science 10

1. Gases
REVIEW
Properties
and
Measuring
Variables

2. Properties of Gases
Gases and Its Mass

3. Properties of Gases
Gases and Its Mass

4. Properties of Gases
Gases and Its Volume

5. Properties of Gases
Gases and Its Volume

6. Properties of Gases
Gases and Its Temperature
initial
temperature of
the air just
above the
water level.

7. Properties of Gases
Gases and Its Temperature
temperature of
the air just
above the
COLD water
level.

8. Properties of Gases
Gases and Its Temperature
temperature of
the air just
above the HOT
water level.

9. Properties of Gases
Gases and Its Pressure
inflated balloon
on the mouth of
the Erlenmeyer
flask
with hot water.

10. Properties of Gases
Gases and Its Pressure
deflated balloon
on the mouth of
the Erlenmeyer
flask on top of
alcohol lamp.

11. Properties of Gases
Based on experiments we
can summarized that Gases
have .
MASS
VOLUME
TEMPERATURE
PRESSURE

12. Properties of Gases
MASS- quantitative measure of
inertia, a fundamental property of
all matter.
UNIT: g or kg
MASS NOT EQUAL TO WEIGHT
W=mg
= mass x gravity

13. Properties of Gases
VOLUME- the amount of space
that a substance or object
occupies, or that is enclosed
within a container
UNIT: mL or L
1L=1 000mL

14. Properties of Gases
TEMPERATURE- is how hot or
cold a thing is
UNIT: C (degrees Celsius)
convert to K (Kelvin)

15. 5
9

F32
C K = ºC + 273
Always use absolute temperature
(Kelvin) when working with gases.
ºF
-459 32 212
ºC
-273 0 100
K
0 273 373

16. Properties of Gases
PRESSURE -is defined as force
per unit area.

17. Pressure
area
pressure 
force
Which shoes create the most pressure?

18. Pressure
Barometer
• measures atmospheric pressure
Mercury Barometer
Aneroid Barometer

19. Pressure
Manometer
• measures contained gas pressure
U-tube Manometer Bourdon-tube gauge

20. Pressure
m2
kPa 
N
KEY UNITS AT SEA LEVEL
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
14.7 psi

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

22. The LAWS O
GASES

23. A. Boyle’s Law
Activity 2
Boyle’s Law
Objective:
•Investigate the relationship
between volume and pressure of
gases at constant temperature.

24. A. Boyle’s Law
Activity 2
Boyle’s Law

25. A. Boyle’s Law
Activity 2
Boyle’s Law

26. A. Boyle’s Law
Activity 2
Boyle’s Law
Q1. What happens to the volume
of the syringe as the set of
weights is
added on top of it?
Q2. What happens to the pressure
on the syringe when the set of
weights
is added?

27. A. Boyle’s Law
Robert Boyle
during the 16th century
When pressure is
applied on the gas,
the molecules move
closer to one
another which result
in the decrease in
volume.

28. A. Boyle’s Law
P
V
PV = k

29. A. Boyle’s Law
States that the pressure
and volume of a gas are
inversely related at
constant temperature &
mass
P
V
PV = k

30. A. Boyle’s Law
Boyle’s law can be expressed
mathematically as P 1/V at
constant temperature
To remove the proportionality
sign, we introduce the constant k P
= k 1/V

31. A. Boyle’s Law
Therefore, k=PV If the same gas is brought
into different pressures, it will give two
different volumes, with the same value for k.
Then the equation will become P1V1=P2V2
Where: P1 = initial pressure P2 = final
pressure V1 = initial volume V2 = final volume
Any unit of pressure and volume may be
used. However, uniformity of units must be

32. A. Boyle’s Law

33. A. Boyle’s Law
Sample Problem:
Two hundred cubic centimeter of
gas (200 cm3) is contained in a
vessel under a pressure of 850 mm
Hg. What would be the new volume
of the gas if the pressure is changed
to 1000 mm Hg? Assume that the
temperature remains constant.

34. A. Boyle’s Law
Given: 𝑃1 = 850 mm Hg
𝑉1 = 200 cm3
𝑃2 = 1000 mm Hg
𝑉2 = x
Formula: P1V1 = P2V2 𝑉2 = 𝑃1𝑉1 / 𝑃2
Solution: 𝑉2 = (850 𝑚𝑚 𝐻𝑔) (200 𝑐𝑚
3) /1000 𝑚𝑚 𝐻𝑔
𝑽𝟐 = 170 𝒄𝒎3

35. A. Boyle’s Law
Activity 2. Let’s solve! Directions: Answer the following problem.
Write your answers on a separate sheet of paper.
1. The inflated balloon that slipped from the hand of Renn has
a volume of 0.50 L at sea level (1.0 atm) and it reached a
height of approximately 8 km where the atmospheric
pressure is approximately 0.33 atm. Assuming that the
temperature is constant, compute for the final volume of the
balloon.
2. Oxygen gas inside a 1.5 L-gas tank has a pressure of 0.95 atm.
Provided that the temperature remains constant, how much
pressure is needed to reduce its volume by ½ ?

36. T
V
 k
V
T
B. Charles’ Law

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

38. 1. A gas occupies 900.0 mL at a
temperature of 27.0 °C. What is the
volume at 132.0 °C?
2. 2. Calculate the decrease in temperature
when 2.00 L at 21.0 °C is compressed
to 1.00 L?

39. T
P
 k
P
T
C. Gay-Lussac’s Law

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

41. P
V
P
V
P
T
V = k
D. Combined Gas Law
T
=
P1V1 P2V2
T
1 2
P1
V1
T2=P2
V2
T1

42. T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
E. Gas Law Problems
A gas occupies 473 cm3 at 36°C.
Find its volume at 94°C.
CHARLES’ LAW
GIVEN: T V WORK:
V1 = 473 cm3
P1
V1
T2= P2
V2
T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3

43. V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
E. Gas Law Problems
A gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN: P V WORK:
P1
V1
T2= P2
V2
T1
(150.kPa)(100.mL)=(200.kPa)V2
2
V = 75.0 mL

44. GIVEN:
V1 = 7.84 cm3
P = 71.8 kPa
1
T1 = 25°C = 298 K
V2 = ?
P2 = 101.325 kPa
T2 = 273 K
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=(101.325 kPa) V2 (298 K)
V2 = 5.09 cm3
E. Gas Law Problems
P T V
A gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
COMBINED GAS LAW

45. P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
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
GIVEN: P T WORK:
P1
V1
T2= P2
V2
T1
(765 torr)T2 = (560. torr)(309K)
2
T = 226 K = -47°C

46. Ideal Gas Law
The mole

47. n
V
 k
V
n
A. Avogadro’s Principle

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

49. P
V
V
B. Ideal Gas Law
UNI VERSAL GAS
CONSTANT
R=0.0821 Lat m/ molK
R=8.315 dm3kPa/ molK
n
n
T
T = R
k

50. B. Ideal Gas Law
UNI VERSAL GAS
CONSTANT
R=0.0821 Lat m/ molK
R=8.315 dm3kPa/ molK
PV=nRT

51. 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

52. GIVEN:
P= 104.5 kPa
R = 8.315 dm3kPa/molK
B. Ideal Gas Law
V = ?
n = 85 g = 2.7 mol
T= 25°C = 298 K
WORK:
85 g 1 mol = 2.7 mol
32.00 g
PV = nRT
(104.5)V=(2.7) (8.315) (298)
kPa mol dm3kPa/molK K
V = 64 dm3
Find the volume of 85 g of O2 at 25°C
and 104.5 kPa.
IDEAL GAS LAW

53. Reactions
IV. Gas Stoichiometry
at Non-STP Conditions

54. 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.

55. 100.09
g
Problem
5.25 g of
5.25g
CaCO3
1mol 1mol
CaCO3 CO2
1mol
CaCO3
What volume of CO2 forms from
CaCO3 at 103 kPa & 25ºC?
CaCO3  CaO + CO2
5.25g ?L
non-STP
Looking for liters: Start with stoich
and calculate moles of CO2.
=1.26molCO2
Plug this into the Ideal
Gas Law to find liters.

56. WORK:
PV = nRT
(103kPa)V
=(1mol)(8.315dm3kPa/molK)(298K)
V = 1.26 dm3 CO2
Problem
5.25
What volume of CO2 forms from
g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
P = 103 kPa
V = ?
n = 1.26 mol
T = 25°C = 298 K
R = 8.315 dm3kPa/molK

57. PV = nRT
(97.3kPa) (15.0L)
=n(8.315dm3kPa/molK) (294K)
n = 0.597 mol O2
Problem
How many grams of Al2O3 are formed from
15.0 L of O2 at 97.3 kPa & 21°C?
GIVEN:
P = 97.3 kPa
V = 15.0 L
n = ?
T = 21°C = 294 K
R = 8.315 dm3kPa/molK
15.0L
non-STP
WORK:
4 Al + 3 O2  2 Al2O3
?g
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
NEXT 

58. 2mol
Al2
O3
3mol O2
Problem
How many grams of Al2O3are formed
from 15.0 L of O2 at 97.3 kPa & 21°C?
0.597
mol O2
=40.6gAl2
O3
101.96
g
Al2O3
1mol
Al2O3
15.0L
non-STP
4 Al + 3 O2  2 Al2O3
Use stoich to convert moles ?g
of O2 to grams Al2O3.

59. A. Dalton’s Law
The total pressure of a mixture
of gases equals the sum of the
partial pressures of the
individual gases.
Ptotal = P1 + P2 + ...
Patm = PH2 + PH2O

60. PH2 = ?
Ptotal = 94.4 kPa
GIVEN: WORK:
Ptotal = PH
2+ PH
2
O
94.4 kPa = PH2 + 2.72 kPa
H
2
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.
PH
2
O= 2.72 kPa
Look up water-vapor pressure
on p.899 for 22.5°C.
P = 91.7 kPa
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 H2 and water vapor.

61. Pgas = ?
Ptotal = 742.0 torr
Ptotal = Pg
a
s+ PH
2
O
742.0 torr = PH2 + 42.2 torr
g
a
s
DALTON’S LAW
PH
2
O= 42.2 torr
Look up water-vapor pressure
on p.899 for 35.0°C.
P = 699.8 torr
Sig Figs: Round to least number
of decimal places.
A. Dalton’s Law
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?
The total pressure in the collection bottle is equal to barometric
pressure and is a mixture of the “gas” and water vapor.
GIVEN: WORK:

62. 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

63. B. Graham’s Law
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…
KE = ½mv2

64. B. Graham’s Law
Graham’s Law
• Rate of diffusion of a gas is
inversely related to the square root
of its molar mass.
vA mB
vB mA

Ratio of gas
A’s speed to
gas B’s speed

65. Determine the relative rate of diffusion
for krypton and bromine.
Kr diffuses 1.381 times faster than Br2.
vKr
vBr
mBr
2
mKr
2


159.80 g/mol
 1.381
83.80 g/mol
vA mB
vB mA

B. Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate mean find the ratio “vA/vB”.

66. 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?
vB mA
vA mB

2
2
2
2
mH
mO
vO
vH

32.00 g/mol
2.02 g/mol
12.3 m/s

vH
2
B. Graham’s Law
12.3 m/s
 3.980
vH
2
vH  49.0 m/s
2
Put the gas with
the unknown
speed as
“Gas A”.

67. An unknown gas diffuses 4.0 times faster than
O2. Find its molar mass.
16 
32.00 g/mol
mA
vB mA
vA mB

mA
m
vA O2
vO
2

mA

32.00 g/mol
16
2





A
m
32..00 g//moll 
4..0 
 2.0 g/mol
B. Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “vA/vB” is 4.0.
Square both
sides to get rid
of the square
root sign. 