In the previous section we completed a discussion on ideal gas. In this section, we will see kinetic molecular theory of gases. Later in this section, we will see real gases also
We have seen the three gas laws:
♦ At constant T and n, $\mathbf\small{\rm{V =k_1 \times \frac{1}{p}}}$ (Boyle's law)
♦ At constant p and n, V = K2 × T (Charles' law)
♦ At constant p and T, V = K2 × n (Avogadro law)
• All the above laws were formulated, based on experimental observations. No theoretical calculations were involved
• Let us write analysis about the relation between 'experimental methods' and 'theoretical methods'. It can be written in 15 steps:
1. Consider Boyle’s law: At constant T and n, $\mathbf\small{\rm{V =k_1 \times \frac{1}{p}}}$
2. We can find the value of k1 very easily:
♦ Apply a certain pressure p1 and measure the corresponding volume V1.
♦ Input those values in the equation and calculate k1.
3. To confirm, take another set of readings (p2, V2) at the same temperature
♦ Input p2 and V2 in the equation. We will get the same k1.
4. Several trials are done like this so that, the value of k1 can be reported with confidence
5. In the same way, trials are done in the laboratory to find the values of:
♦ k2 (Charles' law)
♦ k3 (Avogadro law)
6. So it is clear that, the three laws are based on experiments
♦ k3 (Avogadro law)
6. So it is clear that, the three laws are based on experiments
♦ If we perform the experiments with utmost care and precision, we will get accurate values
♦ Such accurate values will give sufficient proofs for the laws
7. Once the laws were proved, scientists began to think about the next steps
• They wanted to find the reason for those behavior of gases
• That is., scientists wanted to know these:
♦ When pressure increases, why does volume decrease?
♦ When temperature increases, why does volume increase?
♦ When the number of moles increases, why does volume increase?
8. After much research and discussions, scientists put forward the kinetic molecular theory
• This theory tries to give proper explanation for the ‘experimental observations’
• We will now see the main postulates of the theory:
♦ The word 'postulate' has the following meaning:
✰ 'Some thing which is suggested' so as to get a good basis for a reasoning or discussion
✰ 'Some thing which is assumed to be true' so as to get a good basis for a reasoning or discussion
✰ The dictionary meaning can be seen here.
9. Postulate 1:
This postulate is about 'size of particles'. It can be explained in 7 steps:
(i) Take a sample of any gas
♦ There will be a large number of particles in that sample
(ii) These particles may be atoms or molecules
• For example:
♦ If it is a neon sample. the particles will be Ne atoms
♦ If it is a carbon dioxide sample, the particles will be CO2 molecules
(iii) All the particles in a sample will be identical
♦ All those particles will be very small
♦ All those particles will be very far apart
(iv) We have to make a careful comparison between ‘very small’ and ‘very far apart’
Let us see an example:
(iv) We have to make a careful comparison between ‘very small’ and ‘very far apart’
Let us see an example:
• Consider two cricket balls placed at a distance of 15 cm apart
♦ That is., the distance between the centers of the two balls is 15 cm
✰ The center of the first ball is at A
✰ The center of the second ball is at B
✰ The distance AB is 15 cm
• We would say that: 'the two balls are close to each other'
• Now, remove the two balls
✰ Place a N2 molecule at A
✰ Place another N2 molecule at B
• We would say: 'the two molecules are very far apart'
• This is because, compared to the 15 cm distance, the size of the N2 molecules (radius of a N2 molecule is 155 pm) is very very small
♦ Two nitrogen molecules at a distance of 15 cm apart
♦ is equivalent to
♦ Two cars at a distance of 18,75,000 km apart
■ So in a gas sample, the particles are very far apart
(v) Also, there is another important assumption related to this ‘large distance’
(v) Also, there is another important assumption related to this ‘large distance’
■ When compared to the large distance between them, the particles are so small that, their volumes are ignored
• That is., the particles are considered as ‘point masses’
• Their lengths or volumes are not included in the mathematical calculations
(vi) So this postulate gives us a satisfactory explanation for the ‘large compressibility’ of gases
(vi) So this postulate gives us a satisfactory explanation for the ‘large compressibility’ of gases
♦ Gases are highly compressible because, there is enough space available between particles
♦ In solids and liquids, the particles are already closely packed. We cannot compress them further
(vii) Note that, for our present discussion, we are considering ‘pure samples only’
(vii) Note that, for our present discussion, we are considering ‘pure samples only’
• That is., all the particles in the sample must be identical. The sample must not contain impurities like dust particles or molecules of water (water vapour)
10. Postulate 2:
This postulate is about 'interaction between particles'. It can be explained in 3 steps:
This postulate is about 'interaction between particles'. It can be explained in 3 steps:
(i) There is no force of attraction between the particles of a gas
(ii) This is readily proved because, if there was any such attraction, the particles would prefer to stay close to each other, forming a group
(iii) But we see that, the gas particles travel away from one another and occupy every corner of the container
11. Postulate 3:
This postulate is about 'motion of particles'. It can be explained in 5 steps:
(ii) This is readily proved because, if there was any such attraction, the particles would prefer to stay close to each other, forming a group
(iii) But we see that, the gas particles travel away from one another and occupy every corner of the container
11. Postulate 3:
This postulate is about 'motion of particles'. It can be explained in 5 steps:
(i) Particles of a gas are always in constant and random motion
(ii) We think that, we already know this postulate. However, we have to pay special attention to the two words: ‘constant’ and ‘random’
(iii) ‘Constant motion’ tells us that, we will never find any particle which is at rest. All the particles will be ‘always in motion’
(ii) We think that, we already know this postulate. However, we have to pay special attention to the two words: ‘constant’ and ‘random’
(iii) ‘Constant motion’ tells us that, we will never find any particle which is at rest. All the particles will be ‘always in motion’
• If, even some of the particles were able to take rest for small intervals of time, we would observe a ‘some what definite shape’ during those small intervals. In reality, we do not see such shapes
(iv) ‘Random motion’ tells us that, there is is no specific direction. Particles can travel in any possible directions
(iv) ‘Random motion’ tells us that, there is is no specific direction. Particles can travel in any possible directions
(v) It may also be noted that, the path taken by any particle will be linear. We will not see any particle travelling along curved paths
12. Postulate 4:
This postulate is about 'collision between particles'. It can be explained in 3 steps:
12. Postulate 4:
This postulate is about 'collision between particles'. It can be explained in 3 steps:
(i) We have seen that, particles of a gas move randomly. That is., in all possible directions
(ii) During this random motion,
♦ They collide with each other
♦ They collide with the walls of the container
(iii) The 'pressure experienced by the walls of the container' is due to the second collision mentioned above
(ii) During this random motion,
♦ They collide with each other
♦ They collide with the walls of the container
(iii) The 'pressure experienced by the walls of the container' is due to the second collision mentioned above
♦ That is., the 'collision with the walls of the container'
13. Postulate 5:
13. Postulate 5:
This postulate is about 'elasticity of collision'. It can be explained in 7 steps:
(i) All the collisions occurring in a gas sample are perfectly elastic
• Elastic collision and non-elastic collision can be explained as follows:
(ii) Consider two particles colliding with each other
• During the collision, both of them will deform a bit
(iii) For deformation to occur, some energy is required
• For example, energy is required to stretch a rubber band
(iv) That means, during collision, some energy is lost
(v) But after the collision, the particles (if elastic) will soon regain their original shapes
• For example, if we let go off a stretched rubber band, it will regain the original shape
(vi) When a particle revert back to the original shape, energy will be released
(vii) Now consider the two energies (A and B):
A. Energy used up during deformation
B. Energy released when original shape is regained
■ If the collision is perfectly elastic, A and B will be equal
■ So the net effect is that, there is no loss of energy
• Elastic collision and non-elastic collision can be explained as follows:
(ii) Consider two particles colliding with each other
• During the collision, both of them will deform a bit
(iii) For deformation to occur, some energy is required
• For example, energy is required to stretch a rubber band
(iv) That means, during collision, some energy is lost
(v) But after the collision, the particles (if elastic) will soon regain their original shapes
• For example, if we let go off a stretched rubber band, it will regain the original shape
(vi) When a particle revert back to the original shape, energy will be released
(vii) Now consider the two energies (A and B):
A. Energy used up during deformation
B. Energy released when original shape is regained
■ If the collision is perfectly elastic, A and B will be equal
■ So the net effect is that, there is no loss of energy
(Some details about elastic collisions can be seen here)
(viii) We have got enough evidence that, collisions in the gas are perfectly elastic. This can be written in steps:
♦ If there was loss of energy, the particles will gradually begin to move less and less vigorously
♦ After some time they will stop moving
♦ We will see that, the gas has settled down
♦ But we never observe such a situation in real life
14. Postulate 6:
This postulate is about the 'speed of particles'. It can be explained in 9 steps:
(viii) We have got enough evidence that, collisions in the gas are perfectly elastic. This can be written in steps:
♦ If there was loss of energy, the particles will gradually begin to move less and less vigorously
♦ After some time they will stop moving
♦ We will see that, the gas has settled down
♦ But we never observe such a situation in real life
14. Postulate 6:
This postulate is about the 'speed of particles'. It can be explained in 9 steps:
(i) Let us observe a gas sample for a time interval ‘t’
(ii) Consider any instant t1 during that time interval
(iii) Consider any one particle at that instant. Note down it’s velocity ‘v(t1)’ at that instant
♦ That ‘v(t1)’ will be different from the velocities of all other particles
♦ That means, at any instant, the particles will be having different velocities
(iv) Also, the ‘v(t1)’ that we noted down, will change at the very next instant
(iv) Also, the ‘v(t1)’ that we noted down, will change at the very next instant
♦ That means, the velocities of all the particles are continuously changing
(v) This ‘continuous change in speed’ is due to the ‘continuous collisions’
• When two particles collide, their original speeds will change
(vi) In physics classes, we will see some more details about such collisions. Here we will write some basics in steps:
• Consider any instant t1
(v) This ‘continuous change in speed’ is due to the ‘continuous collisions’
• When two particles collide, their original speeds will change
(vi) In physics classes, we will see some more details about such collisions. Here we will write some basics in steps:
• Consider any instant t1
♦ Note down the temperature T of the sample
♦ Let the particles be numbered as: 1, 2, 3, 4, . . .
♦ Write down the individual speeds (v1(t1), v2(t1), v3(t1), . . . ) of each of the particles at that instant
♦ Write down the individual speeds (v1(t1), v2(t1), v3(t1), . . . ) of each of the particles at that instant
♦ A special type of ‘mathematical average’ of those speed values is calculated
✰ This 'mathematical average' is denoted as: $\mathbf\small{\bar{v}}$
✰ So at the instant t1, we can denote it as: $\mathbf\small{\bar{v}_{(t1)}}$
♦ This $\mathbf\small{\bar{v}}$ is applicable to all the particles
✰ That means., $\mathbf\small{\bar{v}}$ is a characteristic value of the sample as a whole
(vii) Remember that, the speed values (v1(t1), v2(t1), v3(t1), . . . ) were written down at a particular instant t1. At any other instant, the particles will be having different velocities from these
✰ That means., $\mathbf\small{\bar{v}}$ is a characteristic value of the sample as a whole
(vii) Remember that, the speed values (v1(t1), v2(t1), v3(t1), . . . ) were written down at a particular instant t1. At any other instant, the particles will be having different velocities from these
(viii) Consider any other instant (t2)
♦ The temperature must be the same T at the first instant (t1)
♦ Note down the velocities (v1(t2), v2(t2), v3(t2), . . . )
♦ Calculate $\mathbf\small{\bar{v}_{(t2)}}$
(ix) If the two temperatures are the same, $\mathbf\small{\bar{v}_{(t1)}}$ will be equal to $\mathbf\small{\bar{v}_{(t2)}}$
(ix) If the two temperatures are the same, $\mathbf\small{\bar{v}_{(t1)}}$ will be equal to $\mathbf\small{\bar{v}_{(t2)}}$
■ So it is clear that:
Though the individual speeds continuously change, the $\mathbf\small{\bar{v}}$ remains constant at a particular temperature
Though the individual speeds continuously change, the $\mathbf\small{\bar{v}}$ remains constant at a particular temperature
• We will see details about this $\mathbf\small{\bar{v}}$ in physics classes
15. Postulate 7:
This postulate is about the 'energy of particles'. It can be explained in 6 steps:
(i) We saw that the velocity of any particle changes continuously
♦ So the kinetic energy will also change continuously
(ii) But we have seen that, if the temperature is constant, $\mathbf\small{\bar{v}}$ will be a constant
(i) We saw that the velocity of any particle changes continuously
♦ So the kinetic energy will also change continuously
(ii) But we have seen that, if the temperature is constant, $\mathbf\small{\bar{v}}$ will be a constant
(iii) So, if instead of using the individual velocities, we use $\mathbf\small{\bar{v}}$, we will get constant kinetic energy
♦ The kinetic energy calculated using $\mathbf\small{\bar{v}}$ is called average kinetic energy
(iv) So it is clear that:
♦ If temperature remains constant, the average kinetic energy of the sample will be a constant
(v) From this, we can write:
• Each temperature has a particular value of 'average kinetic energy' associated with it
♦ If the temperature increases, the average K.E increases
♦ If the temperature decreases, the average K.E decreases
(vi) So, when temperature increases, the particles will hit the walls of the container with greater force
The walls will experience greater pressure
• We can write:
♦ When temperature increases, the pressure exerted by the gas increases
♦ When temperature decreases, the pressure exerted by the gas decreases
• So we have completed a discussion on all the postulates of the kinetic theory of gases
♦ We have seen three gas laws in the previous sections
♦ All three of them can be derived theoretically using the kinetic molecular theory
• Scientists have made the comparison between the two items:
♦ Results of the experiments performed in the labs
♦ Results obtained by theoretical calculations using kinetic molecular theory
■ The two results are found to be the same
♦ Since the two results are the same, we can say with confidence that, the kinetic model is correct
Deviation from ideal gas behaviour
• Most gases obey Boyle’s law at normal pressures
♦ But if we increase the pressure, the gases begin to show deviations
• Let us first see what those deviations are. It can be written in 3 steps:
1. The deviation can be visualized if we plot the pV vs p graph
1. The deviation can be visualized if we plot the pV vs p graph
• The significance of this graph can be written in 5 steps:
(i) We know that, according to Boyle’s law:
(i) We know that, according to Boyle’s law:
♦ For all values of p, the product pV will be a constant k1.
(ii) So, if we plot pV along the y axis, and p along the x-axis, the graph will be a horizontal line
♦ This horizontal line will pass through ‘k1 on the y-axis’
♦ This horizontal line will pass through ‘k1 on the y-axis’
(iii) So, to test the behavior of a gas, scientists plot the pV vs p graph of that gas
This is shown in fig.5.20 below:
This is shown in fig.5.20 below:
 |
| Fig.5.20 |
(iv) We see that
♦ For H and He, pV increases when p increases
♦ For CO and CH4, pV decreases initially
♦ For CO and CH4, pV decreases initially
✰ They decrease up to certain minimum values
✰ After that, they increase
(v) So it is clear that, real gases do not obey gas laws under all conditions
✰ After that, they increase
(v) So it is clear that, real gases do not obey gas laws under all conditions
2. Now a question arises:
■ While doing the experiments, did Robert Boyle notice these deviations?
• The answer can be written in 5 steps:
(i) Consider the graph shown in fig.5.21(a) below:
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| Fig.5.21 |
On the x-axis, we see pressure values: 200, 400, 600, . . . so on . . .
(ii) Consider the graphs shown in fig.5.21(b) above
On the x-axis, we see pressure values: 2, 4, 6, . . . so on . . .
(iii) That means, the two graphs are drawn in different scales
(iv) When the experiments are done at low pressure values, the graphs are very close to the horizontal dashed line
• Robert Boyle did the experiments at low pressures. He would not see 'appreciable deviations'
(v) All the values in fig.b are present inside fig.a
• But since the 'scale of fig.a' is large, we get the impression that, the graphs 'deviate quickly' from the horizontal dashed line
3. Next, we will see another method for visualizing the 'deviation':
• This method uses the p vs V graph. It is shown in fig.5.22(a) below:
 |
| Fig.5.22 |
The significance of this graph can be written in 5 steps:
(i) The red curve is plotted using the equation: $\mathbf\small{\rm{p=k_1 \times \frac{1}{V}}}$
♦ That means, it is the theoretical curve
(ii) The blue curve is plotted using data obtained in experiments
♦ That means, it is the experimental curve
(iii) Mark a point p1 on the y-axis. This is shown in fig.b
♦ That means, it is the theoretical curve
(ii) The blue curve is plotted using data obtained in experiments
♦ That means, it is the experimental curve
(iii) Mark a point p1 on the y-axis. This is shown in fig.b
♦ p1 is high up on the y-axis. That means, p1 is a high pressure value
• We want the volumes corresponding to p1
• For that, we draw a horizontal dashed line through p1
♦ This dashed line meets the red curve at A
♦ This dashed line meets the blue curve at A'
(v) We draw vertical dashed lines through A and A'
♦ The vertical dashed line through A meets the x-axis at V1
♦ This dashed line meets the blue curve at A'
(v) We draw vertical dashed lines through A and A'
♦ The vertical dashed line through A meets the x-axis at V1
♦ The vertical dashed line through A' meets the x-axis at V1’
• That means,
♦ V1 is the ideal volume corresponding to the pressure p1
♦ V1 is the ideal volume corresponding to the pressure p1
♦ V1’ is the real volume corresponding to the pressure p1
• We see that, V1’ is greater than V1
■ That means, the actual volume is greater than the ‘volume calculated theoretically’
(iv) Mark a point p2 on the y-axis
(iv) Mark a point p2 on the y-axis
♦ p2 is low down on the y-axis. That means, p2 is a low pressure value
• We want the volumes corresponding to p2
• For that, we draw a horizontal dashed line through p2
♦ This dashed line meets the red curve at C
♦ This dashed line meets the blue curve at D
• We draw vertical dashed lines through C and D
♦ The vertical dashed line through C meets the x-axis at V2
♦ This dashed line meets the blue curve at D
• We draw vertical dashed lines through C and D
♦ The vertical dashed line through C meets the x-axis at V2
♦ The vertical dashed line through D meets the x-axis at V2’
• That means,
♦ V2 is the ideal volume corresponding to the pressure p2
♦ V2 is the ideal volume corresponding to the pressure p2
♦ V2’ is the real volume corresponding to the pressure p2
• We see that, V2’ is nearly equal to V2
■ That means, the actual volume is nearly equal to the ‘volume calculated theoretically’
(v) So we can write:
♦ At low pressures, real gases obey Boyle’s law
♦ At high pressures, real gases deviate from Boyle’s law
(v) So we can write:
♦ At low pressures, real gases obey Boyle’s law
♦ At high pressures, real gases deviate from Boyle’s law
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