moles that is A, but without the % or x 100.

24. = 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

25. 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)

26. 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

27. 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

28. 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

29. 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
KEHe = KEAr
(at same T)
½ mv2 = ½ mv2
M & v are __________
proportional
inversely
↑ M , ↓ v
↓ M , ↑ v

30. 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
HW p.437 #8,74,76a
SO2

31. 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)
(weaker IMAFs)
(high KE)
(high Vtotal)
(negligible)
WHY?
Non-Ideal: (Low T) (High P)
(stronger IMAFs)
(low KE)
(low Vtotal)
(not negligible)

32. 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

33. n2a V2 ) (V ) = nRT (P + − nb Van der Waal Equation:
a & b are two constants.
a: measures how strongly the gas particles are attached together.
b: a measure of finite volume occupied by the molecules.
a & b generally increase with increasing molecular mass ??
Larger, massive molecules have large volumes and tend to have greater IMF.
n2a V2
) (V ) = nRT
(P +
− nb