 ## Key Concepts

Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.
• So, at constant temperature and pressure, the volume, V, of a sample of gas is proportional to the number of molecules, N, in the gas sample:

Mathematically we can express this as:

volume of gas ∝ number of molecules

substituting V for gas volume and N for number of molecules:

V ∝ N

using a constant of proportionality we can convert this to an equation:

V = N × "a constant"

which means the ratio of V to N is a constant:

V ÷ N = "a constant"

• Since a "mole" of gas represents 6.02 × 1023 particles,

number of particles ÷ 6.02 × 1023 = moles of particles (n)

Then, the volume of gas (V) must be proportional to the moles of gas (n):

V ∝ n

using a constant of proportionality we can also write this relationship as an equation:

V = n × "a constant"

which we can rearrange to get:

V ÷ n = "a constant"

• This relationship tells us that:

(a) If the quantity of gas (n) increases, then at the same temperature and pressure the volume the gas occupies (V) must also increase.

(b) If the quantity of gas (n) decreases, then at the same temperature and pressure the volume the gas occupies (V) must also decrease.

• The relationship can be also be written as

 V1n1 = V2n2

in which n1 moles of gas 1 has a volume of V1, while at the same temperature and pressure, n2 moles of the same or a different gas1 will have a volume of V2.
• V1/n1 = V2/n2 can be rearranged to find either V1, n1, V2 or n2

 To find V1: V1 = V2 × n1/n2 To find V2: V2 = V1 × n2/n1 To find n1: n1 = n2 × V1/V2 To find n2: n2 = n1 × V2/V1

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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 × 1022 particles of helium gas.
Balloon 2 contains 2.4 × 1022 particles of nitrogen gas.
Balloon 3 contains 2.4 × 1022 particles of of argon gas.

That is, 1 L of gas at 25°C and 100 kPa pressure contains 2.4 × 1022 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 × 1023 24.79
1.20 × 1024 49.58
1.81 × 1024 74.37
2.41 × 1024 99.16

We can see that as we increase the number of gas particles from 6.02 × 1023 to 2.41 × 1024 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 × 1023 24.79 24.79 ÷ (6.02 × 1023) = 4.1 × 10-23
1.20 × 1024 49.58 49.58 ÷ (1.20 × 1024) = 4.1 × 10-23
1.81 × 1024 74.37 74.37 ÷ (1.81 × 1024) = 4.1 × 10-23
2.41 × 1024 99.16 99.16 ÷ (2.41 × 1024) = 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 × 1023 particles of that substance.
So, n moles of gas contains n × 6.02 × 1023 gas particles.
Therefore, the number of gas particles (N) divided by 6.02 × 1023 will be equal to the moles of gas (N(gas)):

n(gas) = N ÷ 6.02 × 1023

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 × 1023 6.02 × 1023 ÷ 6.02 × 1023
= 1 mole
24.79
1.20 × 1024 1.20 × 1024 ÷ 6.02 × 1023
= 2 mole
49.58
1.81 × 1024 1.81 × 1024 ÷ 6.02 × 1023
= 3 mole
74.37
2.41 × 1024 2.41 × 1024 ÷ 6.02 × 1023
= 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 y-intercept) = 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 = n1
initial volume of gas = V1
final moles of gas = n2
final volume of gas = V2

then
 V1n1 = "a constant" = V2n2

In other words:
 V1n1 = V2n2

For our balloon experiment:
initial moles of gas = n1 = 1 mol
initial volume of gas = V1 = 20 L
final moles of gas = n2 = 0.5 mol
final volume of gas = V2 = ? L

Substituting these values into the equation,
 V1n1 = V2n2 20 1 = V20.5 20 = V20.5

Multiply both sides of the equation by 0.5 and solve for n2:
 20 × 0.5 = V2 × 0.50.5 10 = V2

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 = n1 = 0.1 mol
volume of helium gas = V1 = 6.0 L
moles of argon gas = n2 = 0.3 mol
volume of argon gas = V2 = ? L

 V1n1 = V2n2

Substituting the values into the equation:
 6.0 0.1 = V2 0.3 60 = V2 0.3

Multiply both sides of the equation by 0.3 and solve for V2:
 60 × 0.3 = V2 × 0.3 0.3 18 L = V2

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.

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## Worked Example: Changing Quantity of Gas

Question: 10 moles of carbon dioxide gas has a volume of 245 L at 25oC and 1 atm pressure.
If 5 moles of the carbon dioxide is removed at the same temperature and pressure, what volume will the gas now occupy?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate new volume of gas
V2 = ? L

2. What data (information) have you been given in the question?

Extract the data from the question:

Conditions: constant temperature and pressure

n1 = initial moles of gas = 10 mol
V1 = initial volume of gas = 245 L

moles gas removed = 5 mol
n2 = initial moles gas - moles gas removed
n2 = 10 - 5 = 5 mol

3. What is the relationship between what you know and what you need to find out?
Because the experiment is done at constant temperature and pressure we can use Avogadro's Principle:

 V1n1 = V2n2

Which we can rearrange to find V2:

 V2 = V1n1 × n2

4. Substitute the values into the equation and solve for V2

 V2 = V1n1 × n2 = 24510 × 5 = 122.5

Consider what happens to the volume of gas as some of the gas is removed:
n1 moles of gas initially had a volume of V1 L
When some of this gas is removed while the temperature is held constant, the pressure should drop because there will be fewer collisions between the gas molecules and the container walls.
BUT, we have been told that the pressure remains constant, therefore the volume of the gas must decrease in order to maintain the same pressure.
That is, by making a smaller volume in which the gas particles can move we increase the number of collisions between the gas molecules and the walls of the container.
Since the final volume (V2 = 122.5 L) is smaller than the intial gas volume (V1 = 245 L), we are reasonably confident that our answer is plausible.
6. State your solution to the problem "calculate new volume of gas":

V2 = 122.5 L

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## Worked Example: Changing Volume of Gas

Question: At 0oC and 1 atm pressure, 0.5 mol of oxygen gas has a volume of 11.2 L.
If the volume of gas is expanded to 22.4 L by allowing more oxygen gas into the system while maintaining the same temperature and pressure, what is the final quantity of gas in moles?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate the final amount of O2(g) in moles
n2 = ? mol

2. What data (information) have you been given in the question?

Extract the data from the question:

Conditions: constant temperature (0°C) and pressure (1 atm)

n1 = 0.5 mol

V1 = 11.2 L

V2 = 22.4 L

3. What is the relationship between what you know and what you need to find out?
Because the experiment is done at constant temperature and pressure we can use Avogadro's Principle:

 V1n1 = V2n2

Which we can rearrange to find n2:

 n2 = V2V1 × n1

4. Substitute in the values and solve for n2

 n2 = V2V1 × n1 = 22.4 11.2 × 0.5 = 1.0 mol

Consider what happens when we add more O2(g).
In order to maintain the same temperature and pressure, the volume occupied by the gas must change.
Adding more particles would increase the number of collisions with the container walls (at constant temperature), that is, if the volume was constant the pressure would increase.
But, if the volume can increase, this will decrease pressure, that is the volume of the container must expand in order to maintain a constant temperature and pressure.
If we double the number of gas particles, we expect the volume to double.
Since our calculated value for n2 (1.0 mol) is greater than the initial amount of gas, n1 (0.5 mol), we are reasonaby confident that our answer is correct.
6. State your solution to the problem "moles of O2(g) after addition of more gas ":

n2 = 1.0 mol

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## Worked Example: Comparing Different Gases at the Same Temperature and Pressure

Question : At a certain temperature and pressure, a 500 mL flask contains 25 mol of nitrogen gas.
A different flask at the same temperature and pressure contains 100 mL of helium gas.
How many moles of helium are in the second flask?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate the moles of helium in the second flask
n(He(g))= ? mol

2. What data (information) have you been given in the question?

Extract the data from the question:

Conditions: same temperature and pressure for both gas samples (that is, constant temperature and pressure)

n(N2(g)) = 25 mol

V(N2(g)) = 500 mL

V(He(g)) = 100 mL

3. What is the relationship between what you know and what you need to find out?
Because the experiment is done at constant temperature and pressure we can use Avogadro's Principle:

 V(N2(g))n(N2(g)) = V(He(g))n(He(g))

Which we can rearrange to find n(He(g)):

 n(He(g)) = V(He(g))V(N2(g)) × n(N2(g))

4. Substitute in the values and solve for n(He(g))

 n(He(g)) = V(He(g))V(N2(g)) × n(N2(g)) = 100 500 × 25 = 5 mol

Consider that 25 moles of gas occupies a volume of 500 mL.
If we maintain the same temperature and reduce the volume of gas to 100 mL then the number of collisions between the gas particles and the walls of the container would increase IF there are the same number of gas particles in the container.
But, if some of the gas particles can escape from the container this would reduce the pressure.
So, if the volume of the container decreases from 500 mL to 100 mL while maintaining constant temperature and pressure, this means that the smaller volume of gas must contain fewer gas particles.
Since 500 mL container contained 25 moles of gas, 100 mL container must contain less than 25 moles of gas.
Since our calculated value for n(He(g)) (5 mol) is less than n(N2(g)) (25 mol) we are reasonably confident our answer is correct.
6. State your solution to the problem "moles of He(g) ":

n(He(g)) = 5 mol

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1. For this to be true, the gases should be ideal gases.