The document discusses gas laws and the unique properties of gases, emphasizing their low density, ability to diffuse, and behavior under varying pressure and temperature conditions as governed by fundamental laws such as Boyle's, Charles's, and Gay-Lussac's laws. It explains the kinetic molecular theory and provides practical applications and exercises related to these gas laws. The combined gas law is also introduced, alongside standard temperature and pressure (STP) for experimental measurements.

1. GAS LAWS
Rajeev G. Bayan

2. OVERVIEW
Gases are all around us. We live in a “sea of gases.”
Gases have characteristic behaviors that affect us
and the environment.
Therefore, it is important to understand some
fundamental laws that govern the behavior of
gases.
In this module, the following questions will be
answered:
1. What makes gases unique?
2. How do you explain the behavior of gases?
3. How do gas laws affect the daily routine of
people and environment?

3. General Behavior of Gases
Gases exhibit the following observable behaviors:
1. Gases have very low densities as compared to
liquids and solids;
2. Gases diffuse easily;
3. Gases have neither definite shape nor volume;
4. Gases exert pressure as measured in gauge
(tire pressure) or in millimeter (barometer);
and
5. Gases are easily compressible and expandable
as may be observed by twisting, bending and
reshaping.

4. Properties of Gases
The properties of gases can be explained by the kinetic molecular
theory (KMT) that observe the ff. postulates:
1. Gases are composed of molecules where size is negligible
compared to the average distance between them, resulting in
the absence of any attractive or repulsive forces;
2. Molecules move in randomly straight lines at different
speeds;
3. The force of attraction between gas molecules is very weak or
negligible;
4. Molecules collide with one another and unto the walls of the
container with no net energy loss. Collisions are elastic; and
5. The average velocity of a molecule is directly proportional to
the absolute temperature. When the temperature increases,
the energy possessed by the molecules increases. Hence
molecules move faster and their kinetic energy increases.

5. Particle Model of Gases
Properties of
Gases
KMT
Low density Molecules are far apart and occupy a large
space.
Diffuses easily Negligible forces between molecules;
random motion in all directions
No definite shape
and volume
Particles move randomly in all directions.
Exerts Pressure Molecules collide with the walls of the
container and towards other molecules.
Compressible Large distance between molecules, and
with negligible attractive and repulsive
forces between molecules, may be
squeezed/ compressed to become closer to
each other.

6. Gas Laws
Gas laws are examples of scientific laws or statements
that describe, summarize, and predict some aspects in
nature. The following are characteristics of scientific
laws:
 Are strongly supported by empirical evidences that re
repeatedly verified with no contradicting
observations;
 Can be mathematically formulated in one or several
statements or equations; and
 Are constant because they summarize whet have been
observed.
Some gas laws are that of Boyle’s, Charle’s, Gay –
Lussac’s, Graham’s, Avogadro’s and Dalton’s.

7. Boyle’s Law: Pressure and Volume
Robert Boyle (1627 – 1691) studied the effects of
pressure to the volume of gas in the sealed end of the
tube by adding mercury on the open end.
He noticed that the volume of a definite quantity of a dry
gas is inversely proportional to the pressure, provided
that the temperature remains constant.
P ⍺ 1/V
P = k/V where the
constant (k) is
temperature.
(P)(V) = (k/V)(V)
PV = k
P1V1 = P2V2

8. Boyle’s Law: Pressure and Volume
Boyle’s law explains the following phenomena:
1. The bubbles exhaled by the scuba diver increase in
size as he/she approaches the surface of the ocean.
The pressure exerted by the weight of water on the
bubble decreases as one goes up the surface. Hence,
the size of the bubbles increases.
2. Deep – sea fish die when brought to the surface. The
pressure decreases thereby causing the volume of
gases in their bodies to increase in size. This causes
the membranes of their bladders and cells to pop.
3. An inflated balloon bursts if squeezed too hard.
Squeezing reduces the volume of the balloon, thus,
increasing the pressure. If the balloon cannot
withstand the pressure, it bursts.

9. Boyle’s Law: Pressure and Volume
4. Syringe is used in drawing blood sample or giving
injection. If the plunger of the syringe is pulled back,
the volume of the syringe is increased and the
pressure inside is decreased. To balance the effect of
low pressure, air or blood is sucked in through the
needle. This equalizes the pressure inside and outside
the syringe.
5. Ears pop at high altitude. Most of the time, you feel
pain in your ear while in ascending or a descending
plane. In these situations, there is pressure imbalance
inside and outside your ear (increasing and decreasing
the volume of air inside your ear) that strains your
eardrum. To relieve the discomfort, swallow hard to
equalize the pressure on either side of the eardrum.

10. Practice Exercises
1. A sample of gas occupies a volume of 3.2 L at 1.8 atm.
What would be its volume if the pressure is: (a)
doubled? (b) reduced by one – half?
2. The atmospheric pressure at the peak of Mt. Everest
is 150 mmHg. If a climber carries 10 L tank with an
internal pressure of 3.04x104 mm Hg, what will be the
volume of the gas when it is released from the tank?
3. A 12 – L gas confined in a container exerts 3.8x104
mmHg. If the pressure is reduced to 760 mmHg, what
will be the volume of the gas?
4. A sample of neon gas occupies a volume of 2.8 L at 1.8
atm. What would be its volume at 1.2 atm?

11. Solutions at to Practice Exercises
1. A sample of gas occupies a volume of 3.2 L at 1.8 atm. What
would be its volume if the pressure is: (a) doubled? (b) reduced
by one – half?
Given:
P1 = 1.8 atm V1 = 3.2 L
(a) P2 = 2(1.8 atm) (b) P2 = (0.5)(1.8 atm)
Find: V2
Solution:
P1V1 = P2V2
(a) (1.8 atm)(3.2 L) = (3.6 atm) V2
5.76 atm•L = 3.6 atm • V2
5.76 atm ∙ L
3.6 atm
=
3.6atm ∙ V2
3.6 atm
1.6 L = V2
P1V1 = P2V2
b) (1.8 atm)(3.2 L) = (0.9 atm) V2
5.76 atm•L = 0.9 atm • V2
5.76 atm ∙ L
0.9 atm
=
0.9atm ∙ V2
0.9 atm
6.4 L = V2

12. 2. The atmospheric pressure at the peak of Mt.
Everest is 150 mmHg. If a climber carries 10 L
tank with an internal pressure of 3.04x104 mm Hg,
what will be the volume of the gas when it is
released from the tank?
Given:
P1 = 150 mmHg V1 = 10 L
P2 = 3.04 X 104mmHg
Find: V2
Solution:
P1V1 = P2V2
(150 mmHg)(10 L) = (3.04X104 mmHg) V2
1, 500 mmHg•L = 3.04X104mmHg• V2
1,500 mmHg ∙ L
3.04X104mmHg
=
3.04X104mmHg ∙ V2
3.04X104mmHg
0.049 L = V2
3. A 12 – L gas confined in a
container exerts 3.8x104 mmHg. If
the pressure is reduced to 760
mmHg, what will be the volume of
the gas?
Answer: 600 L
4. A sample of neon gas occupies a
volume of 2.8 L at 1.8 atm. What
would be its volume at 1.2 atm?
Answer: 4.2 L

13. Charles’s Law: Volume and Temperature
Jacques Alexandre Charles (1746 – 1823) duplicated the
experiment done by Joseph and Etienne Montgolfier on
June 4, 1783 in Annonay, France. He used fire to inflate a
spherical balloon about 30 feet in diameter that traveled
about 1.5 miles before coming back to Earth.
Charles found out that the volume of the gas is directly
proportional to its absolute temperature under constant
pressure.

14. Charles’s Law: Volume and Temperature
Application of Charles’s Law:
1. As one flies in a hot air balloon. The air inside the
balloon, when heated causes the air to expand; thus,
becomes lighter and so it rises.
2. An inflated balloon shrinks when placed inside the
refrigerator.
3. During cold weather, bicycle tires become flat
because the air inside the bicycle tires shrink.
4. Smoke is less dense than the surrounding air, hence,
it rises. Heat from the flame causes smoke to move
rapidly, increasing the volume it occupies, thus
lowering its density.

15. Practice Exercises
1. At constant pressure, find the final volume of
the following gas if the temperature is
changed to 0OC:
a. 4.50 L at 21OC
b. 100 L at 100OC
2. A balloon when cooled to 33OC has a volume
of 120 mL. What is the initial temperature in
OC of a 200 mL balloon?
3. If you inhale 2.0 L of air at a temperature of
20OC, what would be the volume of the gas if
it heats to 38OC in your lungs?

16. Solutions at to Practice Exercises
1. At constant pressure, find the final volume of the following gas if
the temperature is changed to 0OC:
a. 4.50 L at 21OC b. 100 L at 100OC
Given:
T2 = 0OC = 273 K
(a) T1 = 21OC =294 K V1 = 4.50 L
(b) T1 = 100OC = 373K V1 =100 L
Find: V2
Solution: V1T2 = V2T1
(a) (4.50L)(273K) = V2 (294K)
1, 228.5 K•L = 294 K• V2
1,228.5 K ∙ L
294 K
=
294K ∙ V2
294 K
4.179 L = V2
V1T2 = V2T1
b) (100L)(273K) = V2 (373K)
27, 300K•L = 373 K• V2
27,300 K ∙ L
373 K
=
373K ∙ V2
373 K
73.190 L = V2

17. Your Turn !
2. A balloon when cooled to 33OC has a volume
of 120 mL. What is the initial temperature in
OC of a 200 mL balloon?
Answer: T1= 237 OC
3. If you inhale 2.0 L of air at a temperature of
20OC, what would be the volume of the gas if
it heats to 38OC in your lungs?
Answer: V2 = 2.123 L

18. Gay-Lussac’s Law: Pressure and Temperature
When the temperature of a sample of gas in a rigid container
is increased, the pressure of the gas increases as well. The
increase in kinetic energy results in the molecules of gas
striking the walls of the container with more force, resulting in
a greater pressure. The French chemist Joseph Gay-Lussac
(1778 - 1850) discovered the relationship between the
pressure of a gas and its absolute temperature. Gay-Lussac's
Law states that the pressure of a given mass of gas varies
directly with the absolute temperature of the gas, when the
volume is kept constant. Gay-Lussac's Law is very similar to
Charles's Law, with the only difference being the type of
container. Whereas the container in a Charles's Law
experiment is flexible, it is rigid in a Gay-Lussac's Law
experiment.

19. Gay-Lussac’s Law: Pressure and Temperature
Gay – Lussac’s Law (Amontons’s Law in other books) explains the
following phenomena/ applications:
 There are more flat tires during summer. Temperature is high
during summer, causing an increased pressure inside the tires.
 Throwing an aerosol can into a fire can cause it to explode. High
temperature causes gas in the aerosol can to exert more
pressure.
 Pressure is adjusted in tires before the trip. The flexing of tires
while driving increases the temperature of the gas inside the
tire, thus increases the pressure.
 Dipping distended ping-pong balls in warm water. Hot water
raises the temperature of air inside the ping-pong ball thereby
causing the gas molecule to move faster, which in turn increases
pressure.

20. Practice Exercises
1. Gas exerts 1.0 atm pressure at 27OC. What is the
pressure if the temperature increased by 50OC?
2. A gas at 300 K has a pressure of 740 mmHg. If the
previous pressure is 760 mmHg, what was its original
temperature in OC?
3. A gas at 500OC has a pressure of 720 mmHg. What is
its pressure if its temperature is reduced in half?
4. A gas with 2.0 atm has a temperature of 250 K.
calculate the pressure if its temperature…
… increases to 300 K.
…decreases by 150 K.

21. Solutions to Practice Exercises
1. Gas exerts 1.0 atm pressure at 27OC. What
is the pressure if the temperature
increased by 50OC?
Given: P1=1 atm T1=27OC=300 K
T2=50OC=323 K
Find: P2
Solution: P1T2 = P2T1
(1 atm) (323 K) = P2 (300 K)
323 atm•K = 300K•P2
323 atm ∙ K
300K
=
300K ∙ P2
300K
1.077 atm = P2
2. A gas at 300 K has a pressure of
740 mmHg. If the previous
pressure is 760 mmHg, what was
its original temperature in OC?
Given: P2 =740 mmHg T2=300 K
P1=760 mmHg
Find: T1
Solution: P1T2 = P2T1
(760 mmHg) (300 K) = (740 mmHg) T1
228, 000 mmHg•K = 740 mmHg•T1
228,000 mmHg ∙ K
740 mmH𝑔
=
740 mmHg ∙ P2
740 mmHg
308.108 K= T1

22. Your Turn!!
3. A gas at 500OC has a pressure of 720 mmHg. What is
its pressure if its temperature is reduced in half?
Answer: P2= 487.141 mmHg
4. A gas with 2.0 atm has a temperature of 250 K.
calculate the pressure if its temperature…
… increases to 300 K.
…decreases by 150 K.
Answers: P2 = 2.4 atm
P2 = 1.2 atm

23. Combined Gas Law: Pressure,
Volume and Temperature
When combined the three gas laws can be expressed as:
PV/T = k
By comparing the same substance under two different sets of
conditions,
P1V1/T1=P2V2/T2
There is a standard set of conditions for experimental
measurement to enable computations between sets of data.
This is referred to as standard temperature and pressure (STP).
IUPAC established the standard temperature of 0OC and an
absolute pressure of 100kPa (or 0.987atm1 atm or 14.504psi
or 1 bar).

24. Practice Exercises
1. One hundred millimeters of a gas is measured
at 27OC and 700 mmHg. What is its volume at
STP?
2. A sample of O2 occupies a volume of 200 mL
at 25OC and 700 mmHg. If the size of a
container is reduced to 100 mL, what will be
the pressure exerted by the gas if its
temperature is increased to 90OC?
3. A 500 mL gas is at STP. What will be its
temperature in OC if the volume is increased
to 800mL and the pressure is reduced to
500mmHg?

25. Solution to Practice Exercises
1. One hundred milliliters of a gas is measured at 27OC and 700 mmHg.
What is its volume at STP?
Given: P1=700 mmHg T1=27OC=300K V1= 100 mL
P2=760 mmHg T2=273 K
Find: V2
Solution: P1V1T2=P2V2T1
(700 mmHg)(100mL)(273K)=(760mmHg) V2 (300 K)
19, 110, 000 mmHg•K•mL=228,000 mmHg•K•V2
19, 110, 000 mmHg•K•mL
228,000 mmHg•K
=
228,000 mmHg•K•V2
228,000 mmHg•K
83.816 mL = V2

26. Your Turn !!!
2. A sample of O2 occupies a volume of 200 mL at 25OC and 700
mmHg. If the size of a container is reduced to 100 mL, what will
be the pressure exerted by the gas if its temperature is
increased to 90OC?
Answer: P2= 1, 705.369 mmHg
3. A 500 mL gas is at STP. What will be its temperature in OC if the
volume is increased to 800mL and the pressure is reduced to
500mmHg?
Answer: 14.368OC

27. Avogadro’s Law: Volume and Amount
Amadeo Avogadro (1776 – 1856) proposed that equal
volumes of gases at the same temperature and pressure
contain the same number of molecules regardless of their
chemical nature and physical properties.
This is known as Avogadro’s law.
Mathematically, it is expressed as V n.
V/n = k
V1/n1 = V2/n2
One mole of any gas consists of
6.02X1023 molecules known
Avogadro’s number.
It is the number of molecules of any
gas present in a volume of 22.4L.

28. Practice Exercises
1. A 6.0 L sample at 25°C and 2.00 atm of pressure
contains 0.5 mole of a gas. If an additional 0.25 mole
of gas at the same pressure and temperature are
added, what is the final total volume of the gas?
2. If 1.0mol of helium gas (He) at standard temperature
and pressure (STP) has a volume of 22.4L, how many
moles of carbon tetrachloride gas (CCl4) will be
present in a container with an increase of 22.4L?
3. A 6.00 L sample at 25.0 °C and 2.00 atm contains
0.500 mol of gas. If we add 0.250 mol of gas at the
same pressure and temperature, what is the final
total volume of the gas?

29. Solution to Practice Exercises
1. A 6.0 L sample at 25°C and 2.00 atm of pressure
contains 0.5 mole of a gas. If an additional 0.25
mole of gas at the same pressure and
temperature are added, what is the final total
volume of the gas?
Given: V1= 6.0 L n1=0.5 mol
n2=0.5 mol + 0.25 mol= 0.75 mol
Find: V2
Solution: n1V2 = n2V1
(0.5 mol)V2=(0.75 mol)(6.0 L)
(0.5 mol)V2
0.5 mol
=
(0.75 mol)(6.0 L)
0.5 mol
V2=9 L
2. If 1.0mol of helium gas (He) at
standard temperature and pressure
(STP) has a volume of 22.4L, how
many moles of carbon tetrachloride
gas (CCl4) will be present in a
container with an increase of 22.4L?
Answer: 1 mol
3. A 6.00 L sample at 25.0 °C and 2.00
atm contains 0.500 mol of gas. If we
add 0.250 mol of gas at the same
pressure and temperature, what is
the final total volume of the gas?
Answer: 9 mol

30. Ideal Gas Law
Recall the following:
Boyle’s Law: P  1/V
Charles’s Law: V  T
Avogadro’s Law: V  n
These relationship can be combined in a
single relationship:
V  nT/P
V = R nT/P
Rearranging, PV = nRT
The value of R has a computed value
0.0821 L-atm/mol-K
An ideal gas is defined as one in which
all collisions between atoms and
molecules are perfectly elastic and
there are no IMF’s.
At normal conditions such as STP, most
real gases behave qualitatively like
ideal gases.
Ideal gas model fails at very low
temperature and very high pressure
when IMF’s and molecular sizes
become important.

31. Practice Exercises
1. A 500 mL bottle contains H2 at 600 mmHg at 20OC.
How many moles of hydrogen gas does the bottle
contain?
2. 6.2 liters of an ideal gas is contained at 3.0 atm and 37
°C. How many moles of this gas are present?
3. The balloon that Charles used for his initial flight in
1783 was destroyed, but we can estimate that its
volume was 31,150 L (1100 ft3), given the dimensions
recorded at the time. If the temperature at ground
level was 86°F (30°C) and the atmospheric pressure
was 745 mmHg, how many moles of hydrogen gas
were needed to fill the balloon?

32. Solution to Practice Exercises
1. A 500 mL bottle contains H2 at 600 mmHg at 20OC. How many moles of hydrogen gas does the bottle
contain?
Given: P=600mmHg=0.789atm V=500mL=0.5L
T=20OC=293K R=0.0821 L•atm/mol•K
Find: n
Solution: PV=nRT
0.789atm 0.5L = n(0.0821
L•atm
mol•K
)(293K)
0.395L ∙ atm = 24.055
L•atm
mol
∙ n
0.395L ∙ atm
24.055
L•atm
mol
=
24.055
L•atm
mol
∙ n
24.055
L•atm
mol
0.016 mol = n

33. Your Turn!!!!!
2. 6.2 liters of an ideal gas is contained at 3.0 atm and 37 °C. How many moles of this
gas are present?
Answer: 0.731 mol
3. The balloon that Charles used for his initial flight in 1783 was destroyed, but we
can estimate that its volume was 31,150 L (1100 ft3), given the dimensions
recorded at the time. If the temperature at ground level was 86°F (30°C) and the
atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were
needed to fill the balloon?
Answer: 1, 227.481 mol

34. Dalton’s Law of Partial Pressures
John Dalton (1766 – 1844) found that the pressure
exerted by the mixture of gases is equal to the sum of the
partial pressures of the gases present.
This is known as the Dalton’s law of partial pressures.
In equation, the pressure of a mixture can be
summarized as:
PTOTAL= P1 + P2 + P3 + … Pn

35. Dalton’s Law of Partial Pressures
In a laboratory, gases are obtained through water
displacement method.
To obtain the pressure of gas collected over water, the
partial pressure of water must be taken into
consideration.
As the gas bubbles, it picks up water vapor.
The amount of water vapor depends on the
temperature.
PTOTAL = Pgas + Pwater vapor
The pressure of the dry gas in the container is:
Pgas = PTOTAL – Pwater vapor
The pressure of water vapor is shown in the table on
the right.

36. Graham’s Law of Diffusion and Effusion
Diffusion is the gradual mixing of gases due to the motion of their component
particles even in the absence of mechanical agitation such as stirring.
The result is a gas mixture with uniform composition. Diffusion is also a property
of the particles in liquids and liquid solutions and, to a lesser extent, of solids and
solid solutions.
The related process, effusion, is the escape of gaseous molecules through a small
(usually microscopic) hole, such as a hole in a balloon, into an evacuated space.
The phenomenon of effusion had been known for thousands of years, but it was
not until the early 19th century that quantitative experiments related the rate of
effusion to molecular properties. The rate of effusion of a gaseous substance is
inversely proportional to the square root of its molar mass. This relationship is
referred to as Graham’s law, after the Scottish chemist Thomas Graham (1805–
1869). The ratio of the effusion rates of two gases is the square root of the inverse
ratio of their molar masses:
rate of effusion A
rate of effusion B
=
M𝐵
M𝐴

37. That is it!
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38. References: