In the previous section, we saw how to convert 'number of moles' into mass. We also saw how to calculate the mass of a single atom or molecule. In this section we will see the relation between 'mole' and 'Standard temperature and pressure' in the case of gases.
• In solids and liquids, the molecules are packed close to each other. The distance between any two molecules is very small.
♦ So the volume of a solid or a liquid can be considered as the total volume of the molecules
• In gases, the molecules are not packed close to each other. The distance between any two molecules is very large when compared to the size of the molecules.
♦ So the volume of a gas cannot be considered as the total volume of it's molecules.
• Also, the molecules of the gases are always in motion. So they occupy the entire space of it's container.
♦ So volume of a gas can be considered as the volume of it's container.
1. Let us take a sample from each of the two gases. Let both the samples contain the equal number of molecules.
2. If the number of molecules are the same, naturally, we will think that the Gas A will have greater volume because, it has large size molecules.
3. But that is not the case. Let us see the reason:
• The molecules with lesser mass will move with a larger speed than those with heavier mass.
♦ So the lighter molecules are able to reach greater distances.
♦ That means, they will be able to occupy a greater volume, even if they are small in size.
4. In fact, it is seen that, if any two gases gases have the same number of molecules, they will occupy the same volume.
5. But, to occupy the same volume, there is one condition:
• The temperature and pressure experienced by the two gases should be the same.
6. This is because:
• If any one gas is subjected to a greater temperature than the other, it's molecules will begin to move with a greater speed. So that gas will occupy a greater volume
♦ For example, if the temperature around an inflated balloon is increased, it will expand
• If any one gas is subjected to a greater pressure than the other, it's volume will decrease.
♦ For example, if we apply pressure on an inflated balloon, it will contract.
• Thus it is clear that, the gases under consideration must be experiencing the same temperature and pressure. Only then we will get the same volume for equal number of molecules
7. Consider the words: 'equal number of molecules'.
• Why not take the 'mole' as the equal number?
Let us analyse:
• We know that, a mole will contain NA number of molecules. It will definitely be convenient to take it as the 'equal number of molecules'.
8. So take a mole each of Gas A and Gas B.
• Let both the gases be subjected to the same temperature: 273 K
• Let both the gases be subjected to the same pressure: 1 atm
■ In such a condition, each of the gases A and B, will occupy a volume of 22.4 L
9. 273 K temperature and 1 atm pressure are known as standard temperature and pressure. It is abbreviated as STP
Note:
Units of Temperature:
• In day to day life, we use the celsius scale for measuring temperature.
• For scientific and engineering purposes, another scale is also used. It is the 'kelvin scale'. It's symbol is 'K'.
• They can be easily converted into each other
Units of Pressure:
• We know that pressure is 'Force per unit area'. In day to day life, we use pascal as the unit for pressure. It is 'newton per square metre'. It's symbol is 'pa'.
• For scientific and engineering purposes, another scale is also used. It is the 'atmosphere scale'. It's symbol is 'atm'
• They can be easily converted into each other
Units of Volume
• In day to day life we use 'Litre' to measure volume. It's symbol is 'L'.
• For scientific and engineering purposes, another unit is also used. It is the 'Cubic centimetre'. It's symbol is cm3
• They can be easily converted into each other using the relation: 1 L = 1000 cm3
We will learn more details about such units in higher classes
■ Take one mole of molecules of any gas. If those molecules are subjected to 273 K and 1 atm, they will occupy a volume of 22.4 L
■ Conversely, take 22.4 L of any gas. If the temperature and pressure experienced by that volume is 273 K and 1 atm, then there will be one mole of molecules in that volume.
So we can modify the presentation that we saw earlier. The modified presentation is given in fig.10.8 below:
Let us see an example:
A gas occupies a volume of 224 L when it is subjected to 273 K temperature and 1 atm pressure. How many molecules are present in that volume?
Solution:
1. Temperature is 273 K and pressure is 1 atm. So the gas is at STP
2. At STP, 22.4 L of any gas will contain 1 mole of it's molecules
3. So 1 L will contain (1⁄22.4) mole of it's molecules
4. Thus, no. of moles in 224 L = 224 × 1⁄22.4 = 10
5. No. of molecules in 10 moles = 10 × 6.022×1023
Solved example 10.21
Calculate the number of molecules in each of the following two samples:
(a) 44.8 L of NH3 at STP (b) 67.2 L of N2 at STP
Solution:
(a) 1 mol of any gas will occupy 22.4 L at STP
• So 1 L at STP will contain (1⁄22.4) mol
• So 44.8 L will contain [44.8 × (1⁄22.4)] = 2 mol
• 1 mol will contain 6.022×1023 molecules
• So 2 mol will contain (2 × 6.022×1023) molecules
(b) 1 mol of any gas will occupy 22.4 L at STP
• So 1 L at STP will contain (1⁄22.4) mol
• So 67.2 L will contain [67.2 × (1⁄22.4)] = 3 mol
• 1 mol will contain 6.022×1023 molecules
• So 3 mol will contain (3 × 6.022×1023) molecules
Solved example 10.22
Calculate the volume of 170 g ammonia at STP
Solution:
1. The GMM of ammonia is [(1×14)+(3×1)] = [14+3] = 17
2. So 1 mole of ammonia will have a mass of 17 g
3. So 1 g will contain (1⁄17) mole
4. So 170 g will contain [170×(1⁄17) ] = 10 mole
5. 1 mole of any gas will have a volume of 22.4 L at STP
So 170 g (10 moles) of ammonia will have a volume of 22.4×10 = 224 L
Consider the balanced chemical equation for the manufacture of ammonia:
N2 + 3H2 → 2NH3
• One molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia.
• So if we take 1 molecule of nitrogen, it is compulsory to take 3 molecules of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get 2 molecules of ammonia.
• If we take 2 molecules of nitrogen, it is compulsory to take 6 molecules of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get 4 molecules of ammonia.
• If we take 15 molecules of nitrogen, it is compulsory to take 45 molecules of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get 30 molecules of ammonia.
So on . . .
• Thus, if we take NA molecules of nitrogen, it is compulsory to take 3NA molecules of hydrogen. After the reaction, we will get 2NA molecules of ammonia.
■ But NA molecules is 1 mole molecules. So we can write:
If we take 1 mole of nitrogen, it is compulsory to take 3 moles of hydrogen. After the reaction, we will get 2 moles of ammonia.
• 1 mole of hydrogen molecules will have a mass of GMM of hydrogen, which is equal to 2 grams
♦ so 3 moles of hydrogen will have a mass of (3 × 2) = 6 grams
■ If we take 28 grams of nitrogen, it is compulsory to take 6 grams of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get (28+6) = 34 grams of ammonia.
Check:
• 1 GMM of ammonia = [(1×14)+(3×1)] = [14+3] = 17
• The product is 2 moles of ammonia, which is 2 GMM of ammonia, which is equal to (2 × 17) = 34 grams of ammonia.
• So, if we take 22.4 L of nitrogen gas at STP, it is compulsory to take (3×22.4) = 67.2 L of hydrogen at STP. When the reaction is complete, we will get (2×22.4) = 44.8 L of ammonia at STP.
The above results can be written in a tabular form as shown below:
Solved example 10.23
Calculate the amount of hydrogen required to react with 500 g of nitrogen during the manufacture of ammonia.
Solution:
1. The balanced equation is: N2 + 3H2 → 2NH3.
2. So for one mole of nitrogen, three moles of hydrogen is required
• One mole of nitrogen is 28 grams.
• One mole of hydrogen is 2 grams
♦ Three moles of hydrogen is (3×2) = 6 grams
3. So for 28 grams of nitrogen, 6 grams of hydrogen is required
4. So for 1 gram of nitrogen, (6⁄28) gram of hydrogen is required
5. Thus for 500 grams of nitrogen, amount of hydrogen required = 500 × (6⁄28) = 107.14 grams
Solved example 10.24
How many grams of N2 is required to obtain 1000 L of ammonia at STP?
Solution:
1. One mole of ammonia at STP will have a volume of 22.4 L
2. So 1 L of ammonia at STP will have (1⁄22.4) mole
3. So 1000 L of ammonia at STP will have 1000 × 1⁄22.4 = 1000⁄22.4 mole
4. The balanced equation for the manufacture of ammonia is: N2 + 3H2 → 2NH3.
5. 2 moles of ammonia can be obtained from one mole of nitrogen
6. So one mole of ammonia can be obtained from 1⁄2 mole of nitrogen
7. So 1000⁄22.4 mole of ammonia can be obtained from 1000⁄22.4 × 1⁄2 = 500⁄22.4 mole of nitrogen
8. One mole of nitrogen is 28 grams
9. So 500⁄22.4 mole of nitrogen is (500⁄22.4 × 28) grams
In the next section we will see Molarity of solutions.
• In solids and liquids, the molecules are packed close to each other. The distance between any two molecules is very small.
♦ So the volume of a solid or a liquid can be considered as the total volume of the molecules
• In gases, the molecules are not packed close to each other. The distance between any two molecules is very large when compared to the size of the molecules.
♦ So the volume of a gas cannot be considered as the total volume of it's molecules.
• Also, the molecules of the gases are always in motion. So they occupy the entire space of it's container.
♦ So volume of a gas can be considered as the volume of it's container.
• Now consider any two gases: Gas A & Gas B.
• Let the molecules of Gas A have a greater mass than those of Gas B. This is shown in the fig.10.7 below:
• Let the molecules of Gas A have a greater mass than those of Gas B. This is shown in the fig.10.7 below:
1. Let us take a sample from each of the two gases. Let both the samples contain the equal number of molecules.
2. If the number of molecules are the same, naturally, we will think that the Gas A will have greater volume because, it has large size molecules.
3. But that is not the case. Let us see the reason:
• The molecules with lesser mass will move with a larger speed than those with heavier mass.
♦ So the lighter molecules are able to reach greater distances.
♦ That means, they will be able to occupy a greater volume, even if they are small in size.
4. In fact, it is seen that, if any two gases gases have the same number of molecules, they will occupy the same volume.
5. But, to occupy the same volume, there is one condition:
• The temperature and pressure experienced by the two gases should be the same.
6. This is because:
• If any one gas is subjected to a greater temperature than the other, it's molecules will begin to move with a greater speed. So that gas will occupy a greater volume
♦ For example, if the temperature around an inflated balloon is increased, it will expand
• If any one gas is subjected to a greater pressure than the other, it's volume will decrease.
♦ For example, if we apply pressure on an inflated balloon, it will contract.
• Thus it is clear that, the gases under consideration must be experiencing the same temperature and pressure. Only then we will get the same volume for equal number of molecules
7. Consider the words: 'equal number of molecules'.
• Why not take the 'mole' as the equal number?
Let us analyse:
• We know that, a mole will contain NA number of molecules. It will definitely be convenient to take it as the 'equal number of molecules'.
8. So take a mole each of Gas A and Gas B.
• Let both the gases be subjected to the same temperature: 273 K
• Let both the gases be subjected to the same pressure: 1 atm
■ In such a condition, each of the gases A and B, will occupy a volume of 22.4 L
9. 273 K temperature and 1 atm pressure are known as standard temperature and pressure. It is abbreviated as STP
Units of Temperature:
• In day to day life, we use the celsius scale for measuring temperature.
• For scientific and engineering purposes, another scale is also used. It is the 'kelvin scale'. It's symbol is 'K'.
• They can be easily converted into each other
Units of Pressure:
• We know that pressure is 'Force per unit area'. In day to day life, we use pascal as the unit for pressure. It is 'newton per square metre'. It's symbol is 'pa'.
• For scientific and engineering purposes, another scale is also used. It is the 'atmosphere scale'. It's symbol is 'atm'
• They can be easily converted into each other
Units of Volume
• In day to day life we use 'Litre' to measure volume. It's symbol is 'L'.
• For scientific and engineering purposes, another unit is also used. It is the 'Cubic centimetre'. It's symbol is cm3
• They can be easily converted into each other using the relation: 1 L = 1000 cm3
We will learn more details about such units in higher classes
In the above discussion, we considered two gases A and B. In fact, it is valid for any number of gases. That is:
• Take samples of 'n' gases. Where 'n' can be any number greater than 1: 2, 3, 4, 5 so on...
♦ Each sample should be subjected to a temperature of 273 K
♦ Each sample should be subjected to a pressure of 1 atm
• In other words, each sample should be at STP
• Further, each sample should contain exactly NA number of it's molecule
■ Then each of those n samples will occupy a volume of 22.4 L
• NA (Avogadro number) of molecules of a gas means 'one mole of molecules' of that gas.
So we can write:■ Take one mole of molecules of any gas. If those molecules are subjected to 273 K and 1 atm, they will occupy a volume of 22.4 L
■ Conversely, take 22.4 L of any gas. If the temperature and pressure experienced by that volume is 273 K and 1 atm, then there will be one mole of molecules in that volume.
So we can modify the presentation that we saw earlier. The modified presentation is given in fig.10.8 below:
Let us see an example:
A gas occupies a volume of 224 L when it is subjected to 273 K temperature and 1 atm pressure. How many molecules are present in that volume?
Solution:
1. Temperature is 273 K and pressure is 1 atm. So the gas is at STP
2. At STP, 22.4 L of any gas will contain 1 mole of it's molecules
3. So 1 L will contain (1⁄22.4) mole of it's molecules
4. Thus, no. of moles in 224 L = 224 × 1⁄22.4 = 10
5. No. of molecules in 10 moles = 10 × 6.022×1023
Solved example 10.21
Calculate the number of molecules in each of the following two samples:
(a) 44.8 L of NH3 at STP (b) 67.2 L of N2 at STP
Solution:
(a) 1 mol of any gas will occupy 22.4 L at STP
• So 1 L at STP will contain (1⁄22.4) mol
• So 44.8 L will contain [44.8 × (1⁄22.4)] = 2 mol
• 1 mol will contain 6.022×1023 molecules
• So 2 mol will contain (2 × 6.022×1023) molecules
(b) 1 mol of any gas will occupy 22.4 L at STP
• So 1 L at STP will contain (1⁄22.4) mol
• So 67.2 L will contain [67.2 × (1⁄22.4)] = 3 mol
• 1 mol will contain 6.022×1023 molecules
• So 3 mol will contain (3 × 6.022×1023) molecules
Solved example 10.22
Calculate the volume of 170 g ammonia at STP
Solution:
1. The GMM of ammonia is [(1×14)+(3×1)] = [14+3] = 17
2. So 1 mole of ammonia will have a mass of 17 g
3. So 1 g will contain (1⁄17) mole
4. So 170 g will contain [170×(1⁄17) ] = 10 mole
5. 1 mole of any gas will have a volume of 22.4 L at STP
So 170 g (10 moles) of ammonia will have a volume of 22.4×10 = 224 L
Mole concept – In balanced chemical equations
In this topic,we will learn how the mole concept is related to balanced chemical equations. We have already learned about balanced chemical equations. Details here.Consider the balanced chemical equation for the manufacture of ammonia:
N2 + 3H2 → 2NH3
• One molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia.
• So if we take 1 molecule of nitrogen, it is compulsory to take 3 molecules of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get 2 molecules of ammonia.
• If we take 2 molecules of nitrogen, it is compulsory to take 6 molecules of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get 4 molecules of ammonia.
• If we take 15 molecules of nitrogen, it is compulsory to take 45 molecules of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get 30 molecules of ammonia.
So on . . .
• Thus, if we take NA molecules of nitrogen, it is compulsory to take 3NA molecules of hydrogen. After the reaction, we will get 2NA molecules of ammonia.
■ But NA molecules is 1 mole molecules. So we can write:
If we take 1 mole of nitrogen, it is compulsory to take 3 moles of hydrogen. After the reaction, we will get 2 moles of ammonia.
Once we write the quantities in terms of moles, we will be able to write the quantities in terms of grams:
• 1 mole of nitrogen molecules will have a mass of GMM of nitrogen, which is equal to 28 grams• 1 mole of hydrogen molecules will have a mass of GMM of hydrogen, which is equal to 2 grams
♦ so 3 moles of hydrogen will have a mass of (3 × 2) = 6 grams
■ If we take 28 grams of nitrogen, it is compulsory to take 6 grams of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get (28+6) = 34 grams of ammonia.
Check:
• 1 GMM of ammonia = [(1×14)+(3×1)] = [14+3] = 17
• The product is 2 moles of ammonia, which is 2 GMM of ammonia, which is equal to (2 × 17) = 34 grams of ammonia.
In terms of litres:
• 1 mole of any gas = 22.4 L at STP• So, if we take 22.4 L of nitrogen gas at STP, it is compulsory to take (3×22.4) = 67.2 L of hydrogen at STP. When the reaction is complete, we will get (2×22.4) = 44.8 L of ammonia at STP.
| Nitrogen | Hydrogen | Ammonia | |
|---|---|---|---|
| Mole | 1 | 3 | 2 |
| Mass | 28 grams | 6 grams | 34 grams |
| Volume | 22.4 L | 67.2 L | 44.8 L |
Calculate the amount of hydrogen required to react with 500 g of nitrogen during the manufacture of ammonia.
Solution:
1. The balanced equation is: N2 + 3H2 → 2NH3.
2. So for one mole of nitrogen, three moles of hydrogen is required
• One mole of nitrogen is 28 grams.
• One mole of hydrogen is 2 grams
♦ Three moles of hydrogen is (3×2) = 6 grams
3. So for 28 grams of nitrogen, 6 grams of hydrogen is required
4. So for 1 gram of nitrogen, (6⁄28) gram of hydrogen is required
5. Thus for 500 grams of nitrogen, amount of hydrogen required = 500 × (6⁄28) = 107.14 grams
Solved example 10.24
How many grams of N2 is required to obtain 1000 L of ammonia at STP?
Solution:
1. One mole of ammonia at STP will have a volume of 22.4 L
2. So 1 L of ammonia at STP will have (1⁄22.4) mole
3. So 1000 L of ammonia at STP will have 1000 × 1⁄22.4 = 1000⁄22.4 mole
4. The balanced equation for the manufacture of ammonia is: N2 + 3H2 → 2NH3.
5. 2 moles of ammonia can be obtained from one mole of nitrogen
6. So one mole of ammonia can be obtained from 1⁄2 mole of nitrogen
7. So 1000⁄22.4 mole of ammonia can be obtained from 1000⁄22.4 × 1⁄2 = 500⁄22.4 mole of nitrogen
8. One mole of nitrogen is 28 grams
9. So 500⁄22.4 mole of nitrogen is (500⁄22.4 × 28) grams
In the next section we will see Molarity of solutions.
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