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Theory: Avogadro′s Principle
Imagine you have 3 identical balloons all at the same temperature and pressure (say 25°C and 100 kPa pressure).
Balloon 1 contains 1 L of helium gas.
Balloon 2 contains 1 L of nitrogen gas.
Balloon 3 contains 1 L of argon gas.
If you could count each individual gas particle in each balloon you would find the number of gas particles in each of the balloons is the same:
Balloon 1 contains 2.4 × 10^{22} particles of helium gas.
Balloon 2 contains 2.4 × 10^{22} particles of nitrogen gas.
Balloon 3 contains 2.4 × 10^{22} particles of of argon gas.
That is, 1 L of gas at 25°C and 100 kPa pressure contains 2.4 × 10^{22} particles of gas.
When put in general terms, this becomes Avogadro's Principle (or Avogadro's Hypothesis):
Equal volumes of gases at the same temperature and pressure contain the same number of gas molecules.
Avogadro's Principle (Avogadro's Hypothesis) is extremely useful when we need to know the amount or volume of gas at constant temperature and pressure.
Imagine you are blowing up a balloon using helium gas.
The temperature of the gas in the balloon will be constant, that is it will be the same as the temperature in the room.
The gas pressure inside in the balloon will also be constant, since it will also be the same as the pressure inside the room.
As you add more helium gas particles to the balloon, the volume of the balloon expands.
Imagine we could add a set number of helium gas particles to a very large balloon and then measure its volume.
The results of the experiment are shown in the table below:
Constant Temperature and Pressure 
Number of He_{(g)} atoms 
Volume of Balloon (L) 
6.02 × 10^{23} 
24.79 
1.20 × 10^{24} 
49.58 
1.81 × 10^{24} 
74.37 
2.41 × 10^{24} 
99.16 
We can see that as we increase the number of gas particles from 6.02 × 10^{23} to 2.41 × 10^{24} the volume of the balloon increases from 24.79 L to 99.17 L.
In order to describe the relationship between the number of gas particles and the volume of the balloon, we could take the ratio N(He_{(g)}):V(He_{(g)}), or, put another way, we could divide the volume of gas by the number of the gas particles, that is V(He_{(g)}) ÷ N(He_{(g)})
The results of these calculations has been added to the table below:
Constant Temperature and Pressure 
Number of He_{(g)} atoms 
Volume of Balloon (L) 
Volume ÷ Number 
6.02 × 10^{23} 
24.79 
24.79 ÷ (6.02 × 10^{23}) = 4.1 × 10^{23} 
1.20 × 10^{24} 
49.58 
49.58 ÷ (1.20 × 10^{24}) = 4.1 × 10^{23} 
1.81 × 10^{24} 
74.37 
74.37 ÷ (1.81 × 10^{24}) = 4.1 × 10^{23} 
2.41 × 10^{24} 
99.16 
99.16 ÷ (2.41 × 10^{24}) = 4.1 × 10^{23} 
For this experiment, Volume of gas ÷ Number of gas particles = 4.1 × 10^{23}
If we performed exactly the same experiment, but at a different temperature, V ÷ N would still be constant, but not the same constant.
We can express this mathematically by saying that, at constant temperature and pressure, the volume of a gas (V) is directly proportional to the number of gas particles (N):
V ∝ N
We can use a constant of proportionality to turn this into an equation:
V = "a constant" × N
or
V ÷ N = "a constant"
Because the number of gas particles in a measurable volume of gas tends to be very, very large, Chemists prefer to work with "moles" of gases rather than number of gas particles.
Recall that 1 mole of a substance contains 6.02 × 10^{23} particles of that substance.
So, n moles of gas contains n × 6.02 × 10^{23} gas particles.
Therefore, the number of gas particles (N) divided by 6.02 × 10^{23} will be equal to the moles of gas (N(gas)):
n(gas) = N ÷ 6.02 × 10^{23}
Let′s calculate the moles of gas present in each stage of our helium balloon experiment above:
Constant Temperature and Pressure 
Number of He_{(g)} atoms 
Moles of He_{(g)} atoms 
Volume of Balloon (L) 
6.02 × 10^{23} 
6.02 × 10^{23} ÷ 6.02 × 10^{23} = 1 mole 
24.79 
1.20 × 10^{24} 
1.20 × 10^{24} ÷ 6.02 × 10^{23} = 2 mole 
49.58 
1.81 × 10^{24} 
1.81 × 10^{24} ÷ 6.02 × 10^{23} = 3 mole 
74.37 
2.41 × 10^{24} 
2.41 × 10^{24} ÷ 6.02 × 10^{23} = 4 mole 
99.16 
And now we will divide the volume of gas by the moles of gas in order to establish if there is still a relationship between these:
Constant Temperature and Pressure 
Moles of He_{(g)} atoms (mol) 
Volume of Balloon (L) 
Volume ÷ Moles 
1 
24.79 
24.79 ÷ 1 = 24.79 
2 
49.58 
49.58 ÷ 2 = 24.79 
3 
74.37 
74.37 ÷ 3 = 24.79 
4 
99.16 
99.16 ÷ 4 = 24.79 
Once again we see that gas volume (V) ÷ moles of gas (n) = "a constant"
At 25°C and 100 kPa:
V ÷ n = 24.79
or
V = 24.79 × n
You might also recall that the equation for a straight line graph is y = mx + b
If b (the yintercept) = 0
The equation of the straight line becomes y = mx
So, if y = Volume of gas (V)
and x = moles of gas (n)
the equation becomes V = mn
and a graph of V against n should be straight line with slope (or gradient) = m
the results of our helium balloon experiment are graphed below:
volume (L) 
Relationship between Amount and Volume of Gas
moles of gas (mol)

We can calculate the slope of the line between the points (0,0) and (4.0,99.16)
(99.16  0) ÷ (4.0  0) = 24.79
So the equation for this line is V = 24.79 × n
The graph clearly shows the relationship between volume of gas and the number of gas particles:
 If you increase the moles of gas in a container while maintaining a constant temperature and pressure, the volume the gas occupies will increase.
 If you decrease the moles of gas in a container while maintaining a constant temperature and pressure, the volume the gas occupies will decrease.
This relationship allows to calculate volume (or moles) of any gas at any constant temperature and pressure as long as we can establish the slope (gradient) of the line.
We can do this because, at constant temperature and pressure, V ÷ n = "a constant"
Let′s say that another helium balloon experiment is conducted at constant temperature and pressure.
In this experiment, 1 mole of helium gas had a volume of 20 L.
Next, 0.5 moles of helium gas escaped and the volume of the balloon decreased.
We can calculate the new volume of the balloon because the temperature and pressure are constant!
First we know that V ÷ n = "a constant"
We can use the information about the first part of the experiment in which
n = 1 mole
V = 20 L
to calculate the value of the "constant"
V ÷ n = "a constant"
20 ÷ 1 = 20
So, at this temperature and pressure, the equation becomes:
V ÷ n = 20
then we can use this value of constant to calculate the new gas volume because we know that n = 0.5 mole:
V ÷ n = 20
V ÷ 0.5 = 20
V = 20 × 0.5 = 10 L
If we let
initial moles of gas = n_{1}
initial volume of gas = V_{1}
final moles of gas = n_{2}
final volume of gas = V_{2}
then
V_{1} n_{1} 
= "a constant" = 
V_{2} n_{2} 
In other words:
For our balloon experiment:
initial moles of gas = n_{1} = 1 mol
initial volume of gas = V_{1} = 20 L
final moles of gas = n_{2} = 0.5 mol
final volume of gas = V_{2} = ? L
Substituting these values into the equation,
V_{1} n_{1} 
= 
V_{2} n_{2} 
20 1 
= 
V_{2} 0.5 
20 
= 
V_{2} 0.5 
Multiply both sides of the equation by 0.5 and solve for n_{2}:
20 × 0.5 
= 
V_{2} × 0.5
0.5 
10 
= 
V_{2} 
This relationship can also be used to calculate the volume or moles of different gases, as long as all the gases are at the same constant temperature and pressure, because equal volumes of different gases at the same temperature and pressure contain the same number of gas particles.
If I have two identical balloons at the same temperature and pressure in which balloon 1 contains 0.1 moles of helium gas in a volume of 6.0 L while balloon 2 contains 0.3 moles of argon gas, I can calculate the volume of argon gas:
moles of helium gas = n_{1} = 0.1 mol
volume of helium gas = V_{1} = 6.0 L
moles of argon gas = n_{2} = 0.3 mol
volume of argon gas = V_{2} = ? L
Substituting the values into the equation:
6.0 0.1 
= 
V_{2} 0.3 
60 
= 
V_{2} 0.3 
Multiply both sides of the equation by 0.3 and solve for V_{2}:
60 × 0.3 
= 
V_{2} × 0.3 0.3 
18 L 
= 
V_{2} 
Which means that, at the same temperature and pressure as 0.1 moles of helium gas with a volume of 6.0 L, 0.3 moles of argon gas has a volume of 18 L.