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

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• 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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