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 5.4 - Ideal Gas

In the previous section we completed a discussion on Avogadro law. We saw the details about STP also. In this section, we will see ideal gases

• A gas that follows Boyle's law, Charles' law and Avogadro law strictly is called an ideal gas
• From those three gas laws, we have three relations:
    ♦ At constant T and n, $\mathbf\small{\rm{V \propto \frac{1}{p}}}$ (Boyle's law)
    ♦ At constant p and n, V ∝ T (Charles' law)
    ♦ At constant p and T, V ∝ n (Avogadro law)
• Combining the three, we get: $\mathbf\small{\rm{V \propto \frac{nT}{p}}}$
■ Thus we get:
Eq.5.1: $\mathbf\small{\rm{V=R \frac{nT}{p}}}$
    ♦ R is the proportionality constant
          ✰ This R is called gas constant
    ♦ It is same for all gases
          ✰ So it is called Universal gas constant
• Eq.5.1 is called ideal gas equation

We want to find the value of R. It can be done in 4 steps:
1. To find R, we input all the known values into Eq.5.1
• Let us first write those known values:
• Consider one mole of a gas at STP
(i) Since there is only one mole, we can put n = 1
(ii) Since the gas is at STP, we can put:
    ♦ T = 273 K
(iii) Since the gas is at STP, we can put:
    ♦ p = 1 bar
(iv) We know that, one mole of a gas at STP will occupy 22.71 L
(v) Thus, R is the only unknown
2. We can substitute the known values in Eq.5.1 and obtain R
• But before that, we must convert all units into SI system
(i) n is a number. It can be written as such
(ii) T = 273 K is already in SI units
(iii) 1 bar is equal to 10⁵ Nm^-2
    ♦ So we must put p = 10⁵
(iv) 1 L is equal to 10^-3 m³
    ♦ So we must put V = 22.71 × 10^-3
3. Thus we get: $\mathbf\small{\rm{R=\frac{10^5 \times 22.71 \times 10^{-3}}{1 \times 273.15}}}$

= 8.314 Nm^-2 m³ K^-1 mol^-1

= 8.314 Pa m³ K^-1 mol^-1  (∵ 1 Nm^-2 = 1 Pa)

= 8.314 × 10^-2 bar L K^-1 mol^-1 (∵ [Pa m³] = [(10^-5 bar) × (10³ L)] = 10^-2 bar L)

= 8.314 J K^-1 mol^-1. (∵ [Pa m³] = [(Nm^-2) × (m³)] = [N m] = J)
4. We see that, there are four possible SI units
• If any of the units written in (2) are changed, the value '8.314' will also change
    ♦ For example, if we put '32 ^oFahrenheit' in the place of '273 K', we will not get '8.314'
■ So we can write:
The value of R depends upon the units in which p, V and T are measured

---

• The Eq.5.1 is a relation between four variables
• Using that equation, we can describe the state of any gas
• So Eq.5.1 is called equation of state 

Next, we will discuss about combined gas law. It can be written in 4 steps:
1. Let us rearrange the Eq.5.1 as: $\mathbf\small{\rm{\frac{pV}{T}=nR}}$
• If we perform experiments on the same sample of a gas, n will be a constant
• Then the 'right side as a whole' will become a constant
2. In such a situation, we measure the initial pressure, initial volume and initial temperature: p₁, V₁ and T₁
• These values will satisfy the rearranged equation in (1)
• So we get: $\mathbf\small{\rm{\frac{p_1V_1}{T_1}=nR}}$
3. Change the pressure to a new value p₂
• Measure the corresponding V₂ and T₂
• These values will also satisfy the rearranged equation in (1)
• So we get: $\mathbf\small{\rm{\frac{p_2V_2}{T_2}=nR}}$
4. Change the pressure to a new value p₃
• Measure the corresponding V₃ and T₃
• These values will also satisfy the rearranged equation in (1)
• So we get: $\mathbf\small{\rm{\frac{p_3V_3}{T_3}=nR}}$
so on . . .
4. All results are equal to nR
• So we can write: $\mathbf\small{\rm{\frac{p_1 V_1}{T_1}=\frac{p_2V_2}{T_2}=\frac{p_2V_2}{T_2}\;.\;.\;.\;so\;on\;.\;.\;.}}$
• From this we get:
Eq.5.2: $\mathbf\small{\rm{\frac{p_1 V_1}{T_1}=\frac{p_2V_2}{T_2}}}$
■ This equation is known as Combined gas law

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We can bring density and molar mass also into the calculations. It can be written in 3 steps
1. Eq.5.1 can be rearranged as: $\mathbf\small{\rm{\frac{n}{V}=\frac{p}{RT}}}$
2. If m is the mass of the sample and M is the molar mass, we get:
• Number of moles (n) = $\mathbf\small{\rm{\frac{m}{M}}}$
• Substituting this in (1), we get: $\mathbf\small{\rm{\frac{m}{MV}=\frac{p}{RT}}}$
3. But $\mathbf\small{\rm{\frac{m}{V}}}$ is the density (d). So the result in (2) becomes: $\mathbf\small{\rm{\frac{d}{M}=\frac{p}{RT}}}$
• Rearranging this, we get:
Eq.5.3: $\mathbf\small{\rm{M=\frac{dRT}{p}}}$

Dalton’s law of partial pressures

• John Dalton did extensive research works on mixture of gases
• His findings were published as Dalton’s law of partial pressures
■ The law states that:
The total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases

• This can be explained in 8 steps:

1. In fig.5.19(a) below, a gas mixture is occupying a certain volume
• The 'occupied volume' is indicated by the grey rectangle

Fig.5.19

2. The gas mixture consists of three gases
• Gas 1, Gas 2 and Gas 3
    ♦ All molecules of Gas 1 are yellow in color
    ♦ All molecules of Gas 2 are red in color 
    ♦ All molecules of Gas 3 are green in color
3. All molecules of Gas 1 are separated from the mixture
    ♦ Those separated yellow molecules are put in a second container 
    ♦ This is shown in fig.b
4. All molecules of Gas 2 are separated from the mixture
    ♦ Those separated red molecules are put in a third container 
    ♦ This is shown in fig.c
5. All molecules of Gas 3 are separated from the mixture
    ♦ Those separated green molecules are put in a fourth container 
    ♦ This is shown in fig.d
6. The following three conditions must be satisfied:
(i) The gases in the mixture in fig.a must not react with each other
(ii) All the four containers must have the same volume
(iii) All the four containers must be at the same temperature
7. Relation between pressures:
    ♦ Let the pressure exerted by the mixture in fig.a be p_Total

    ♦ Let the pressure exerted by Gas 1 in fig.b be p₁

    ♦ Let the pressure exerted by Gas 2 in fig.c be p₂

    ♦ Let the pressure exerted by Gas 3 in fig.d be p₃

■ Then, according to Dalton’s law, we get:

p_Total = p₁ + p₂ + p₃ (at constant T, V)

■ This relation is applicable to any number of gases. So we can write:

Eq.5.4: p_Total = p₁ + p₂ + p₃ + . . . (at constant T, V)

• This is the mathematical form of Dalton's law of partial pressures

8. p₁, p₂, p₃. . . , which are the 'pressures exerted by individual gases in the mixture' are called partial pressures

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Let us see a practical application of the law. It can be written in 5 steps:

1. Sometimes gases like CO₂, CH₃ etc., gets collected above water surface

• Then the 'gas collected above the water surface', will be a mixture of the original gas and water vapour

2. We are able to measure the 'pressure exerted by the mixture'

• But that measured pressure will include the pressure exerted by water vapour also

• We want the pressure exerted by the original gas alone

3. For that, we apply Dalton’s law of partial pressure

• We have:  p_Total = p_{Water vapour} + p_{Original gas}.

    ♦ So, if we subtract p_{Water vapour} from p_Total, we will get p_{Original gas}.

    ♦ That is., p_{Original gas} = p_Total - p_{Water vapour} 

4. So our next task is to find p_{Water vapour}. It can be written in 5 steps:

(i) The air can hold a ‘certain maximum quantity’ of water vapour

■ When this maximum quantity is available in the air, we say that:

The air is saturated with water vapour

(ii) The saturation depends on temperature

    ♦ If the temperature is high, the air is able to hold more water vapour

          ✰ Consequently, more water vapour is required to make the air saturated

    ♦ If the temperature is low, the air is able to hold only less water vapour

          ✰ Consequently, less water vapour is sufficient to make the air saturated

(iii) We encounter different temperatures like 0 ^oC, 14 ^oC, 27 ^oC, 32 ^oC etc.,

• At each of those temperatures, air requires a ‘unique quantity of water vapour’ to become saturated

• Consequently, at each of those temperatures, there will be a ‘unique pressure’ exerted by the water vapour in the saturated air

■ This ‘unique pressure’ is called aqueous tension

(iv) Aqueous tension at various temperatures can be obtained from the data book

• Let us see some examples:

Example 1:

• Aqueous tension at 0 ^oC is 0.0060 bar

• That means, if air is saturated at 0 ^oC, the water vapour in that air will be exerting a pressure of 0.0060 bar

Example 2:

• Aqueous tension at 15 ^oC is 0.0168 bar

• That means, if air is saturated at 15 ^oC, the water vapour in that air will be exerting a pressure of 0.0168 bar

(v) Thus, once we know the temperature, we can obtain p_{Water vapour} from the data book

5. Now we can use the equation in (3) to obtain p_{Original gas}

• Note that in p_{Original gas}, there will not be even a single water molecule

    ♦ Because, p_{Water vapour} is already deducted

• So p_{Original gas} is also called p_{Dry gas}

■ Thus we get:

Eq.5.5: p_{Dry gas} = p_Total - p_{Water vapour} 

Partial pressure in terms of mole fraction

This can be explained in 7 steps:

1. Consider fig.5.19 again. We have seen that:

• Pressure contributed by Gas 1 in fig.a = Pressure exerted by Gas 1 in fig.b

    ♦ But ‘pressure exerted by Gas 1 in fig.b’ = $\mathbf\small{\rm{p_1=\frac{n_1 RT}{V}}}$

          ✰ Where n₁ is the number of moles of Gas 1

• Pressure contributed by Gas 2 in fig.a = Pressure exerted by Gas 2 in fig.c

    ♦ But ‘pressure exerted by Gas 2 in fig.c’ = $\mathbf\small{\rm{p_2=\frac{n_2 RT}{V}}}$

          ✰ Where n₂ is the number of moles of Gas 1

• Pressure contributed by Gas 3 in fig.a = Pressure exerted by Gas 3 in fig.d

    ♦ But ‘pressure exerted by Gas 3 in fig.d’ = $\mathbf\small{\rm{p_3=\frac{n_3 RT}{V}}}$

          ✰ Where n₃ is the number of moles of Gas 3 

2. So the total pressure p_Total will be given by:

p_Total = p₁ + p₂ + p₃ = $\mathbf\small{\rm{\frac{n_1 RT}{V}+\frac{n_2 RT}{V}+\frac{n_3 RT}{V}}}$

$\mathbf\small{\rm{\Rightarrow\;p_{Total}=(n_1+n_2+n_3)\frac{ RT}{V}}}$

3. Let us divide p₁ by p_Total

• We get: $\mathbf\small{\rm{\frac{p_1}{p_{Total}}=\frac{n_1}{(n_1+n_2+n_3)}=\frac{n_1}{n}=X_1}}$

$\mathbf\small{\rm{\Rightarrow p_1=X_1 \times  p_{Total}}}$

    ♦ Where:

          ✰ n = n₁ + n₂ + n₃.

          ✰ X₁ = mole fraction of the first gas

4. In a similar way, if we divide p₂ by p_Total, we will get:

$\mathbf\small{\rm{p_2=X_2 \times  p_{Total}}}$

          ✰ X₂ = mole fraction of the second gas

5. In a similar way, if we divide p₃ by p_Total, we will get:

$\mathbf\small{\rm{p_3=X_3 \times  p_{Total}}}$

          ✰ X₃ = mole fraction of the third gas

6. In general, we can write:

Eq.5.6: $\mathbf\small{\rm{p_i=X_i \times  p_{Total}}}$

          ✰ X_i = mole fraction of the i^th gas

7. So if we know the total pressure, we can use Eq.5.6 to find the pressure exerted by individual gases

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Now we will see some solved examples

Solved example 5.5 to 5.15

Solved example 5.16 to 5.24

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• In the next section, we will see kinetic molecular theory of gases

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