Chapter 5.5 - Real gas

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 = K₃ × 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

    ♦ 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:

• 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’

■ 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

    ♦ 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’

• 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:

(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:

(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’

• 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

(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:

(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

    ♦ That is., the 'collision with the walls of the container'
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

(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:

(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

    ♦ 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

    ♦ Note down the temperature T of the sample

    ♦ Let the particles be numbered as: 1, 2, 3, 4, . . .
    ♦ Write down the individual speeds (v_1(t1), v_2(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 (v_1(t1), v_2(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 (v_1(t2), v_2(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 $\mathbf\small{\bar{v}_{(t2)}}$

■ So it is clear that:
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

(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

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• 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

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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

• The significance of this graph can be written in 5 steps:
(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’

(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:

Fig.5.20

(iv) We see that

    ♦ For H and He, pV increases when p increases
    ♦ 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

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:

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

    ♦ 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

    ♦ 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 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

    ♦ 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

    ♦ 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 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

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• In the next section, we will see the reasons for the deviation

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Chapter 5.1 - Boyle's Law

In the previous section, we saw the basic details about inter molecular forces and thermal energy. In this section, we will see gaseous state

• Consider the elements like hydrogen, helium, oxygen, nitrogen etc.,
    ♦ They exist in the gaseous state even when temperature and pressure are normal
    ♦ In this section, we will be discussing about such gases
• By altering the normal temperature and pressure, some liquids and solids can be converted into gaseous state. We will not be discussing them here

---

■ The Anglo-Irish scientist Robert Boyle did extensive research on the properties of gases. He published his findings as the Boyle’s Law
■ The law states that:

At constant temperature, the pressure of a fixed amount of gas varies inversely with it’s volume

• We can write an explanation in 29 steps:

1. Take a sample of a gas

• Note down the number of moles present in it

Some examples:

• If the gas is N₂, and if ‘m’ grams of N₂ is present in that sample, then:

    ♦ There will be $\mathbf\small{\rm{\frac{m}{28}}}$ moles of N₂ in that sample

• If the gas is CO₂, and if ‘m’ grams of CO₂ is present in that sample, then:

    ♦ There will be $\mathbf\small{\rm{\frac{m}{44}}}$ moles of CO₂ in that sample

2. Note down the temperature T of the sample

    ♦ Let it be 200 K

3. Apply a pressure p₁ on that sample

    ♦ Measure the resulting volume V₁

• Apply another pressure p₂ on that sample at the same temperature

    ♦ Measure the new volume v₂

4. A number of pressure-volume readings can be taken in this way

• We get a set of readings: (p₁, V₁), (p₂, V₂), (p₃, V₃), . . .

■ For all the readings, the temperature T must be the same as noted in (2)

■ The number of moles should also be the same

    ♦ It is not difficult to keep the ‘number of moles same’ 

    ♦ Because, the same sample is used for all readings

5. Boyle found out that, there is a definite relation between the applied pressure and resulting volume

• He found out that, the pressure is inversely proportional to volume

('Inversely proportional' means, when one quantity increases, the other quantity decreases and vice versa)

• That is: $\mathbf\small{\rm{p \propto \frac{1}{V}}}$

• This can be written as: $\mathbf\small{\rm{p=k_1 \times  \frac{1}{V}}}$

    ♦ Where k₁ is the constant of proportionality

6. Applying the equation to the first reading, we get: $\mathbf\small{\rm{p_1=k_1 \times  \frac{1}{V_1}}}$

$\mathbf\small{\rm{\Rightarrow p_1 V_1=k_1}}$

• Applying the equation to the first reading, we get: $\mathbf\small{\rm{p_2=k_1 \times  \frac{1}{V_2}}}$

$\mathbf\small{\rm{\Rightarrow p_2 V_2=k_1}}$

• Applying the equation to the third reading, we get: $\mathbf\small{\rm{p_3=k_1 \times  \frac{1}{V_3}}}$

$\mathbf\small{\rm{\Rightarrow p_3 V_3=k_1}}$

so on . . .
7. Since all results are k₁, we get: $\mathbf\small{\rm{p_1 V_1=p_2 V_2=p_3 V_3\;.\;.\;.\;so\;on\;.\;.\;.}}$

8. Let us plot the relation: $\mathbf\small{\rm{p=k_1 \times  \frac{1}{V}}}$

    ♦ We can plot V₁, V₂, V₃, . . . along the x-axis

    ♦ We can plot p₁, p₂, p₃, . . . along the y-axis

• The resulting graph is the blue curve shown in fig.5.4(a) below

Fig.5.4

■ In this graph, all the readings were taken when the temperature of the sample was 200 K. If there is any change in this temperature, while any of the readings are taken, we will not get this shape

■ Also, if there is any change in the 'number of moles in the sample', we will not get this shape
9. The shape of the graph in fig.5.4(a) seems to be peculiar

• But it is a common shape, seen very often in science and engineering

    ♦ It is the graph of: $\mathbf\small{\rm{y= A\;constant \times  \frac{1}{x}}}$
    ♦ Note the similarity between [$\mathbf\small{\rm{y= A\;constant \times  \frac{1}{x}}}$] and [$\mathbf\small{\rm{p=k_1 \times  \frac{1}{V}}}$]

• The graph of [$\mathbf\small{\rm{y= 1 \times  \frac{1}{x}}}$] is plotted in fig.5.4(b). Here, the value of the 'constant' is 1. Note the similarity between the two graphs

• Whenever two quantities have a 'inverse proportional relationship', the graph will have this shape  

10. Next we repeat the experiment on the same sample

    ♦ That is., we take the readings (p₁, V₁), (p₂, V₂), (p₃, V₃), . . . again

    ♦ But this time, the temperature must be another convenient value (say 400 K) all the while

    ♦ This time, the graph is the green curve in fig.5.5 (a) below:

Fig.5.5

11. We repeat the experiment one more time

    ♦ That is., we take the readings (p₁, V₁), (p₂, V₂), (p₃, V₃), . . . one more time

    ♦ Again this time, the temperature must be another convenient value (say 600 K) all the while

    ♦ This time, the graph is the red curve in fig.5.5 (a) above

12. So in fig.5.5(a):

    ♦ All the readings in the blue curve are taken when the temperature is 200 K

    ♦ All the readings in the green curve are taken when the temperature is 400 K    

    ♦ All the readings in the red curve are taken when the temperature is 500 K

■ Since the temperature is constant for any curve, each curve in the p-V graph in fig.5.5(a) is known as an isotherm

13. The constants will be different

    ♦ The constant k₁ of the blue curve

          ✰ will be different from

    ♦ The constant k₁ of the green curve

          ✰ will be different from

    ♦ The constant k₁ of the red curve

• We obtain different curves in the fig.5.5(a) because, the constants are all different

    ♦ If the constants were the same, the curves would overlap

14. It is possible to obtain different curves with [$\mathbf\small{\rm{y= A\;constant \times  \frac{1}{x}}}$] also. This is shown in fig.5.5(b)

• In fig.5.5(b), we see that:

    ♦ For the blue curve, the constant is 1

    ♦ For the green curve, the constant is 2

    ♦ For the green curve, the constant is 3

15. Now we know the details about the 'graphical representation of Boyle's Law'

• Let us see another method for making the graphical representation

• The next seven steps from (15) to (21) explains this method:

16. Consider the readings that we obtained in (4):

(p₁, V₁), (p₂, V₂), (p₃, V₃), . . .

• Before plotting them, we convert them into the form:

(p₁, V'₁), (p₂, V'₂), (p₃, V'₃), . . .

    ♦ Where: $\mathbf\small{\rm{V'_1 = \frac{1}{V_1},\;\;\;\;V'_2 = \frac{1}{V_2},\;\;\;\;V'_3 = \frac{1}{V_3},\;.\;.\;.\;so\;on\;.\;.\;.}}$

• That means, we have to pick out each volume and write it's reciprocal

17. Now, when we apply the Boyle's law, we get:

$\mathbf\small{\rm{p\;=\;k_1 \times \frac{1}{V}\;=\;k_1 \times V'}}$

$\mathbf\small{\rm{\Rightarrow\;p\;\;=\;\;k_1 \times V'}}$

18. So we plot a p vs V' graph

    ♦ We can plot V'₁, V'₂, V'₃, . . . along the x-axis

    ♦ We can plot p₁, p₂, p₃, . . . along the y-axis

• The resulting graph is the blue line shown in fig.5.6(a) below:

Fig.5.6

■ In this graph, all the readings were taken when the temperature of the sample was 200 K. If there is any change in this temperature, while any of the readings are taken, we will not get this shape

■ Also, if there is any change in the 'number of moles in the sample', we will not get this shape
19. The shape of the graph in fig.5.6(a) is: straight line

• This is expected because:

    ♦ [$\mathbf\small{\rm{p\;\;=\;\;k_1 \times V'}}$] is the equation of a straight line

    ♦ It is of the form: [y = mx] that we commonly see in analytical geometry classes

20. Next we take the readings obtained in (9)

    ♦ We convert those readings into the form: (p₁, V'₁), (p₂, V'₂), (p₃, V'₃), . . .

    ♦ This time, the graph is the green line fig.5.6 (b) above

21. Next we take the readings obtained in (10)

    ♦ We convert the readings into the form: (p₁, V'₁), (p₂, V'₂), (p₃, V'₃), . . .

    ♦ This time, the graph is the red line fig.5.6 (b) above

22. So we have seen two methods of plotting the relation between pressure and volume

■ The 'straight line method' has an advantage. It can be explained in 3 steps:

(i) If we obtain straight lines as in fig.5.6 above, we can say that the sample obeys Boyle’s law

(ii) In practice, it is seen that, at high pressures, the ‘straight line character’ is not obtained

(iii) That means, at high pressure, gases deviate from Boyle’s law

23. Let us see some actual readings obtained during an experiment

• The readings are given in the table 5.1 below

    ♦ The experiment was done on 0.09 mol of CO₂

    ♦ All the readings were taken when the sample was at a temperature of 300 K

Table 5.1

The following 4 details can be written about the table:

(i) The first column gives the pressure values

• The actual values are: 2 × 10⁴, 2.5 × 10⁴ etc.,

• Those are very large values

• For convenience of putting them in the table, each of those values are divided by 10⁴

(ii) The second column gives the volume values

• The actual values are: 112 × 10^-3, 89.2 × 10^-3 etc., 

• Those are very small values

• For convenience of putting them in the table, each of those values are divided by 10^-3

(iii) The third column gives the reciprocals of the values in the second column

• For example: The reciprocal of 112 × 10^-3 of is 8.9

(iv) The fourth column gives the product of first and second columns

• The actual values are: 22.4 × 10², 22.3 × 10² etc.,

• Those are large values

• For convenience of putting them in the table, each of those values are divided by 10²

24. Based on the table, we can plot two graphs: p-V graph and p-1/V graph

• The p-V graph is shown in fig.5.7 below:

Fig.5.7

• Note that, only the actual values should be used for plotting

    ♦ The factored values in the table should not be used

• However, suitable scales can be used while plotting

    ♦ The 'scales' need not be same as the 'factors used in the table'

    ♦ The scales must be clearly mentioned at the top right corner of the graph

25. The p-1/V graph is shown in fig.5.8 below:

Fig.5.8

• If the line is extended backwards, it will pass through the origin

    ♦ This is shown by the yellow dashed line

• Indeed, if a line of the form [y = mx] is extended, it will pass through the origin

26. Let us analyse the p-V graph:

• It can be done in 4 steps:

(i) On the y-axis, mark any two convenient points P and Q as shown in fig.5.9 below:

Fig.5.9

(ii) From P, draw a horizontal dashed line to meet the curve at P'

    ♦ From P’, draw a vertical dashed line to meet the x-axis

    ♦ This vertical line meets the x-axis at P’’

    ♦ So the volume at P’’ is the ‘volume corresponding to the pressure at P’

(iii) Repeat the same procedure for the other point Q

    ♦ Q’’ is the ‘volume corresponding to the pressure at Q’

(iv) We see that:

    ♦ Q is higher up above P

    ♦ Q’’ is to the left of P’’

■ So it is clear that, when pressure increases, volume decreases

27. Let us see the effect of ‘decrease in volume’. It can be written in 5 steps

(i) We increase the pressure of a gas by compressing it

• when the gas is compressed, the volume decreases

■ So we can write:

When pressure increases, volume decreases

(ii) Note that, there is no change in the mass of the gas before and after compression

• That means, after compression, the mass occupies a lesser space

(iii) We have: $\mathbf\small{\rm{Density=\frac{Mass}{Volume}}}$

• From this, we get: $\mathbf\small{\rm{d_1=\frac{m}{V_1}\;\;and\;\;d_2=\frac{m}{V_2}}}$

    ♦ Where:

          ✰ m is the unchanging mass

          ✰ d₁ and V₁ are the initial density and volume 

          ✰ d₂ and V₂ are the final density and volume

(iv) We see that, the numerator m has no change

• But the denominator V₂ is less than V₁

• So d₂ will be larger than d₁

(v) That means, when we compress a gas, it becomes denser

• This is so because, after compression, the molecules have a lesser space to occupy

28. Modification of the Boyle’s law equation:

• We can modify the equation by including 'density' in it

• It can be written in 4 steps:

(i) We have the original equation: $\mathbf\small{\rm{p=k_1 \times  \frac{1}{V}}}$

(ii) Also we have: $\mathbf\small{\rm{d=\frac{m}{V}}}$

$\mathbf\small{\rm{\Rightarrow V=\frac{m}{d}}}$ 

(iii) Substituting for V in (i), we get:

$\mathbf\small{\rm{p=k_1 \times  \frac{1}{\frac{m}{d}}}}$

$\mathbf\small{\rm{\Rightarrow p=k_1 \times  \frac{d}{m}=\left(\frac{k_1}{m} \right)d}}$

$\mathbf\small{\rm{\Rightarrow d=\left(\frac{m}{k_1} \right)p}}$

(iv) Both m and k₁ are constants. So we can put a new constant k’

■ Thus we get: d = k'p

29.  Based on the modified equation in (27), we can write the following points:

(i) Take a sample of a gas

    ♦ Note down the temperature T of the sample

    ♦ Note down the mass (no. of moles) of the sample

(ii) Keeping T and m constant, apply various pressures: p₁, p₂, p₃, . . .

(iv) Determine the corresponding densities: d₁, d₂ d₃, . . .

(v) Thus we get various pairs: (p₁, d₁), (p₂, d₂), (p₃, d₃), . . . 

(vi) We will see that, each pair will satisfy the relation: d = k'p

• That means, the constant k’ is applicable to all pairs

(vii) So it is clear that, density is directly proportional to pressure

In other words: When pressure increases, density also increases

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• In the next section, we will see Charles' law

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