Gas_Laws_powerpoint_notes.ppt. slides for grade 7

KMT and Gas Laws for Grade 10

1. I. Physical
Properties
Ch. 11 -Gases

2. A. Kinetic Molecular Theory
Particles in an ideal gas…
•have no volume.
•have elastic collisions.
•are in constant, random, straight-
line motion.
•don’t attract or repel each other.
•have an avg. KE directly related
to Kelvin temperature.

3. B. Real Gases
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

4. C. Characteristics of Gases
Gases expand to fill any container.
•random motion, no attraction
Gases are fluids (like liquids).
•no attraction
Gases have very low densities.
•no volume = lots of empty space

5. C. Characteristics of Gases
Gases can be compressed.
•no volume = lots of empty space
Gases undergo diffusion & effusion.
•random motion

6. D. Temperature
ºF
ºC
K
-459 32 212
-273 0 100
0 273 373 32FC
9
5

K = ºC + 273
Always use absolute temperature
(Kelvin) when working with gases.

7. E. Pressurearea
force
pressure
Which shoes create the most pressure?

8. E. Pressure
Barometer
•measures atmospheric pressure
Mercury Barometer
Aneroid Barometer

9. E. Pressure2
m
N
kPa
KEY UNITS AT SEA LEVEL
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
14.7 psi

10. F. STP
Standard Temperature & Pressure
0°C 273 K
1 atm 101.325 kPa
-OR-
STP

11. II. The Gas
Laws
BOYLES
CHARLES
GAY-
LUSSAC
Gas Laws

12. A. Boyle’s Law
P
V
PV = kVolume
(mL)
Pressure
(torr)
P·V
(mL·torr)
10.0 760.0 7.60 x 10
3

20.0 379.6 7.59 x 10
3

30.0 253.2 7.60 x 10
3

40.0 191.0 7.64 x 10
3

13. A. Boyle’s Law
The pressure and volume
of a gas are inversely
related
•at constant mass & temp
P
V
PV = k

14. k
T
V
 V
T
B. Charles’ LawVolume
(mL)
Temperature
(K)
V/T
(mL/K)
40.0 273.2 0.146
44.0 298.2 0.148
47.7 323.2 0.148
51.3 348.2 0.147

15. k
T
V
 V
T
B. Charles’ Law
The volume and absolute
temperature (K) of a gas
are directly related
•at constant mass &
pressure

16. k
T
P
 P
T
C. Gay-Lussac’s LawTemperature
(K)
Pressure
(torr)
P/T
(torr/K)
248 691.6 2.79
273 760.0 2.78
298 828.4 2.78
373 1,041.2 2.79

17. k
T
P
 P
T
C. Gay-Lussac’s Law
The pressure and absolute
temperature (K) of a gas
are directly related
•at constant mass &
volume

18. = kPV
P
T
V
T
PV
T
D. Combined Gas Law
P
1V
1
T
1
=
P
2V
2
T
2
P
1V
1T
2 =P
2V
2T
1

19. GIVEN:
V
1= 473 cm
3
T
1= 36°C = 309K
V
2= ?
T
2= 94°C = 367K
WORK:
P
1V
1T
2 = P
2V
2T
1
E. Gas Law Problems
A gas occupies 473 cm
3
at 36°C.
Find its volume at 94°C.
CHARLES’ LAW
TV
(473 cm
3
)(367 K)=V
2(309 K)
V
2= 562 cm
3

20. GIVEN:
V
1= 100. mL
P
1= 150. kPa
V
2= ?
P
2= 200. kPa
WORK:
P
1V
1T
2 = P
2V
2T
1
E. Gas Law Problems
A gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
BOYLE’S LAW
PV
(150.kPa)(100.mL)=(200.kPa)V
2
V
2= 75.0 mL

21. GIVEN:
V
1=7.84 cm
3
P
1=71.8 kPa
T
1=25°C = 298 K
V
2=?
P
2=101.325 kPa
T
2=273 K
WORK:
P
1V
1T
2= P
2V
2T
1
(71.8 kPa)(7.84 cm
3
)(273 K)
=(101.325 kPa)V
2 (298 K)
V
2= 5.09 cm
3
E. Gas Law Problems
A gas occupies 7.84 cm
3
at 71.8 kPa &
25°C. Find its volume at STP.
PTV
COMBINED GAS LAW

22. GIVEN:
P
1= 765 torr
T
1= 23°C = 296K
P
2= 560. torr
T
2= ?
WORK:
P
1V
1T
2 = P
2V
2T
1
E. Gas Law Problems
A gas’ pressure is 765 torr at 23°C.
At what temperature will the
pressure be 560. torr?
GAY-LUSSAC’S LAW
PT
(765 torr)T
2 = (560. torr)(296K)
T
2= 217 K = -56.3°C

23. k
n
V
 V
n
A. Avogadro’s PrincipleGas
Volume
(mL)
Mass
(g)
Moles, n
V/n
(L/mol)
O2 100.0 0.122 3.81  10
-3
26.2
N2 100.0 0.110 3.93  10
-3
25.5
CO2 100.0 0.176 4.00  10
-3
25.0

24. k
n
V
 V
n
A. Avogadro’s Principle
Equal volumes of gases contain
equal numbers of moles
•at constant temp & pressure
•true for any gas

25. PV
T
V
n
PV
nT
B. Ideal Gas Law
= k
UNIVERSAL GAS
CONSTANT
R=0.0821 Latm/molK
R=8.315 dm
3
kPa/molK
= R

26. B. Ideal Gas Law
UNIVERSAL GAS
CONSTANT
R=0.0821 Latm/molK
R=8.315 dm
3
kPa/molK
PV=nRT

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

28. GIVEN:
V=?
n=85 g
T=25°C = 298 K
P=104.5 kPa
R=8.315dm
3
kPa/molK
B. Ideal Gas Law
Find the volume of 85 g of O
2at 25°C
and 104.5 kPa.
= 2.7 mol
WORK:
85 g 1 mol = 2.7 mol
32.00 g
PV = nRT
(104.5)V=(2.7) (8.315) (298)
kPa mol dm
3
kPa/molKK
V = 64 dm
3
IDEAL GAS LAW

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

30. 1 mol
CaCO
3
100.09g
CaCO
3
B. Gas Stoichiometry Problem
What volume of CO
2forms from 5.25 g
of CaCO
3at 103 kPa & 25ºC?
5.25 g
CaCO
3
= 0.052 mol CO
2
CaCO
3CaO + CO
2
1 mol
CO
2
1 mol
CaCO
3
5.25 g ? L
non-STP
Looking for liters: Start with stoich
and calculate moles of CO
2.
Plug this into the Ideal
Gas Law to find liters.

31. WORK:
PV = nRT
(103 kPa)V
=(.052mol)(8.315dm
3
kPa/molK)(298K)
V = 1.26 dm
3
CO
2
B. Gas Stoichiometry Problem
What volume of CO
2forms from
5.25 g of CaCO
3at 103 kPa & 25ºC?
GIVEN:
P=103 kPa
V = ?
n=.052 mol
T=25°C = 298 K
R=8.315dm
3
kPa/molK

32. WORK:
PV = nRT
(97.3 kPa) (15.0 L)
= n (8.315dm
3
kPa/molK) (294K)
n = 0.597 mol O
2
B. Gas Stoichiometry Problem
How many grams of Al
2O
3are formed from
15.0 L of O
2at 97.3 kPa & 21°C?
GIVEN:
P=97.3 kPa
V = 15.0 L
n=?
T=21°C = 294 K
R=8.315dm
3
kPa/molK
4 Al + 3 O
22 Al
2O
3
15.0 L
non-STP ? g
Given liters: Start with
Ideal Gas Law and
calculate moles of O
2.
NEXT 

33. 2 mol
Al
2O
3
3 mol O
2
B. Gas Stoichiometry Problem
How many grams of Al
2O
3are formed
from 15.0 L of O
2at 97.3 kPa & 21°C?
0.597
mol O
2
= 40.6 g Al
2O
3
4 Al + 3 O
22 Al
2O
3
101.96 g
Al
2O
3
1 mol
Al
2O
3
15.0L
non-STP
? gUse stoich to convert moles
of O
2to grams Al
2O
3.

34. A. Dalton’s Law
The total pressure of a mixture
of gases equals the sum of the
partial pressures of the
individual gases.
P
total= P
1+ P
2+ ...
P
atm= P
H2+
P
H2O

35. GIVEN:
P
H2= ?
P
total= 94.4 kPa
P
H2O= 2.72 kPa
WORK:
P
total= P
H2 + P
H2O
94.4 kPa = P
H2+ 2.72 kPa
P
H2= 91.7 kPa
A. Dalton’s Law
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.
Look up water-vapor pressure
on p.859 for 22.5°C.
Sig Figs: Round to least number
of decimal places.
The total pressure in the collection bottle is equal to atmospheric
pressure and is a mixture of H
2and water vapor.

36. GIVEN:
P
gas= ?
P
total= 742.0 torr
P
H2O= 42.2 torr
WORK:
P
total= P
gas + P
H2O
742.0 torr = P
H2+ 42.2 torr
P
gas= 699.8 torr
A gas is collected over water at a temp of 35.0°C
when the barometric pressure is 742.0 torr.
What is the partial pressure of the dry gas?
DALTON’S LAW
Look up water-vapor pressure
on p.859 for 35.0°C.
Sig Figs: Round to least number
of decimal places.
A. Dalton’s Law
The total pressure in the collection bottle is equal to barometric
pressure and is a mixture of the “gas” and water vapor.

37. 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

38. B. Graham’s Law
KE = ½mv
2
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 because…

39. B. Graham’s Law
Graham’s Law
•Rate of diffusion of a gas is
inversely related to the square root
of its molar mass.A
B
B
A
m
m
v
v

Ratio of gas
A’s speed to
gas B’s speed

40. Determine the relative rate of diffusion
for krypton and bromine.1.381
Krdiffuses 1.381 times faster than Br
2.Kr
Br
Br
Kr
m
m
v
v
2
2
 A
B
B
A
m
m
v
v
 g/mol83.80
g/mol159.80

B. Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate mean find the ratio “v
A/v
B”.

41. 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?A
B
B
A
m
m
v
v
 2
2
2
2
H
O
O
H
m
m
v
v
 g/mol 2.02
g/mol32.00
m/s 12.3
v
H

2
B. Graham’s Law3.980
m/s 12.3
v
H

2 m/s49.0 v
H

2
Put the gas with
the unknown
speed as
“Gas A”.

42. An unknown gas diffuses 4.0 times faster than
O
2. Find its molar mass.Am
g/mol32.00
16 A
B
B
A
m
m
v
v
 A
O
O
A
m
m
v
v
2
2
 Am
g/mol32.00
4.0 16
g/mol32.00
m
A 2









A
m
g/mol32.00
4.0 g/mol2.0 
B. Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “v
A/v
B” is 4.0.
Square both
sides to get rid
of the square
root sign.