This pdf deals with Boyle's Law , Charles' Law and Kinetic Molecular theory of gases

1. MARYLE D. MEJOS
LUGAIT NHS

2. Boyle’s Law
What is Boyle’s Law?
Boyle’s law was put forward by the Anglo-Irish chemist Robert
Boyle in the year 1662.
When a gas is under pressure it takes up less space
The higher the pressure the smaller the volume
Boyle’s law is a gas law which states that the pressure exerted
by a gas (of a given mass, kept at a constant temperature) is
inversely proportional to the volume occupied by it.
, the pressure and volume of a gas are inversely proportional to
each other as long as the temperature and the quantity of gas
are kept constant.

3. Boyle’s Law
https://www.youtube.com/watch?v=4R-gjxPUj_M
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4. Boyle’s Law
For a gas, the relationship between volume and pressure (at constant
mass and temperature) can be expressed mathematically as follows.
P ∝ (1/V)
Where P is the pressure exerted by the gas and V is the
volume occupied by it. This proportionality can be
converted into an equation by adding a constant, k.
P = k*(1/V) ⇒ PV = k

5. Boyle’s Law
This law can be expressed mathematically as follows:
P1V1 = P2V2
Where,
•P1 is the initial pressure exerted by the gas
•V1 is the initial volume occupied by the gas
•P2 is the final pressure exerted by the gas
•V2 is the final volume occupied by the gas

6. Boyle’s Law
Units of Pressure
•atm
•torr
•mmHg
•N/m2
Units of Volume
•L (liter)
•mL
•cm3
•dm3

7. Boyle’s Law
Boyle’s Law apparatus in breathing……
Boyle’s Law Application

8. Boyle’s Law
The pressure v/s volume curve for a fixed amount of gas kept at
constant temperature is illustrated below.
It can be observed that a straight line is obtained when the pressure exerted by the
gas (P) is taken on the Y-axis and the inverse of the volume occupied by the gas (1/V)
is taken on the X-axis.

9. Boyle’s Law
Problem:
A deep sea diver is working at
a depth where pressure is 3.0
atmospheres. He is breathing
out air bubbles is 2 cm3 . At
the surface the pressure is 1
atmosphere. What is the
volume of each bubble when
it reaches the surface?

10. Boyle’s Law
How we work this out?
1. Write the given and the unknown.
P1=3 atm
V1=2 cm3
V2= 1 cm3
P2=?
2. The formula needed to solve the
problem
P1V1=P2V2
P2= P1V1
V2
3. Substitute the given value and do
the operation
P2=(3 atm)(2 cm3)
1 cm3
P2= 6 atm
Thus, P2 increases when the volume
V2 is reduced by half.

11. Charles Law
French Chemist Jacques Charles discovered
that the volume of a gas at constant pressure
changes with temperature
As the temperature of the gas increases, so
does its volume, and as its temperature
decreases, so does its volume

12. Charles Law
The law says that at constant pressure the volume of a fixed number of
particles of gas is directly proportional to the absolute (kelvin) temperature,
mathematically expressed as:
V=kT
Where
V= volume
k=Charles’ Law constant of Proportionality
T= Temperature in Kelvin

13. Charles Law
Explanation
 Raising the tempaerature of a
gas causes the gas to fill a
greater volume as long as
pressure remains constant.
 Gases expand at a constant
rate as temperature increases,
and the rate of expansion is
similar for all gases

14. Charles Law
This law describes how a gas expands as the temperature increases;
conversely, a decrease in temperature will lead to a decrease in volume. For
comparing the same substance under two different sets of conditions, the
law can be written as:
V1 V2
T1 T2
= V1 T2 =V2T1or
Where
V1- initial volume
T1- initaial temperature in Kelvin scale
V2-final volume
T2-final temparature
Note:
to convert 0C to 0K
0K = 0C t 273

15. Charles Law
Demonstration
An animation demonstrating the relationship
between volume and temperature

16. Charles Law
Demonstration

17. Charles Law
https://www.youtube.com/watch?v=7ZpuMBkf1Ss
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18. Charles Law
Practical Applications

19. Charles Law
Sample Problem
A 600.0 mL sample of nitrogen is warmed from 77.0 °C to
86.0 °C. Find its new volume if the pressure remains
constant.
Solution:
Given
V1= 600.0 mL
T1= 77.00C + 273= 350.00K
T2= 86.00C + 273= 359.00K
V2=?
Formula V2= V1T2/T1
V2= (600.0 mL)(359.00K)/350.00K
V2= 615.4mL
Thus, the volume increase to 615.43 as the
temperature was increase from 77.0
o
C to 86.0
o
C

20. Kinetic Molecular theory of gases

21. Kinetic Molecular theory of gases

22. Kinetic Molecular theory of gases

23. Kinetic Molecular theory of gases

24. Kinetic Molecular theory of gases

25. Kinetic Molecular theory of gases

26. Kinetic Molecular theory of gases

27. Kinetic Molecular theory of gases

28. Kinetic Molecular theory of gases
https://www.youtube.com/watch?v=B0uxOH4Bm2M
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29. Kinetic Molecular theory of gases

31. This law can be expressed mathematically as follows: