PCO2 Ptotal = PN2 + PO2 + PCO2 2.50 atm = 0.90 atm + 0.60 atm + PCO2 PCO2 = 2.50 atm - 0.90 atm - 0.60 atm PCO2 = 1 atm So the partial pressure of CO2 is 1 atm.

1. Combined Gas
Law

2. P = k
T
V = k
PV = k
Combined Gas
Law
● Combination of Boyle’s Law, Charles’s Law
and Gay-Lussac’s Law.
● Gas Law in which pressure, temperature
and volume of gas changes while the
number of particles is constant.
Boyle’s Law
Charles’s Law
Gay-Lussac’s Law
T
PV = k
T
T
P = k
T
V = k
PV = k
T

3. PV = k
T
V = k P = k
T
PV = k
T
Boyle’s Law Charles’s Law Gay-Lussac’s Law

4. Combined Gas
Law Formula
P1 V1
T1
P2 V2
T2
=

5. ● Also known as standard conditions.
● Common reference points of
temperature and pressure when
comparing volume of gases.
● The standard temperature is 273 K
(0oC or 32oF) and the standard
pressure is 1 atm (760 torr, or 760
mm Hg).
Standard Temperature
and Pressure (STP)
Standard Pressures Standard
Temperatures
1 atm 273 K
760 mm Hg
760 torr 0oC
101.326 kPa
101326 Pa 32oF
14.5 psi
STP Values

6. P2
T1 V2
T1 T2
T1 V2
P2 V2
=
P1 V1 P2
A sample of gas occupies a volume of 2.40 L
at 685 mm Hg at 130oC. What would be the pressure
(in mm Hg) if the volume and temperature were
increased to 5 L and 210oC?
Sample Problem 15.5
V1 =
P1 =
T1 =
V2 =
T2 =
130oC
685 mm Hg
2.40 L
+ 273 = 403 K
5 L 210oC
+ 273 = 483 K
P2
P1 V1
T1
P2 V2
T2
=
P2 =
(483 K) (685 mm Hg) (2.40 L)
(403 K) (5 L)
=
794052 K.mm Hg.L
2015 K.L
= 394.07
mm Hg

7. In a flexible-walled container, 300 L of gas is
prepared at 600 mm Hg and 200oC. The gas is
placed in a tank under high pressure. When the tank
cools to 20oC, the pressure of the gas is 30 atm.
What is the new volume of the gas?
Practice Exercise
15.5
V1 =
P1 =
T1 =
T2 =
P2 =
+ 273 = 473 K
+ 273 = 293 K
V2
300 L
600 mm Hg 200oC
20oC 30 atm
= 0.79 atm
P1 V1T2
P2 T1
V2
=
V2 =
(0.79 atm) (300 L) (293 K)
(30 atm) (473 K)
=
69441 atm.L.K
14190 atm.K
V2 = 4.89 L

8. When measured at STP, a quantity of gas
has a volume of 500 L. What volume will it occupy
at 0oC and 93.3 kPa?
Practice Exercise
V1 =
P1 =
T1 =
T2 =
P2 =
273 K
0oC+ 273 = 273 K
V2
500 L
1 atm
P1 V1T2
P2 T1
V2
=
V2 =
(1 atm) (500 L) (273 K)
(0.92 atm) (273 K)
=
500 atm.L
0.92 atm
V2 = 543.48 L
93.3 kPa = 0.92 atm

9. 1. What is the volume of gas at 2.00 atm
and 200.0 K if its original volume was
300.0 L at 0.250 atm and 400.0 K.
2. A volume of gas collected at 70oC and
1.04 atm occupies 125 ml. What volume
(ml) would it occupy at STP?
Quiz:
Instruction:
Answer the following problems in any
kind of paper. Put your final answers on
the Aralinks and submit the picture of
your solutions in the submission named
“Combined Gas Law Solution” until 1 pm.
1. The volume of the gas that filled the
balloon is 30 L at 40oC and 1148 mm
Hg pressure. What volume will the
balloon have at STP?
Assignment:

11. Avogadro’s Law

12. ● Named after Italian physicist Amedeo
Avogadro.
● States that at constant temperature and
pressure, (V) volume of a gas is directly
proportional to the number of moles
(particles)(n) of the gas present.
Avogadro’s
Law
Amedeo Avogadro
1776-1856
V ∝ n
o
r
V = k
n

13. Avogadro’s
Law

14. Examples:

15. Avogadro’s Law
Formula
V1
n1
V2
n2
=

16. According to Avogadro’s Law, at STP, one
mole of any gas will occupy the same volume
which is 22.4 L. This volume is known as
molar volume of a gas.

17. A 2 L sample at 20oC and 750 torr
contains 0.5 moles of a gas. If 0.1 mole of the
gas at the same temperature and pressure is
added to the sample, what is the new volume
of the gas?
Sample Problem
16.7
V1 =
n1 =
n2 =
V2
2 L
0.5 mol
0.5 mol + 0.1 mol = 0.6 mol
V1 V2
n1 n2
=
V1 V2
n1 n2
= n2
n2
V2 =
(0.6 mol) (2 L)
(0.5 mol)
V2 =
1.2 mol.L
0.5 mol
V2 = 2.4 L

18. If 0.8 g of O2 occupies a volume of 500
ml at STP, what volume will 1.5 g of O2
occupy under the same condition?
Sample Problem
16.7
V1 =
n1 =
n2 =
V2
500 L
0.8 g =
1.5 g =
V1 V2
n1 n2
=
V2 =
(0.047 mol) (500 L)
(0.025 mol)
V2 =
23.5 mol.L
0.025 mol
V2 = 900 L
Conversion of Mass to Mole
0.8 g x 1 mol of O2
n1:
32 g
= 0.025 mol
n2: 1.5 g x 1 mol of O2
32 g
= 0.047 mol
16g x 2 = 32g/mol of O2
grams/1 mol of O2 =

19. 1. If a 10 L balloon contains 0.80 mol of
He, what will be the volume of a balloon
that contains 0.20 mol of He at the same
temperature and pressure.
2. If a balloon initially contains 0.25 mol
of an ideal gas and occupies 5.0 L, how
much gas (in mole) must be placed
inside the balloon to increase its volume
into 6 L?
Assignment:
Answer the following problems in
science journal and submit the
picture of your solutions to the
Aralinks.

20. Ideal Gas Law

22. ● Described by Kinetic
molecular theory as gas with
molecules of negligible size
that exert doesn’t exert
intermolecular forces.
● Gases that doesn’t bond
with other gases within a
given space.
Ideal Gas

23. PV = nRT
V ∝
The relationship of gas laws can be combined and
manipulated to form a more general description and
equation that relates volume, pressure, temperature, and
amount of gas.
The Three Key Relationships
● Boyle’s Law: V ∝
𝟏
𝑷
● Charles’s Law: V ∝ T
● Avogadro’s Law: V ∝ n
T
V ∝ 1
P
V ∝ n
Avogadro’s Law
Boyle’s Law
Charles’s Law
= R
“R” is the proportionality constant
known as the ideal gas constant.
PV = nRT

24. PV = nRT

25. States that the volume of a gas varies
directly with the number of its
molecules and the temperature.
Volume also varies inversely with
pressure.
Ideal Gas Law

26. (1 mol)
T
n
● depends on the unit of P, V, and T.
● common value derived by using the ideal gas equation for 1
mol of a gas at STP which is R = 0.0821.𝐋.𝐚𝐭𝐦
𝐦𝐨𝐥.𝐊
.
R
PV =
Ideal Gas Constant
“R” value
T
n T
n
PV
T
n
=
R
=
R
(1 atm)(22.4 L)
(273 K)
=
22.4 L.atm
273 mol.K
=
R 0.0821
𝐋.𝐚𝐭𝐦
𝐦𝐨𝐥.𝐊

27. ● Ideal gas equation also be used in
getting the molar mass of a mixture
of gases.
● Molar mass is expressed as a mass
per mole, then ”n” can be
substituted in the ideal gas equation.
n =
𝑚
𝑀𝑀
PV =
𝐦
𝐌𝐌
RT or MM =
m𝑅𝑇
PV
● Furthermore since the density (p) of
a gas is p =
𝑚
𝑉
, the equation becomes
MM =
𝑝𝑅𝑇
P
p =
𝑀𝑀 (𝑃)
RT

28. P =
(301 K)
50 L
A 50 L cylinder contains 21.8 g H2
at 28oC. What is the pressure (in atm)
exerted by this gas?
Sample Problems:
Pressure
V = 50 L
m = 21.8 g
T = 28oC+273= 301K
MM H2
= 1g x 2 = 2g/mol
P =
nRT
V
n =
𝑚
𝑀𝑀
=
21.8𝑔
2𝑔/𝑚𝑜𝑙
= 10.9 𝑚𝑜𝑙
(10.9 mol) (0.0821.𝐋.𝐚𝐭𝐦
𝐦𝐨𝐥.𝐊
)
P =
50 L
269.36 L.atm
P = 5.39 atm

29. MM =
(293 K)
(2.13 L)
Calculate the molar mass of
butane gas if 4.96 g occupies 2.13 L at
20oC and 1 atm.
Sample Problems:
Molar mass
MM =
𝐦𝑹𝑻
𝑷𝑽
V = 2.13 L
m = 4.96 g
T = 20oC+273= 293 K
P = 1 atm
(4.96 g) (0.0821.𝐋.𝐚𝐭𝐦
𝐦𝐨𝐥.𝐊
)
MM =
2.13 L.atm
119.31 𝐠.𝐋.𝐚𝐭𝐦
𝐦𝐨𝐥
P = 56.01 g/mol
(1 atm)

30. V =
(273 K)
1. Determine the volume occupied by
4.26 g of CO2 gas at STP.
Practice exercises
Volume
V =
𝒏𝑹𝑻
𝑷
m = 4.26 g
T = 273K
P = 1 atm
MM CO2
=
n =
(0.097 mol)(0.0821𝐋.𝐚𝐭𝐦
𝐦𝐨𝐥.𝐊
)
V =
1 atm
2.17 L.atm
V = 2.17 L
(1 atm)
Molar mass of CO2
C= 12g x 1 = 12 g
O= 16g x 2 = 32g
MM CO2
= 44g/mol of CO2
n =
𝑚
𝑀𝑀
=
4.26 𝑔
44𝑔/𝑚𝑜𝑙
= 𝟎. 𝟎𝟗𝟕 𝒎𝒐𝒍
0.097 mol

31. p =
(325 K)
What is the density of laughing
gas, dinitrogen monoxide, N2O, at a
temperature of 325 K and a pressure of
1.12 atm?
Sample Problems:
density (p)
p =
𝑀𝑀 (𝑃)
RT
MM N2O = 4.96 g/mol
T = 325 K
P = 1.12 atm
(4.96 g/mol)
(0.0821 𝐋.𝐚𝐭𝐦
𝐦𝐨𝐥.𝐊
)
p =
26.68
𝐋.𝐚𝐭𝐦
𝐦𝐨𝐥
5.56 𝐠
𝐦𝐨𝐥
.atm
p = 0.21 g/L
(1.12 atm)

32. 1. At what temperature will 0.731 moles
of Ne gas occupy 10.30 L at 2.50
atm?
2. A portable breathing tank has a
volume of 1.85 L and contains
primarily O2 gas. What is the mass of
the oxygen inside the tank if the
pressure inside is 0.98 atm at 20oC?
Practice exercises Assignment
1. Determine the volume occupied by
0.053 mol of CO2 at STP.
2. A cylinder of O2(g) used in breathing
by emphysema patients has a
volume of 3.00 L at a pressure of
10.0 atm. If the temperature of the
cylinder is 28.0 °C, what mass of
oxygen is in the cylinder?
3. What is the density of Argon gas if
the temperature is 30oC and the
pressure is 1.2 atm?

34. ● Discovered by John Dalton in 1801
● States that the total pressure exerted by a mixture of
gases is equal to the sum of the pressures that the
individual gases would exert alone.
● Partial pressure refers to the pressure exerted by a gas
individually.
Dalton’s Law of
Partial Pressure
Partial
Pressures
Total Pressures

35. Ptotal = P1 + P2 + P3
…….
P1 = pressure exerted by Gas 1
P2 = pressure exerted by Gas 2
P3 = Pressure exerted by Gas 3

36. If an alien planet has an atmosphere
of mixed gas composed of O2 (P=3 atm), Ar
(P= 0.1 atm), N2 (P= 0.8 atm), and
Ne(P=0.01 atm), what is the total pressure
of its atmosphere?
Practice Exercise
Ptotal
Ptotal = PO2
+ PAr + PN2
+ PNe
PO2
= 3 atm
PAr = 0.1 atm
PN2
= 0.8 atm
PNe = 0.01 atm
Ptotal = + + +
3 atm 0.01 atm
0.8 atm
0.1 atm
Ptotal = 3.91 atm

37. An equilibrium mixture contains H2 at
470 torr, N2 at 190 torr, and O2 at 286 torr.
What is the total pressure of the gas in the
system?
Practice Exercise
Ptotal
Ptotal = PH2
+ PN2
+ PO2
PH2
= 470 torr
PN2
= 190 torr
PO2
= 286 torr
Ptotal = + +
470 torr 286 torr
190 torr
Ptotal = 946 torr

38. 2.50 atm =
0.60 atm
0.90 atm
The total pressure inside the container
of N2, O2 and CO2 is 2.50 atm. What is the
atrial pressure of CO2 if the partial pressure
of N2 gas and O2 gas are 0.90 atm and 0.60
atm respectively?
Practice Exercise
PCO2
Ptotal = PH2
+ PN2
+ PO2
Ptotal = 2.50 atm
PN2
= 0.90 atm
PO2
= 0.60 atm
2.50 atm = + +
0.90 atm 0.60 atm
PCO2
= 1 atm
PCO2
PCO2

39. ● The partial pressure of a gas can be calculated by
multiplying its percent composition (in decimal form) to the
total pressure.
Px=(% composition)(Ptotal)
Example
Since 21% of the molecules of the air
is oxygen, it means that 21% of the volume
and pressure of the atmosphere can be
attributed to oxygen. Solve for the following;
a.) The partial pressure of O2 at 760 torr total
atmospheric pressure. (unit should be in
torr)
PO2
=(.21)(760 torr)= 160 torr
b.) The partial pressure of O2 at 1,140 torr
total atmospheric pressure. (unit should be
in torr)
PO2
=(.21)(1,140 torr)= 239.4 torr

40. 1) A gaseous mixture made from 6.00 g O2
and 9.00 g CH4 is placed in a 15.0 L vessel
at 0oC. What is the partial pressure of
each gas, and what is the total pressure in
the vessel?
2) The percent composition of N2 in air is
78%. What is the partial pressure of
Nitrogen gas if the reading from a nearby
barometer is 745 torr? (unit is torr)
Activity