In the previous section, we saw the details about the unit parsec. With that, we have completed a discussion on 'measurement of large distances'. In this section, we will see details about 'small distances'.
• In an earlier section we saw this:
♦ A vernier calipers can be used to measure lengths as small as 10-4 m.
♦ A screw gauge or spherometer can measure lengths as small as 10-5 m.
• To measure lengths of objects which are smaller than these, we first have to observe them.
• Modern techniques are available today, which enable scientists to make observations of very small objects. We will see them in higher classes.
• At present we will see an indirect method to measure the size of a molecule.
• We want to find the diameter of a molecule of oleic acid. We will write it in steps.
1. Consider a circular disc. This disc has three specialties:
(i) It is completely made up of oleic acid molecules.
• That means, there are no other particles. Only oleic acid molecules.
• Those molecules are tightly packed. There is no space in between.
(ii) The disc is in a perfect horizontal position.
(iii) All the molecules are in the same horizontal plane.
• That means, not even a single molecule is stacked above another.
• So thickness of the disc = diameter of one oleic acid molecule.
2. Once we make such a disc, we can do some calculations.
• Volume of the disc can be calculated in the same way as we calculate the volume of a cylinder:
■ Volume of the disc (V) = Area of the disc (A) × thickness of the disc (t)
• If we know the radius r of the disc, we can write area as A = 𝝅r2
• Also t = thickness of the disc = diameter of one molecule
• Thus we get: V = 𝝅r2t.
3. So there are three items: V, r and t.
• If we know the values of V and r, we can easily calculate t.
• How can we find V?
4. Oleic acid is placed in a safe container in the lab.
• We take a small portion from that container to make the disc.
• All the oleic acid taken out from the container does not go into making up the disc.
• That is:
V ≠ Volume taken out from the container.
5. V = Actual volume of oleic acid used for making up the disc.
• To find the right side of this equation, we use a special procedure:
(i) Take out 1 cm3 of oleic acid from the container.
(ii) Dissolve it slowly in alcohol to make up a solution whose volume is 20 cm3.
• Note that, the total volume of the resulting solution is to be exactly 20 cm3.
(iii) Imagine that, this solution is divided into 20 equal parts.
• Then each part will have a volume of 1 cm3.
(iv) Take out one such part. Let us call it P.
• P has a volume of 1 cm3.
• This 1 cm3 of P is made up of oleic acid and alcohol.
(v) How much oleic acid is present in that 1 cm3.
• Obviously, the answer is 1⁄20 = 0.05 cm3.
(vi) Like P, each of the 20 equal parts will contain 0.05 cm3 of oleic acid.
• Since each 1 cm3 contains 0.05 cm3, we can write:
Concentration of that solution is: 0.05 cm3 of oleic acid per cm3.
(vii) Now take another glass jar.
• Dissolve P slowly in alcohol to make up a solution whose volume is 20 cm3.
• Note that, the total volume of the resulting solution is to be exactly 20 cm3.
(ix) Imagine that, this new solution is divided into 20 equal parts.
• Then each part will have a volume of 1 cm3.
(x) Take out one such part. Let us call it Q.
• Q has a volume of 1 cm3.
• This 1 cm3 of Q is made up of oleic acid and alcohol.
(xi) How much oleic acid is present in that 1 cm3.
• Obviously, the answer is 0.05⁄20 = 0.0025 cm3.
(Since P contained 0.05 cm3 of oleic acid)
(xii) Like Q, each of the 20 equal parts will contain 0.0025 cm3 of oleic acid.
• Since each 1 cm3 of this new solution contains 0.05 cm3, we can write:
Concentration of the new solution is: 0.0025 cm3 of oleic acid per cm3.
(xiii) Suppose we take one drop from this solution, how much oleic acid will that contain?
• For that, we need to find the volume of one drop.
• ‘Finding volume of one drop of a solution’ is a simple procedure done in the chemistry lab.
• We will see it later in chemistry lab manual.
• Let us assume that the volume of one drop taken from our second solution is Vd cm3.
(xiv) Each cm3 of our second solution has 0.0025 cm3 of oleic acid.
• So Vd cm3 will contain 0.0025Vd cm3 of oleic acid.
■ Thus we succeeded in finding the volume of oleic acid:
If we take one drop from the solution, we will be having 0.0025Vd cm3 of oleic acid in that drop.
6. Now we enter the final stage of our experiment:
(i) Take some water in a large trough.
(ii) Lightly sprinkle some lycopodium powder on the surface of water.
(iii) Put one drop from the second solution at the center of the lycopodium covered water surface.
• The alcohol present in the drop will dissolve in water.
• But oleic acid will not dissolve in water.
• Also oleic acid is lighter than water.
• So it spreads into a thin film of circular shape.
(iv) The surface of water will always be horizontal.
• So the film will be horizontal.
• All the oleic acid molecules will spread in water.
• No molecule will be stacked above another one.
• So the film will have a thickness of one molecule.
• Also it will be made entirely made up of oleic acid molecules.
■ This circular film is the disc that we wanted in step (1).
(v) Measure the radius (r) of the film.
7. We have put one drop in the water.
• From the result in (5), we have: volume of oleic acid put in water = 0.0025Vd cm3
• If we put n drops, volume of oleic acid put in water = 0.0025nVd cm3
• Then the disc is made up of 0.0025nVd cm3 of oleic acid.
8. From the result in (2), we get: 0.0025nVd = V = 𝝅r2t.
From this we get: $\mathbf\small{t=\frac{0.0025\,n\,V_d}{\pi r^2} \text{cm}}$
Solved example 2.7
Model of an atom is to be made for a science exhibition. The actual size of the nucleus is in the range of 10-15 to 10-14 m. The actual size of the atom is about 10-10 m. In the model, the nucleus should have the size of the tip of a sharp pin. The actual size of the tip of a sharp pin is in the range of 10-5 to 10-4 m. What should be the total size of the model ?
Solution:
1. Actual size of a nucleus = 10-15 to 10-14 m.
• Size of the nucleus in the model = 10-5 to 10-4 m.
2. If we decide to blow up 10-15 to 10-5, we are to multiply 10-15 by 1010
• If we decide to blow up 10-14 to 10-4, then also, we are to multiply 10-14 by 1010
■ So the scale factor is 1010
3. That means, we have to multiply all 'actual lengths' by 1010
• So the size of the atom in the model = (10-10×1010) = 100 m = 1 m.
4. Thus we can write:
■ The model of the atom should be a sphere of 1 m diameter.
■ A very small point (with the size equal to the tip of a sharp pin) at the center of the model, will represent the nucleus.
Note: The doors of rooms in an ordinary residential building have widths equal to 0.9 m. This is 10 cm less than 1 m. So if the model is made while inside the room, we will not be able to take it to the exhibition hall.
Solved example 2.8
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by $\mathbf\small{\mathring{A}}$. (1 $\mathbf\small{\mathring{A}}$ = 10-10 m). The radius of a hydrogen atom is about 0.5 $\mathbf\small{\mathring{A}}$. What is the total atomic volume in m3 of a mole of hydrogen atoms ?
Solution:
1. The radius of a hydrogen atom is given as 0.5 $\mathbf\small{\mathring{A}}$
• So radius of one hydrogen atom in m = 0.5×10-10 m.
• So volume of 1 hydrogen atom in m3 = $\mathbf\small{\frac{4}{3}\pi\,r^3=\frac{4}{3}\pi\,(0.5\times10^{-10})^3=5.236\times10^{-31}\;\text{m}^3}$
2. 1 mole of hydrogen atoms will contain 6.02214 × 1023 hydrogen atoms.
• So total volume = ( 6.02214 × 1023 × 5.236 × 10-31) = 3.15 × 10-7 m3
Solved example 2.9
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the radius of hydrogen molecule to be about 0.5Å). Why is this ratio so large ?
Solution:
1. The radius of a hydrogen molecule is given as 0.5$\mathbf\small{\mathring{A}}$
• So radius of one hydrogen molecule in m = 0.5×10-10 m.
(∵ 1 $\mathbf\small{\mathring{A}}$ = 10-10 m)
• So volume of 1 hydrogen molecule in m3 = $\mathbf\small{\frac{4}{3}\pi\,r^3=\frac{4}{3}\pi\,(0.5\times10^{-10})^3=0.5236\times10^{-30}\;\text{m}^3}$
2. 1 mole of hydrogen molecules will contain 6.02214 × 1023 hydrogen molecules.
• So total volume = ( 6.02214 × 1023 × 0.5236 × 10-30) = 3.1532 × 10-7 m3
3. 22.4 L = 22.4 × 1000 = 22400 cm3 = 22400 × 10-6 m3
4. So the ratio = $\mathbf\small{\frac{\text{molar volume}}{\text{atomic volume}}=\frac{22400 \times 10^{-6}}{3.1532\times 10^{-7}}=7.104\times 10^4}$
5. This is a very large ratio. We can see that the numerator is very large when compared to the denominator. The reason can be written as follows:
(i) The numerator is the molar volume.
At standard temperature and pressure, the hydrogen will be in the gaseous state. There will be large spaces in between it's molecules. So one mole of hydrogen molecules (6.02214 × 1023 hydrogen molecules) at STP will occupy 22.4 L.
(ii) The denominator is the volume when all the 6.02214 × 1023 hydrogen molecules are held together with no space in between. Since the hydrogen molecules are very small, the total volume will be very small even if 6.02214 × 1023 molecules are held together.
(iii) Thus the numerator (22400 × 10-6) is very large when compared to the denominator (3.1532 × 10-7).
We will write about the range in steps:
1. In our day to day life, we come across ordinary lengths such as:
♦ Length of a pen.
♦ Length of a table.
♦ Height of a building.
♦ Distance between two towns etc.,
2. But when we do problems in science, we will have to deal with a range of lengths.
• On the left end of that range, we have very small lengths.
Some examples:
♦ Size of a proton (10-15 m).
♦ Size of a nucleus (10-14 m).
• On the right end of that range, we have very large lengths.
Some examples:
♦ Distance from earth to Andromeda galaxy (2.4 × 1022 m) .
♦ Distance from earth to the end of the observable universe (8.8 × 1026 m).
3. In addition to the above two extremes, we must be able to deal with any given lengths which fall in between them.
• In an earlier section we saw this:
♦ A vernier calipers can be used to measure lengths as small as 10-4 m.
♦ A screw gauge or spherometer can measure lengths as small as 10-5 m.
• To measure lengths of objects which are smaller than these, we first have to observe them.
• Modern techniques are available today, which enable scientists to make observations of very small objects. We will see them in higher classes.
• At present we will see an indirect method to measure the size of a molecule.
• We want to find the diameter of a molecule of oleic acid. We will write it in steps.
1. Consider a circular disc. This disc has three specialties:
(i) It is completely made up of oleic acid molecules.
• That means, there are no other particles. Only oleic acid molecules.
• Those molecules are tightly packed. There is no space in between.
(ii) The disc is in a perfect horizontal position.
(iii) All the molecules are in the same horizontal plane.
• That means, not even a single molecule is stacked above another.
• So thickness of the disc = diameter of one oleic acid molecule.
2. Once we make such a disc, we can do some calculations.
• Volume of the disc can be calculated in the same way as we calculate the volume of a cylinder:
■ Volume of the disc (V) = Area of the disc (A) × thickness of the disc (t)
• If we know the radius r of the disc, we can write area as A = 𝝅r2
• Also t = thickness of the disc = diameter of one molecule
• Thus we get: V = 𝝅r2t.
3. So there are three items: V, r and t.
• If we know the values of V and r, we can easily calculate t.
• How can we find V?
4. Oleic acid is placed in a safe container in the lab.
• We take a small portion from that container to make the disc.
• All the oleic acid taken out from the container does not go into making up the disc.
• That is:
V ≠ Volume taken out from the container.
5. V = Actual volume of oleic acid used for making up the disc.
• To find the right side of this equation, we use a special procedure:
(i) Take out 1 cm3 of oleic acid from the container.
(ii) Dissolve it slowly in alcohol to make up a solution whose volume is 20 cm3.
• Note that, the total volume of the resulting solution is to be exactly 20 cm3.
(iii) Imagine that, this solution is divided into 20 equal parts.
• Then each part will have a volume of 1 cm3.
(iv) Take out one such part. Let us call it P.
• P has a volume of 1 cm3.
• This 1 cm3 of P is made up of oleic acid and alcohol.
(v) How much oleic acid is present in that 1 cm3.
• Obviously, the answer is 1⁄20 = 0.05 cm3.
(vi) Like P, each of the 20 equal parts will contain 0.05 cm3 of oleic acid.
• Since each 1 cm3 contains 0.05 cm3, we can write:
Concentration of that solution is: 0.05 cm3 of oleic acid per cm3.
(vii) Now take another glass jar.
• Dissolve P slowly in alcohol to make up a solution whose volume is 20 cm3.
• Note that, the total volume of the resulting solution is to be exactly 20 cm3.
(ix) Imagine that, this new solution is divided into 20 equal parts.
• Then each part will have a volume of 1 cm3.
(x) Take out one such part. Let us call it Q.
• Q has a volume of 1 cm3.
• This 1 cm3 of Q is made up of oleic acid and alcohol.
(xi) How much oleic acid is present in that 1 cm3.
• Obviously, the answer is 0.05⁄20 = 0.0025 cm3.
(Since P contained 0.05 cm3 of oleic acid)
(xii) Like Q, each of the 20 equal parts will contain 0.0025 cm3 of oleic acid.
• Since each 1 cm3 of this new solution contains 0.05 cm3, we can write:
Concentration of the new solution is: 0.0025 cm3 of oleic acid per cm3.
(xiii) Suppose we take one drop from this solution, how much oleic acid will that contain?
• For that, we need to find the volume of one drop.
• ‘Finding volume of one drop of a solution’ is a simple procedure done in the chemistry lab.
• We will see it later in chemistry lab manual.
• Let us assume that the volume of one drop taken from our second solution is Vd cm3.
(xiv) Each cm3 of our second solution has 0.0025 cm3 of oleic acid.
• So Vd cm3 will contain 0.0025Vd cm3 of oleic acid.
■ Thus we succeeded in finding the volume of oleic acid:
If we take one drop from the solution, we will be having 0.0025Vd cm3 of oleic acid in that drop.
6. Now we enter the final stage of our experiment:
(i) Take some water in a large trough.
(ii) Lightly sprinkle some lycopodium powder on the surface of water.
(iii) Put one drop from the second solution at the center of the lycopodium covered water surface.
• The alcohol present in the drop will dissolve in water.
• But oleic acid will not dissolve in water.
• Also oleic acid is lighter than water.
• So it spreads into a thin film of circular shape.
(iv) The surface of water will always be horizontal.
• So the film will be horizontal.
• All the oleic acid molecules will spread in water.
• No molecule will be stacked above another one.
• So the film will have a thickness of one molecule.
• Also it will be made entirely made up of oleic acid molecules.
■ This circular film is the disc that we wanted in step (1).
(v) Measure the radius (r) of the film.
7. We have put one drop in the water.
• From the result in (5), we have: volume of oleic acid put in water = 0.0025Vd cm3
• If we put n drops, volume of oleic acid put in water = 0.0025nVd cm3
• Then the disc is made up of 0.0025nVd cm3 of oleic acid.
8. From the result in (2), we get: 0.0025nVd = V = 𝝅r2t.
From this we get: $\mathbf\small{t=\frac{0.0025\,n\,V_d}{\pi r^2} \text{cm}}$
Now we will see some solved examples involving very small lengths.
Solved example 2.7
Model of an atom is to be made for a science exhibition. The actual size of the nucleus is in the range of 10-15 to 10-14 m. The actual size of the atom is about 10-10 m. In the model, the nucleus should have the size of the tip of a sharp pin. The actual size of the tip of a sharp pin is in the range of 10-5 to 10-4 m. What should be the total size of the model ?
Solution:
1. Actual size of a nucleus = 10-15 to 10-14 m.
• Size of the nucleus in the model = 10-5 to 10-4 m.
2. If we decide to blow up 10-15 to 10-5, we are to multiply 10-15 by 1010
• If we decide to blow up 10-14 to 10-4, then also, we are to multiply 10-14 by 1010
■ So the scale factor is 1010
3. That means, we have to multiply all 'actual lengths' by 1010
• So the size of the atom in the model = (10-10×1010) = 100 m = 1 m.
4. Thus we can write:
■ The model of the atom should be a sphere of 1 m diameter.
■ A very small point (with the size equal to the tip of a sharp pin) at the center of the model, will represent the nucleus.
Note: The doors of rooms in an ordinary residential building have widths equal to 0.9 m. This is 10 cm less than 1 m. So if the model is made while inside the room, we will not be able to take it to the exhibition hall.
Solved example 2.8
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by $\mathbf\small{\mathring{A}}$. (1 $\mathbf\small{\mathring{A}}$ = 10-10 m). The radius of a hydrogen atom is about 0.5 $\mathbf\small{\mathring{A}}$. What is the total atomic volume in m3 of a mole of hydrogen atoms ?
Solution:
1. The radius of a hydrogen atom is given as 0.5 $\mathbf\small{\mathring{A}}$
• So radius of one hydrogen atom in m = 0.5×10-10 m.
• So volume of 1 hydrogen atom in m3 = $\mathbf\small{\frac{4}{3}\pi\,r^3=\frac{4}{3}\pi\,(0.5\times10^{-10})^3=5.236\times10^{-31}\;\text{m}^3}$
2. 1 mole of hydrogen atoms will contain 6.02214 × 1023 hydrogen atoms.
• So total volume = ( 6.02214 × 1023 × 5.236 × 10-31) = 3.15 × 10-7 m3
Solved example 2.9
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the radius of hydrogen molecule to be about 0.5Å). Why is this ratio so large ?
Solution:
1. The radius of a hydrogen molecule is given as 0.5$\mathbf\small{\mathring{A}}$
• So radius of one hydrogen molecule in m = 0.5×10-10 m.
(∵ 1 $\mathbf\small{\mathring{A}}$ = 10-10 m)
• So volume of 1 hydrogen molecule in m3 = $\mathbf\small{\frac{4}{3}\pi\,r^3=\frac{4}{3}\pi\,(0.5\times10^{-10})^3=0.5236\times10^{-30}\;\text{m}^3}$
2. 1 mole of hydrogen molecules will contain 6.02214 × 1023 hydrogen molecules.
• So total volume = ( 6.02214 × 1023 × 0.5236 × 10-30) = 3.1532 × 10-7 m3
3. 22.4 L = 22.4 × 1000 = 22400 cm3 = 22400 × 10-6 m3
4. So the ratio = $\mathbf\small{\frac{\text{molar volume}}{\text{atomic volume}}=\frac{22400 \times 10^{-6}}{3.1532\times 10^{-7}}=7.104\times 10^4}$
5. This is a very large ratio. We can see that the numerator is very large when compared to the denominator. The reason can be written as follows:
(i) The numerator is the molar volume.
At standard temperature and pressure, the hydrogen will be in the gaseous state. There will be large spaces in between it's molecules. So one mole of hydrogen molecules (6.02214 × 1023 hydrogen molecules) at STP will occupy 22.4 L.
(ii) The denominator is the volume when all the 6.02214 × 1023 hydrogen molecules are held together with no space in between. Since the hydrogen molecules are very small, the total volume will be very small even if 6.02214 × 1023 molecules are held together.
(iii) Thus the numerator (22400 × 10-6) is very large when compared to the denominator (3.1532 × 10-7).
Range of lengths
1. In our day to day life, we come across ordinary lengths such as:
♦ Length of a pen.
♦ Length of a table.
♦ Height of a building.
♦ Distance between two towns etc.,
2. But when we do problems in science, we will have to deal with a range of lengths.
• On the left end of that range, we have very small lengths.
Some examples:
♦ Size of a proton (10-15 m).
♦ Size of a nucleus (10-14 m).
• On the right end of that range, we have very large lengths.
Some examples:
♦ Distance from earth to Andromeda galaxy (2.4 × 1022 m) .
♦ Distance from earth to the end of the observable universe (8.8 × 1026 m).
3. In addition to the above two extremes, we must be able to deal with any given lengths which fall in between them.
In the next section, we will see measurement of mass and time.
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