gases law propared farah khadaad

1. Form three
Physics
Farah yasin( f.khadaad)

2. Kinetic Theory of Gases
 The kinetic theory of gases explains the the relationship between the pressure,
temperature and volume of gases base on the following assumptions:
 The gas consists of very small particles, each of which has a mass.
 These particles are in constant, random motion.
 The rapidly moving particles constantly collide with each other and with the walls of the
container. All these collisions are perfectly elastic.
 There are forces of attraction between particles of matter. These attraction forces will
increase as the distance between the particles becomes closer.
 The average kinetic energy of the gas particles depends only on the temperature of the
system. The higher the temperature, the higher the kinetic energy of the particles.
Pressure and volume relationship of a gas (Boyle's law)
Boyle's law
The relationship of a gas with pressure and volume was developed by the scientist Robert Boyle at
around 1660 and is known as Boyle’s Law.
Boyle’s law states:
"For a fixed mass of gas, at a constant temperature, the product (pressure x
volume) is a constant."
Pressure x Volume =constant
p x V = constant
A sealed cylinder with no leaks contains a fixed
mass of a gas kept at a constant temperature. The
gas pressure is created by the collision of the
moving gas particles with each other and against
the walls of the cylinder.
The relationship between pressure and volume

3. By plotting the recorded values of pressure (p) against volume (V) a curve is produced.
We can see from the values that when the pressure is doubled the volume is halved. If the
pressure was to increase by 3 the volume would decrease to a third.
Thus, the volume is inversely proportional to the pressure. By plotting pressure (p) against the
reciprocal of the volume (1/V) a straight line is obtained the gradient of which is the constant in
Boyle’s Law.
Example:
Using the example of the sealed cylinder above, the volume of gas at the start is 50 cm3with a
pressure of 1.2 x 105 Pascals. The piston is pushed slowly into the syringe until the pressure on
the gauge reads 2.0 x 105 Pascals. What is the volume of gas?

4. Solution:
We KNOW
p x V = constant
therefore,
p1 x V1 = p2 x V2
p1= 1.2 x 105 Pascals
V1 = 50 cm3
p2 = 2.0 x 105 Pascals
V2 = ?
p1 x V1 = p2 x V2
풗ퟐ =
풑
ퟏ풗ퟏ
풑ퟐ
풗ퟐ =
ퟏ ⋅ ퟐ × ퟏퟎퟓ × ퟓퟎ
ퟐ × ퟏퟎퟓ
풗ퟐ = 30푐푚3
Example2
Exercise
A sealed syringe contains 10 x10-6 m3 of air at 1 × 105 Pa. The plunger is pushed until the volume
of trapped air is 4 x 10-6 m3. If there is no change in temperature what is the new pressure of
the gas?
Solution
Therefore,
1. A balloon contains 6m3 of helium at
a pressure of 100kp, as the balloon
rises through the atmosphere the
pressure falls and the balloon
expends, assuming that the
atmosphere does not change, what
is the volume of the balloon when
the pressure has fallen to
a. 50kp
b. 40kp

5. The new pressure in the syringe is 2.5 × 105 Pa
Volume and temperature relationship of a gas (Charles' law)
Charles' law
The relationship between the volume and temperature of a gas was first put forward by the French
scientist Jacques-Alexandre-César Charles at around 1787 and is known as Charles’ Law.
Charles’ law states:
"For a fixed mass of gas, at a constant pressure, the volume (V) is directly
proportional to the absolute temperature (T)."
Volume 휶 Temperature 풗풐풍풖풎풆
푻풆풎풑풆풓풂풕풖풓풆
= 퐜퐨퐧퐬퐭퐚퐧퐭
A sealed cylinder with no leaks contains a fixed
mass. In order to keep the gas pressure constant
the piston is allowed to move freely so that the
internal pressure created by the gas particles can
equal the constant external pressure. If the
internal pressure increases the piston will move
up to allow the pressure to equalize.
The relationship between temperature and volume for a gas

6. By plotting the recorded values of volume (V) against temperature (T) a straight line is
produced. We can see from the values that the gas expands uniformly with temperature.
Converting the recorded temperatures into the Kelvin scale and plotting the volume (V) against the
absolute temperature (T) gives a straight line
This shows the volume of the gas is directly proportional to the absolute temperature of the gas.
Doubling the temperature will double the volume. The gradient of the slope is the constant in Charles’
Law.
Example:
Using the example of the sealed cylinder above, the volume of gas at the start is recorded as 30
cm3 with a temperature of 30°C. The cylinder is heated further till the thermometer records
60°C. What is the volume of gas?
Solution:
We know
V/T = constant
therefore
V1/T1 = V2/T2
V1 = 30 cm3
T1 = 30°C = 30+273 = 303K (remember to convert from Celsius to Kelvin)
T2 = 60°C = 60+273 = 333K
V2 =?
V1/T1 = V2/T2
풗ퟐ =
풗
ퟏ푻ퟐ
푻ퟏ
Exercise
1. A gas in fixed container is at a
pressure of 4 atm and the
temperature of 27oC what will be
its pressure if it is heated to a
temperature of 177oC ?

7. 풗ퟐ =
ퟑퟎ×ퟑퟑퟑ
ퟑퟎퟑ
= 32.97푐푚3
Pressure and temperature relationship of a gas (pressure law)
Imagine a gas is trapped inside a container, which has a fixed size (its volume cannot change).
If the gas is heated the particles will gain kinetic energy which will make them move faster. This
means they will collide with the walls more frequently.
This causes the force on the walls of the container to increase and so the pressure increases.
The pressure law states:
"For a fixed mass of gas, at a constant volume, the pressure (p) is directly
proportional to the absolute temperature (T)."
Pressure α Temperature
풑풓풆풔풔풖풓풆
푻풆풎풑풆풓풂풕풖풓풆
= 풄풐풏풔풕풂풏풕
A sealed cylinder with no leaks contains a fixed
mass. The volume of the gas is kept constant by
using a cylinder with a fixed roof capable of
withstanding high pressures. The gas pressure is
created by the collision of the moving gas particles
with each other and against the walls of the cylinder.

8. The relationship between temperature and pressure for a gas
Heat energy is applied to the cylinder and the temperature of the gas increases. The average
velocity of the gas particles increases resulting in an increase in the rate of collisions and the
average force per collision. Because the areas of the walls are kept constant, the force per unit
area increases resulting in an increase in pressure.
Plotting the pressure (p) against the absolute temperature (T) gives a straight line
This shows the pressure of the gas is directly proportional to the absolute temperature of the
gas. Doubling the temperature will double the volume. The gradient of the slope is the constant
in Charles’ Law.
Example:
Using the example of the sealed cylinder above the pressure of gas is recorded as 1.0 x
105N/m2 at a temperature of 0°C. The cylinder is heated further till the thermometer records
150°C. What is the pressure of the gas?
Solution:
We know
p/T = constant
therefore,
p1/V1 = p2/V2

9. p1 = 1.0 x 105 N/m2
T1 = 0°C = 0+273 = 273K (remember to convert from Celsius to Kelvin)
T2 = 150°C = 150+273 = 423K
p2 =?
p1/T1 = p2/T2
Therefore,
풑ퟐ = ퟏ.ퟎ×ퟏퟎퟓ×ퟒퟐퟑ
ퟐퟕퟑ
1.54× 105N/푚2
Exercise
1. An iron cylinder containing gas with pressure 200kPa when it is kept is a room of temperature
27°C. What is the pressure of the gas when the cylinder is located outdoor where the
temperature is 35°C.
2. A car tyre contains air at 1.25 × 105 Pa when at a temperature of 27°C. Once the car has been
running for a while the temperature of the air in the tyre rises to 42°C. If the volume of the tyre
does not change what is the new pressure of the air in the tyre?
The Gas Equation
The three gas laws give the following equations:
pV = constant (when T is kept constant) Boyle’s Law
V = constant (when p is kept constant)
T
Charles’ Law
P = constant (when V is kept constant)
T
Pressure Law
These 3 equations are combined to give the gas equation:
Any two of the three gas laws can be used to derive a general gas law proceed as follows
Suppose we have fixed mass of gas temperature T1 occupying a volume1 the pressure on the
gas is changed from p1 to p2 at constant temperature we can find the new volume V2 as

10. 풗ퟐ =
풑
ퟏ풗ퟏ
풑ퟐ
(equation1)
The pressure is kept constant at p2, the new volume V2 can be foun as
풗ퟐ =
풗
ퟏ푻ퟐ
푻ퟏ
(equation2)
From equation two , 풗ퟏ =
풗
ퟐ푻ퟏ
푻ퟐ
, substituting in equation1,we obtain
풗ퟐ =
풑
ퟏ푽ퟏ푻ퟐ
풑ퟐ 푻ퟏ
=
풑ퟏ풗ퟏ
푻ퟏ
= 풑ퟐ풗ퟐ
푻ퟐ
Or
풑ퟏ 풗ퟏ
푇1
= constant
Where,
p = the pressure of the gas
V = the volume the gas occupies
T = the gas temperature on the Kelvin scale
From this equation we know that if a fix mass of gas has starting values of p1, V1 and T1, and
then some time later has value p2, V2 and T2, the equation can be written as:
풑ퟏ풗ퟏ
푻ퟏ
= 풑ퟐ풗ퟐ
푻ퟐ
Example1:
Sabah pumps up her front bicycle tyre to 1.7 x 105 Pa. The volume of air in the tyre at this
pressure is 300 cm3. She takes her bike for a long ride during which the temperature of the air
in the tyre increases from 20°C to 30°C. Calculate the new front tyre pressure assuming the tyre
had no leaks and so the volume remained constant?
Solution:
p1 = 1.7 x 105 Pa
T1 = 20°C = 20 + 273 = 293 K
V1 = 300 cm3
p2 = ?
T2 = 30°C = 30 + 273 = 303 K
V2 = 300 cm3

11. Using:
풑ퟏ 풗ퟏ
푻ퟏ
=
풑ퟐ 풗ퟐ
푻ퟐ
Rearranging for P2
풑ퟐ =
풑ퟏ 풗ퟏ푻ퟐ
풗ퟐ 푻ퟏ
We know V1 = V2, therefore the equation can be simplified to
풑ퟐ =
푻ퟐ 풑ퟏ
푻ퟏ
풑ퟐ=
ퟏ.ퟕ×ퟏퟎퟓ×ퟑퟎퟑ
ퟐퟗퟑ
풑ퟐ = 1.76 × ퟏퟎퟓPa
Example2
A gas at 27°C expends from a volume of 5.0 m3 to 7.5 m3 at constant pressure, find its final
temperature?
Solution
풑ퟏ = 풑ퟐ
풗ퟏ = 5.0 m3 and 풗ퟐ = 7.5 m3
T1 = 27°C = 27 + 273 = 300 K
풑ퟏ풗ퟏ
푻ퟏ
= 풑ퟐ풗ퟐ
푻ퟐ
T2= 7.5m3×300 k
T2 = ퟕ.ퟓ×ퟑퟎퟎ 퐤
ퟓ
= 450k or 177°C
Ideal gas law
The gas laws tell us that for fixed amount of ideal gas
PV/T= constant
If we fix the volume and temperature of a container of gas and double the amount of gas in the
container, the pressure doubles because there twice as many particles to collide with the wall s
of the container.

12. Pressure is therefore proportional to the amount of gas, n, measured in moles (Mole)
Pressure α amount of gas
P α n
Since P 훼 1/V and P 훼 T then it follows that P = nT/V
Putting in a constant (R ) we get
P =
풏푹푻
푽
(this equation is called Ideal gas equation)
R is the universal gas constant, and has a value of (8.31jmol-1k-1 or(R= 0.08206L.atm/mol.K )
The number of moles of a substance can be found from its mass(m)in gram and its molar
mass(M) in g/mole
풏 = 풎
푴
Example 1
in an exprement 11.0g of Carbon dioxide( CO2) occupies a volume of 10.0dm3at a temperature
of 350.0K . find the pressure exerted by the gas?
Solution
The molar mass of CO2 = 44.0 g/mole
The number of moles of gas n = 11.0/44.0 =0.250 mol
T= 350 K; V = 0.010m3
P =
풏푹푻
푽
P =
ퟎ.ퟐퟓퟎ×ퟖ.ퟑퟏ×ퟑퟓퟎ
ퟎ.ퟎퟏퟎ
= 72.7kp
Example2
Find the number of moles of a gas occupying a volume of 2L at 25°C and the pressure of the gas
is 2atm.
Solution
Here we use the R value as 0.0821atmL/mol.K
And P is given as 2atm
Volume V is given = 2 L
Temperature of the gas is given in °C, which should be converted in to K.
So 25°C = 25+273 = 298K
Number of moles is to be calculated

13. n = PV/RT
n = 0.163 mol
Example3
If there is 5.0g of CO2 gas in a 10 L cylinder at 25oC, what is the gas pressure within the cylinder?
Solution
We are given the quantity of CO2 in grams but to use the ideal gas law, we must express the
quantity in moles. Therefore we must first convert grams of CO2 to moles, and then use this
value in the ideal gas law.
To convert from grams to moles we use the conversion factor 1 mol CO2 = 44g.
We now use this value in the ideal gas equation to solve for the pressure of the gas.
P = nRTV
P = [0.11 mol CO2][298K]10L × 0.0821L.atmmol.K
P = 0.27atm
Exercise
1. How much gram of nitrogen gas will exert a pressure of 250mm of Hg in a 2 liter
container at 273K (0.84g)
2. One mole of CH4 gas occupies 20.0L at 1.00atm pressure. What is the temperature of
the gas in kelvin? (244K)