Chapter 13.4 - Physical Properties of Alkanes

In the previous section, we completed a discussion on the preparation of alkanes. In this section, we will see the properties of alkanes.

Physical properties of Alkanes

The first physical property that we will discuss, is about the non-polar nature of alkanes. It can be written in 3 steps:
1. We have seen that some molecules are polar in nature. We know the reason for such polarity [see fig.4.223 in section 4.40]
2. In the case of alkanes, there are only two types of bonds: C-C bonds and C-H bonds.
• The difference in electronegativity between C and H atoms is very small. So neither C nor H can pull the electron clouds. Thus both C-C bonds and C-H bonds are non-polar.
3. So we can write:
Alkanes are almost non-polar molecules.

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The second physical property that we will discuss, is about the state of various alkanes. It can be written in 7 steps:
1. Alkanes have weak van der Waal’s forces. We have seen van der Waals’s forces in an earlier section. It is the collection of four forces: (i) London force (ii) Dipole-dipole force (iii) Dipole-induced dipole force (iv) Hydrogen bond [see section 5]
2. When the number of C atoms in the chain is small, there will be a more uniform distribution of charge.
3. If the charge distribution is uniform, there will not be +ve and -ve regions.
• Consequently, there will not be much attraction between the molecules.
4. So, if the number of C atoms is small, the alkanes molecules will be in the gaseous state at room temperature (298 K).
• The first four members (C₁ to C₄) of the alkane series are gases.     
5. When the number of C atoms in the chain is larger, there will be a non- uniform distribution of charge.
6. If the charge distribution is non-uniform, there will be +ve and -ve regions.
• Consequently, there will be attraction between the molecules.
7. So, if the number of C atoms is large, the alkane molecules will be in the liquid or solid states.
• C₅ to C₁₇ are liquids at room temperature (298 K).     
• C₁₈ and higher are solids at room temperature (298 K).

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The third physical property that we will discuss, is about the solubility of alkanes. It can be written in 7 steps:
1. Consider a solution in which the solvent is polar and solute is non-polar.
• Let us name the polar solvent as A and non-polar solute as B
2. Since A is polar, the molecules of A will be attracting each other.
• Consider any two molecules of A, which are close together. They will be acting like a chain.
• B being non-polar, cannot form an electrostatic attraction with the two A molecules.
• So the molecule B will not be able to break the chain and occupy the space between the two A molecules.
• Even if we forcibly put the B between the two A molecules, the B will be expelled.
• So it is clear that, non-polar substances will not dissolve in polar solvents.
3. On the other hand, if B is a polar substance just like A, then B can occupy space in between the two A molecules to form a new chain.
• So if B is polar, it will dissolve in A.
• We can write: Polar substances will dissolve in polar solvents.
4. Consider the situation when both A and B are non-polar.
• In such a situation, neither A or B has permanent polarity. But London forces is applicable to both of them.
• Due to London forces, both A and B have equal chances of forming electrostatic attraction with each other. So B will dissolve in A
5. Thus we see that:
   ♦ Non-polar substances do not dissolve in polar solvents.
   ♦ Polar substances dissolve in polar solvents.
   ♦ Non-polar substances dissolve in non-polar solvents.
• Based on this, we can write:
Like dissolves like.
6. We have seen that alkanes are non-polar. We already know that water is polar.
So it is clear that, alkanes are insoluble in water.
7. Grease is composed of alkanes. Petrol is also composed of alkanes. Both being non-polar, grease will dissolve in petrol. That is why petrol is used in dry cleaning, to remove grease stains from clothes.

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The fourth physical property that we will discuss, is about the boiling points of alkanes. It can be written in 3 steps:
1. The boiling point (b.p) of different alkanes can be obtained from the table in the data book. The table is also shown in the text book.
2. We see that, higher alkanes have higher boiling points.
This can be explained in two steps:
(i) As the number of C atoms increase, the size of the molecules increase. As a result, the magnitude of the inter molecular van der Waals forces will be higher.
(ii) When the magnitude of the forces increase, it will be difficult to separate the molecules from each other.
(iii) Consequently, the b.p increases for higher alkanes.
3. From the table we see another interesting point:
Pentane, 2-Methylbutane and 2,2-Dimethylpropane have the same molecular mass but different boiling points.
• This can be explained in 4 steps:
(i) Pentane, 2-Methylbutane and 2,2-Dimethylpropane are isomers. So they have the same molecular mass.
    ♦ Pentane has a straight chain structure.
    ♦ 2-Methylbutane has one branch.
    ♦ 2,2-Dimethylpropane has two branches.
(ii) When the number of branches increase, the shape of the molecule becomes more and more spherical.
(iii) Spherical shape has lower surface area.
• Due to the lower surface area, the area of contact between the spheres will be small. Consequently, the inter molecular forces will be small.
(iv) If the inter molecular forces are low, the molecules can be easily separated from each other.
• So we can write:
When the shape becomes more and more spherical, the b.p becomes lower and lower.

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The fifth and final physical property needs a mention only:
Alkanes are colourless and odourless

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Chemical properties of Alkanes

• Alkanes are generally inert towards acids, bases, oxidising agents and reducing agents. The reason can be written in 3 steps:
(i) In alkanes, all four valencies of C atoms are satisfied. The single valency of H atoms are also satisfied. So an alkane molecule as a whole, is stable.
(ii) All bonds in alkanes are sigma bonds. We know that, sigma bonds involve linear overlap of orbitals. This creates bonds which are very strong. So it is difficult to break the bonds in alkanes.
(iii) Alkanes are non-polar. So they do not have any additional +ve or -ve charges. This makes it difficult to attack alkane molecules.

• However, alkanes can undergo reactions like substitution, combustion, controlled oxidation etc., We will now see those reactions in detail
I. Substitution reactions
This can be written in 9 steps:
1. In this reaction, one H atom is first removed from the alkane.
• Then a halogen atom takes the place of that H atom.
• Instead of halogen atom, nitro group or sulphonic acid group can also take the place of the H atom.
    ♦ If it is a halogen, the reaction is called halogenation.
    ♦ If it is a nitro group, the reaction is called nitration.
    ♦ If it is a sulphonic acid group, the reaction is called sulphonation.
2. Some times more than one H atoms can be substituted in this way.
3. Halogenation requires higher temperatures (573-773 K)
• However, instead of higher temperatures, we can use diffused sunlight or ultraviolet light also.
• Diffused sunlight is obtained when direct sunlight is scattered by clouds or some artificial means.
• Ultraviolet light can be produced by passing electricity through a suitable gas like mercury vapour.
4. Lower alkanes do not undergo nitration and sulphonation. They can undergo halogenation only.
5. Let us see some examples of halogenation:
(i) $\rm{CH_4~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_3 Cl~+~HCl}$
• CH₃Cl is Chloromethane.
[$h \nu$ represent the energy provided by radiation (in our present case, radiation is sunlight or ultraviolet light). Here h is the plank's constant and 𝜈 is the frequency. We saw those details in chapter 2]
(ii) $\rm{CH_3 Cl~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_2 Cl_2~+~HCl}$
• CH₂Cl₂ is Dichloromethane.
(iii) $\rm{CH_2 Cl_2~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH Cl_3~+~HCl}$
• CHCl₃ is Trichloromethane.
(iv) $\rm{CH Cl_3~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ C Cl_4~+~HCl}$
• CCl₄ is Tetrachloromethane.
(v) $\rm{CH_3 - CH_3~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_3 - CH_2 Cl~+~HCl}$
• CH₃-CH₂Cl is Chloroethane.
6. The rate of reaction of alkanes with the various halogens varies.
   ♦ Rate of reaction with F is the highest.
   ♦ Rate of reaction with I is the lowest.
• The order is: F₂ > Cl₂ > Br₂ > I₂
(Recall that, rate of reaction is the quantity of products formed in unit time)
7. Reaction with F is violent. We will need special equipment to control the reaction.
8. Reaction with I is a very slow reaction. It is a reversible reaction. The equation is:
CH₄ + I₂ ⇌ CH₃I + HI
• We have discussed about reversible reactions in an earlier chapter [see fig.7.3 in section 7.3]
• We can convert more quantities of CH₄ into CH₃I if we remove the HI.
• For that, we use oxidising agents like HIO₃ or HNO₃. The HI will react with the oxidising agent and form I₂. The equation is:
HIO₃ + 5HI ⟶ 3I₂ + 2H₂O
9. We have seen 1^o carbon atom, 2^o carbon atom etc., in a previous section of this chapter [see step 11 below fig.13.8 in section 13.1]
• Based on that, we can now write an important information. It can be written in 2 steps:
(i) During halogenation of alkanes, the H atoms attached to the 3^o carbon atoms are removed more readily than those attached to the 2^o carbon atoms.
(ii) Similarly, the H atoms attached to the 2^o carbon atoms are removed more readily than those attached to the 1^o carbon atoms.

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In the next section we will see the mechanism of halogenation.

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Chapter 7.2 - Activity to Demonstrate Equilibrium

In the previous section, we saw solid-vapour equilibrium and equilibrium in solutions. In this section, we will see the general characteristics of equilibrium

First let us write a summary of what we have learnt so far in the previous two sections. It can be written in 4 steps:

1. Solid-liquid equilibrium
(i) For a solid-liquid reversible process, we write: Solid ⇌ Liquid
    ♦ This process is called melting
(ii) For a given solid and it’s liquid, there exists a particular temperature at which equilibrium is possible
    ♦ This temperature is called melting point
(iii) This temperature depends on the pressure experienced by the system
• So if the experiment is conducted at different pressures, we will obtain different equilibrium temperatures
• In order to avoid such a confusion, we specify that, the experiment must be conducted at 1 atm pressure
(iv) Consider the process: Ice ⇌ Water
• For equilibrium of this process,
    ♦ temperature must be 0 ^оC
    ♦ pressure must be 1 atm
        ✰ If temperature is lower than 0 ^оC, the rate of forward process will increase
        ✰ If temperature is higher than 0 ^оC, the rate of backward process will increase

2. Solid-liquid equilibrium
(i) For a liquid-vapour reversible process, we write: Liquid ⇌ Vapour
    ♦ This process is called vaporization
(ii) For a given liquid and it’s vapour, there exists a particular temperature at which equilibrium is possible
    ♦ This temperature is called boiling point
(iii) This temperature depends on the pressure experienced by the system
• So if the experiment is conducted at different pressures, we will obtain different equilibrium temperatures
• In order to avoid such a confusion, we specify that, the experiment must be conducted at 1 atm pressure
(iv) Consider the process: Water ⇌ Vapour
• For equilibrium of this process,
    ♦ temperature must be 100 ^оC
    ♦ pressure must be 1 atm
        ✰ If temperature is lower than 100 ^оC, the rate of backward process will increase
        ✰ If temperature is higher than 0 ^оC, the rate of forward process will increase

3. Solid-solution equilibrium
(i) For a solid-solution reversible process, we write: Solid ⇌ Solution
    ♦ This process is called dissolution of solid in liquid
(ii) For a given solid and it’s solution, there exists a particular temperature at which equilibrium is possible
(iv) Consider the process: Sugar ⇌ Solution
• At a particular temperature, equilibrium exists between solid sugar and the sugar solution
    ♦ If temperature is lower, then the rate of backward process will increase
        ✰ Solid sugar will precipitate out of the solution
    ♦ If temperature is higher then the rate of forward process will increase
        ✰ Solid sugar will dissolve into the solution

4. Gas-solution equilibrium
(i) For a Gas-solution reversible process, we write: Gas ⇌ Solution
    ♦ This process is called dissolution of gas in liquid
(ii) For a given gas and it’s solution, there exists a particular pressure at which equilibrium is possible
(iv) Consider the process: CO₂(g) ⇌ CO₂(sol)
• At a particular pressure, equilibrium exists between CO₂(g) and the CO₂(sol)
    ♦ If pressure is lower, then the rate of backward process will increase
        ✰ CO₂(g) will escape out of CO₂(sol)
    ♦ If pressure is higher then the rate of forward process will increase
        ✰ CO₂(g) will dissolve into the CO₂(sol)
(v) This equilibrium pressure depends on temperature also
    ♦ If the temperature is lower, lesser pressure is sufficient to cause dissolution
    ♦ If the temperature is higher, greater pressure is required to cause dissolution
• This is because, at lower temperature, the gas molecules will not have sufficient energy to escape out of the solution

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Based on the above summary, we can write 4 points:
1. For the equilibrium of Solid ⇌ Liquid, temperature is the criteria
   ♦ Also the pressure needs to be constant
• The temperature at any instant during the process will tell us:
   ♦ The direction in which, the process is going on
   ♦ The extent up to which the process has reached in that direction
• For example:
   ♦ If the temperature is lower than the specified value
         ✰ we can be sure that, the process will be taking place in the reverse direction
   ♦ If the temperature is much lower than the specified value
         ✰ we can be sure that, greater amount of solid will be present
2. For the equilibrium of Liquid ⇌ Vapour, temperature is the criteria
   ♦ Also the pressure needs to be constant
• The temperature at any instant during the process will tell us:
   ♦ The direction in which, the process is going on
   ♦ The extent up to which the process has reached in that direction
• For example:
   ♦ If the temperature is lower than the specified value
         ✰ we can be sure that, the process will be taking place in the reverse direction
   ♦ If the temperature is much lower than the specified value
         ✰ we can be sure that, greater amount of liquid will be present
3. For the equilibrium of Solid ⇌ Solution, temperature is the criteria
• The temperature at any instant during the process will tell us:
   ♦ The direction in which, the process is going on
   ♦ The extent up to which the process has reached in that direction
• For example:
   ♦ If the temperature is lower than the specified value
         ✰ we can be sure that, the process will be taking place in the reverse direction
   ♦ If the temperature is much lower than the specified value
         ✰ we can be sure that, greater amount of solid will be present
4. For the equilibrium of Gas ⇌ Solution, pressure is the criteria
   ♦ Also the temperature needs to be constant
• The pressure at any instant during the process will tell us:
   ♦ The direction in which, the process is going on
   ♦ The extent up to which the process has reached in that direction
• For example:
   ♦ If the pressure is lower than the specified value
         ✰ we can be sure that, the process will be taking place in the reverse direction
   ♦ If the pressure is much lower than the specified value
         ✰ we can be sure that, greater amount of free gas will be present

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• We know that, equilibrium can occur in both physical processes and chemical reactions
• In both cases, equilibrium is dynamic. That is., although we observe no activity at equilibrium, forward and backward reactions are going on
    ♦ We can use radioactive isotopes to prove this
    ♦ We saw it in the case of sugar solution
• But a procedure using radioactive isotopes is not suitable for school laboratories
• So instead of using isotopes, we can do a simple activity. It can be explained in 11 steps:

1. Take two measuring cylinders, each of 100 mL capacity
    ♦ Mark them as Cylinder-1 and cylinder-2
2. Take two glass tubes, each of length 30 cm
    ♦ Both must have the same diameter of 3 or 4 mm
    ♦ Mark them as tube-1’ and tube-2’
• This is shown in fig.7.2 below:

Fig.7.2

3. Fill about half of cylinder-1 with coloured water
    ♦ Potassium permanganate can be used to color the water
• Keep the measuring cylinder-2 empty
4. Put tube-1’ in cylinder-1 and tube-2’ in cylinder-2
5. Close the upper tip of tube-1’ with finger
    ♦ Transfer this tube-1’ to cylinder-2 and open the tip
    ♦ The water in the tube-1’ will fall into cylinder-2
6. Close the upper tip of tube-2’ with finger
    ♦ Transfer this tube-2’ to cylinder-1 and open the tip
    ♦ The water in the tube-2’ will fall into cylinder-1
7. The quantity transferred in (5) will be greater than quantity transferred in (6)
• This is because:
    ♦ While in cylinder-1, the tube-1’ will take in more water
    ♦ While in cylinder-2, the tube-2’ will take in only less water
8. Keep repeating the steps (5) and (6)
• We will soon see that, the level in the cylinders become equal
• This is shown in fig.7.2(b)
9. Transfer using tube-1’ is the forward process
• In the early stages, the quantity in each of this transfer is high
• But the ‘quantity continuously decreases’ due to the lowering of level in cylinder-1
• This ‘continuous decrease in quantity’ is analogous to the ‘decrease in rate of forward process’
9. Transfer using tube-2’ is the backward process
• In the early stages, the quantity in each of this transfer is low
• But the ‘quantity continuously increases’ due to the rising of level in cylinder-2
• This ‘continuous increase in quantity’ is analogous to the ‘increase in rate of backward process’
10. So the two rates are not the same
    ♦ Rate of the forward process continuously decreases
    ♦ Rate of the backward process continuously increases
• Due to this, the two rates soon become equal
◼ When this happens, we get the equilibrium
◼ The levels becoming equal in step (8) is analogous to equilibrium
11.After reaching equilibrium, we can try repeating the steps (5) and (6)
• There will be no change in the levels
◼ In a real process/reaction also, even after equilibrium, the forward and backward processes/reactions are continuing. But we observe no change

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• In the next section, we will see equilibrium in chemical reactions


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Chapter 7 - Equilibrium

In the previous section, we completed a discussion on thermodynamics. In this section, we will see equilibrium

• We know that, when a chemical reaction takes place, reactants get converted into products
• When the reaction is complete, three types of situation can arise:
Case 1:
• All reactants are converted into products
   ♦ There may be traces of reactants left
   ♦ But it may not be possible even to detect those remaining reactants
• We say that:
   ♦ Quantity of reactants
   ♦ is far less than
   ♦ Quantity of products
Case 2:
• Quantity of products formed is very low
   ♦ Most of the reactants remain as such
• We say that:
   ♦ Quantity of reactants
   ♦ is far greater than
   ♦ Quantity of products
Case 3:
• Reaction stops in the midway
   ♦ Some products are formed
   ♦ Some reactants remain as such
• We say that:
   ♦ Quantity of reactants remaining
   ♦ is comparable to
   ♦ Quantity of products newly formed

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• Consider case 3
• We encounter this case on many occasions in chemical industries and also in labs
• During the chemical reaction, a point is reached at which, we see no activity
   ♦ But at that point, all reactants are not converted into products
   ♦ We call this point: equilibrium
• Let us see the main features of equilibrium. It can be written in 3 steps:
1. At equilibrium, we are inclined to think that, the reaction has stopped. This is because, we see no activity
• But in reality, the reaction is still going on
   ♦ Only that, the reaction is going on in two opposite directions
• We can say:
   ♦ In the forward direction, the reactants are being converted into products
   ♦ In the backward direction, the products are being converted into reactants
2. Then why do we see 'no activity'?
The reason can be written in 3 steps:
(i) There is no change in the quantity of reactants because,
   ♦ The reactants used up in the forward reaction
   ♦ is replenished by
   ♦ The reactants formed from the backward reaction
(ii) There is no change in the quantity of products because,
   ♦ The products used up in the backward reaction
   ♦ is replenished by
   ♦ The products formed from the forward reaction
(iii) Based on (i) and (ii), we can write:
• At equilibrium,
   ♦ Rate of forward reaction
   ♦ is equal to
   ♦ Rate of backward reaction
3. Can we change the equilibrium?
• On many occasions, we may not be satisfied with the quantities obtained at equilibrium
• We may want a greater quantities of products
• This can be achieved by increasing the rate of the forward reaction
• If the rate of forward reaction is increased,
   ♦ The quantities of products obtained from the forward reaction
   ♦ in a given time duration
   ♦ Will be greater than
   ♦ The quantities of reactants formed from the backward reaction
• A reaction can be forced to proceed in a desired direction by two methods:
(i) Making appropriate changes to the temperature, pressure etc.,
(ii) Making appropriate changes to the concentrations of reactants/products

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Solid-Liquid Equilibrium

• Equilibrium occur not only in chemical reactions. It is observed in many physical processes also. Let us see some examples:
• First we will see the equilibrium between solid phase and liquid phase. It can be explained by taking ice and water as an example. It can be written in 8 steps:
1. Consider a mixture of ice and water
   ♦ Let the temperature of the mixture be 273 K
2. The mixture is placed inside a thermos flask
   ♦ So no heat enters or leaves the mixture
3. Using precision instruments, we can measure the individual masses of ice and water in the thermos flask
   ♦ Let the masses of water and ice be m_water and m_ice respectively
4. After a time duration of say 10 minutes, if we measure the masses, we will see that, m_water and m_ice are unchanged
• We can say that, water and ice are in a state of equilibrium
5. We observe no activity
• But in reality, activities are taking place at the interface between water and ice
   ♦ Some water molecules collide with the ice and adhere to it
         ✰ This can be represented as: H₂O(l) → H₂O(s)
   ♦ Some ice molecules escape from the surface of the ice and goes into the liquid
         ✰ This can be represented as: H₂O(s) → H₂O(l)
   ♦ Both the processes are taking place simultaneously
         ✰ So we can combine them as: H₂O(l) ⇌ H₂O(s)
6. We can compare equilibrium and non-equilibrium between ice and water:
Case 1:
• If the pressure is 1 atm and temperature is less than 273 K:
   ♦ The number of water molecules adhering to the ice
   ♦ will be greater than
   ♦ The number of ice molecules escaping into water
• Then the quantity of ice will go on increasing
• Also the quantity of water will go on decreasing
• We can write:
   ♦ In the process: H₂O(l) ⇌ H₂O(s),
   ♦ Both the processes takes place simultaneously
   ♦ But the forward process takes place at a higher rate
   ♦ This is not an equilibrium
Case 2:
• If the pressure is 1 atm and temperature is greater than 273 K:
   ♦ The number of water molecules adhering to the ice
   ♦ will be lesser than
   ♦ The number of ice molecules escaping into water
• Then the quantity of ice will go on decreasing
• Also the quantity of water will go on increasing
• We can write:
   ♦ In the process: H₂O(l) ⇌ H₂O(s),
   ♦ Both the processes takes place simultaneously
   ♦ But the backward process takes place at a higher rate
   ♦ This is not an equilibrium
Case 3:
• If the pressure is 1 atm and temperature is equal to 273 K:
   ♦ The number of water molecules adhering to the ice
   ♦ will be equal to
   ♦ The number of ice molecules escaping into water
• Then the quantity of ice will remain the same
• The quantity of water will also remain the same
• We can write:
   ♦ In the process: H₂O(l) ⇌ H₂O(s),
   ♦ Both the processes takes place simultaneously
   ♦ Both the processes takes place at the same rate
   ♦ This is an equilibrium
7. Case 3 is equilibrium
• At equilibrium,
   ♦ The forward process and the reverse process occur simultaneously
   ♦ The forward process and the reverse process occur at the same rate
8. The above steps gives us a perfect explanation for equilibrium between liquid phase and solid phase. In fact, the equilibrium is used to define melting point/ freezing point:

For any pure substance at atmospheric pressure, the temperature at which the solid and liquid phases are at equilibrium is called normal melting point or normal freezing point of the substance

Liquid-Vapor Equilibrium

• Next we will consider the equilibrium between liquid phase and vapour phase. It can be explained by taking water and water-vapor as an example. It can be written in 8 steps
1. Fig.7.1(a) below shows a transparent box
• It is attached with a U-tube containing mercury
    ♦ An U-tube containing mercury is called manometer
    ♦ It is used to determine pressure

Fig.7.1

2. Place a drying agent like anhydrous calcium chloride or (phosphorus penta-oxide) in the box
• The drying agent will absorb all the gaseous water molecules inside the box
• After a few hours,
    ♦ There will not be any water molecules inside the box
    ♦ Note the level of mercury in the right limb of the manometer
    ♦ Remove the drying agent by tilting the box on one side
    ♦ Quickly place a petri dish containing water inside the box
3. We can observe that:
• The level of mercury in the right limb of the manometer slowly increases
    ♦ This shows that, the pressure inside the box is slowly increasing
• The volume of water in the petri dish decreases
4. Remember that, no gaseous molecules can enter the box because, the mercury acts as a seal
• So the increase in pressure inside the box must be due to the formation of new gaseous molecules
5. How are the new gaseous molecules formed?
• New gaseous molecules are formed due to the evaporation of water in the dish
• Some of the liquid water molecules become gaseous water molecules
• The decrease in volume of water in the dish is clear evidence for this transformation
6. Due to the formation of new gaseous molecules, the pressure inside the box increases
• This excess pressure pushes the mercury
• Thus the level of mercury in the right limb rises
• But this 'increase in pressure' does not continue indefinitely. After some time, the mercury level becomes static
7. Let us analyze the processes taking place inside the box from the moment when the dish of water is placed. It can be written in 6 steps:
(i) When the water is placed inside the box, some liquid water molecules from the surface of the water, escape from the mass of water
• Once those molecules escape, they are in the gaseous state
• This process can be represented using symbols as:
H₂O(l)  → H₂O(vap)
(ii) This process continues and so, the quantity of gas inside the box increases
• At the same time, the volume of water decreases
(iii) The increase in quantity of gas causes increase in pressure
• This excess pressure pushes the mercury upwards
• That is why we see the rise in mercury level
(iv) All the while when this process is taking place, another process is also taking place simultaneously
• It is the condensation of some of the gaseous water molecules back into the liquid state
• This process can be represented using symbols as:
H₂O(vap) → H₂O(l)
(v) Since the two processes are taking place simultaneously, we can write:
H₂O(l) ⇌ H₂O(vap)
• Initially, rate of the forward reaction is greater than rate of backward reaction
• That is.,
   ♦ Number of molecules entering the gaseous phase
   ♦ is greater than
   ♦ Number of molecules leaving the gaseous phase
• So quantity of gas increases, causing the increase in pressure
• The increase in pressure causes the mercury level to rise up
(vi) But after some time, an equilibrium will be reached
• That is.,
   ♦ Number of molecules entering the gaseous phase
   ♦ become equal to
   ♦ Number of molecules leaving the gaseous phase
• So quantity of gas becomes steady, causing no further increase in pressure
• The steady pressure causes the mercury level to remain static
8. So now we know the reason for the two observations:
    ♦ Mercury level rising initially
    ♦ Mercury level becoming static after some time
• Next, slightly increase the temperature of the water
• Now more water molecules will have the required energy to break away from the water mass
• So the number of gaseous water molecules inside the box will increase
• This increases the pressure
• The mercury level in the right limb goes up
• If the new temperature is kept constant, a new equilibrium will be reached

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Let us see the practical application of the above experiment. It can be written in 7 steps
1. In the above experiment, the water is inside a closed container
• So water molecules cannot escape
2. If we place water in a dish which is open to the atmosphere, the water molecules escaping from the water mass will be blown away by air currents
• The water level in the dish gradually decreases until no water is left
◼ We call this process as: evaporation
• Depending on the surrounding temperature, this process may take several hours to a few days
3. Can we obtain equilibrium when water is open to atmosphere?
• Answer can be written in 4 steps:
(i) When the water is open to atmosphere, we can consider the atmospheric pressure as an invisible lid
• This lid will try to suppress the gaseous water from escaping
(ii) But the air currents blow away the few gaseous water molecules
(iii) This decreases the pressure of the vapour
• Thus the equilibrium between liquid and gaseous molecules is disrupted
   ♦ This favors the forward process in H₂O(l) ⇌ H₂O(vap)
• More liquid molecules are able to go into gaseous phase
(iv) Those newly formed molecules are also blown away
• This process continues until no water remains in the dish
◼ So we can write:
Equilibrium between gaseous and liquid molecules cannot be established when water is open to atmosphere
4. Suppose that, there are no air currents. Then there will be equilibrium
• We want to know how such an equilibrium can be disrupted
• It can be written in 8 steps
(i) If there are no air currents, the water level will remain static due to the presence of the invisible lid
(ii) In such a situation, if we increase the temperature of the water, more and more liquid molecules will get sufficient energy to break away from the liquid mass
(iii) At the higher temperature, a new equilibrium will be established
• That is.,
    ♦ Number of molecules leaving the liquid mass increases to a new value
    ♦ The number of gaseous molecules condensing back to the liquid phase also increases to that new value
(iv) Note that, at this stage, due to the presence of a greater number of gaseous molecules, the vapour pressure would have increased
(v) If we go on increasing the temperature, the vapour pressure will become so high that, it can push away the invisible lid
• That means, at high temperature, the atmospheric pressure will no longer be able to suppress the vapour pressure
(vi) While increasing the temperature, there will come a point at which, the vapour pressure becomes equal to the atmospheric pressure
• At that point, the water will begin to boil
(vii) If the atmospheric pressure is 1 atm, this boiling of water usually begins at 100 ^०C
(viii) Upto 100 ^०C, the vapour will be in equilibrium with the liquid. This is because,
    ♦ No gaseous molecules are able to escape
    ♦ They are suppressed by the atmospheric pressure
• So upto 100 ^०C,
    ♦ Number of molecules leaving the liquid
    ♦ will be equal to
    ♦ Number of molecules condensing back
5. So based on the equilibrium between vapour and liquid, we now have a method to define boiling point:

For any pure liquid at one atmospheric pressure (1.013 bar), the  temperature at which the liquid and vapors are at equilibrium is called normal boiling point of the liquid.

6. Remember that, if the atmospheric pressure is low, the vapour will be able to push away that atmospheric pressure more easily
• That means, the vapour can push away that low atmospheric pressure even at a lower temperature
• That is why, at high altitudes, water boil at lower temperatures
7. Let us note an important point. It can be written in 3 steps:
(i) Even if the atmospheric pressure is the same, some liquids have lower boiling points than water
(ii) This is because, such liquids have lower inter molecular attractions
    ♦ When in liquid phase, their molecules can easily break away from each other
    ♦ So they can leave the liquid mass easily
(iii) That is why they boil at lower temperatures

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• In the next section, we will see equilibrium between solid phase and gaseous

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Chapter 5.8 - Vapour Pressure

In the previous section we saw liquefaction of gases. In this section, we will see liquid state

• The study about liquefaction in the previous section, gives us some interesting information

• They can be written in 4 steps:

1. When a gas is liquefied, the molecules remain the same

• For example:

    ♦ When gaseous NH₃ is liquefied, the newly formed liquid ammonia has the same NH₃ molecules

    ♦ When gaseous CO₂ is liquefied, the newly formed liquid carbon dioxide has the same CO₂ molecules

2. If the molecules are the same, what is the difference between the two states?

• The answer is that:

When the gas becomes a liquid, the inter molecular attractions are increased. So the molecules of the liquid remain close together. That means, the liquid can be considered as a dense gas

3. So there is a continuity between gaseous state and liquid state

• The word fluid is used to recognize this continuity

4. In order to turn the gas into liquid, two conditions must be satisfied:

(i) Molecules must be close together

    ♦ We must apply pressure and thus bring the molecules of the gas closer together

(ii) Temperature must be low

    ♦ We must reduce the temperature (to a value below Tc)

    ♦ Then, the molecules will no longer have the kinetic energy to break away from the inter molecular attractions

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Now we will learn about vapor pressure. It can be written in 10 steps:

1. We have seen that, the kinetic energy of the molecules need to be brought down to achieve liquefaction

• Conversely, we can say:

When the molecules are in the liquid state, their kinetic energy will be low

2. But in a liquid sample, even when it is at normal temperature, there will be some molecules that have a ‘high value kinetic energy’

• Those molecules can 'break away' from the other liquid molecules

3. The consequence of such ‘break away’ can be demonstrated using an example. It can be written in 5 steps:

(i) Consider a sample of liquid water placed in an open vessel

• Let it be at room temperature

(ii) Some of the molecules in that sample will be having enough kinetic energy to break away

• Those molecules will escape from the liquid water body and reach a level above the water surface

This is shown in fig.5.29(a) below

Fig.5.29

(iii) Those molecules are no longer attracted to any other molecules. Because, they are in the gaseous state

• So even a small current of air will carry them away

(iv) If the climate is hot, some more molecules will gain the required kinetic energy

• They will also break away from the liquid body and will be carried away by air currents

(v) This process continues and soon all the liquid water will be turned into the gaseous state

• This process is called evaporation

■ So the consequence of the 'break away' is that, there will always be some molecules above the liquid surface

4. Now consider fig.5.29(b) above

• The vessel is closed with a lid

• This time, the ‘molecules which break away’ are confined in a 'definite space'

5. As time passes, more and more molecules will break away from the liquid body

• All those free molecules will be collected and confined in the definite space

    ♦ The lid is the upper boundary of this definite space

    ♦ The surface of the liquid is the lower boundary of this definite space

    ♦ The vessel is the side boundary of this definite space

6. We see that, as time passes, more and more gaseous molecules are added into this definite space

• If this continues, there will not be any liquid left. All liquid will turn into gas

• But such a situation does not occur. The reason can be explained in 8 steps:

(i) The molecules in the definite space will be moving in random directions

    ♦ They will hit each other

    ♦ They will hit the boundaries mentioned in (5)

(ii) As a result, some of those molecules will lose energy and will fall back into the liquid state

(iii) But remember that:

Even when some molecules are falling back to the liquid state, some molecules (as mentioned in step 2), are breaking away from the liquid state and are passing into the gaseous state

(iii) So we have motion in two directions:

    ♦ Some molecules are breaking away from the liquid state into the gaseous state

    ♦ Some molecules are falling back from the gaseous state into the liquid state

(iv) If the ‘number of molecules breaking away’ is greater than the ‘number of molecules falling back’, all the liquid will soon become gas

(v) If the ‘number of molecules falling back’ is greater than the ‘number of molecules breaking away’, all the gas will soon become liquid

(vi) Such situations mentioned in (iv) and (v) do not arise because, an equilibrium will be reached

• That is:

    ♦ The ‘number of molecules breaking away’

    ♦ will soon become equal to

    ♦ The ‘number of molecules falling back’

(vii) It is important to understand the nature of this equilibrium. This can be explained in steps:

    ♦ Even at equilibrium,the molecules are travelling in both directions

    ♦ That is., even at equilibrium,

          ✰ some molecules are breaking away

          ✰ some molecules are falling back

    ♦ We call it an 'equilibrium' because, the numbers are the same

(viii) Once equilibrium is attained, there will be no change in the number of gaseous molecules in the definite space

■ That is the reason why, in fig.5.29(b), all the liquid is not turned into gas

7. So we learnt that:

• At equilibrium:

    ♦ The ‘number of molecules breaking away’

    ♦ will be equal to

    ♦ The ‘number of molecules falling back’

• In other words, at equilibrium, the ‘number of gaseous molecules’ will be a constant

8. A ‘definite space’ is same as ‘constant volume’

• So at equilibrium, we have:

    ♦ A constant volume

    ♦ A constant number of gaseous molecules in that volume

    ♦ A constant temperature

• When those three items are constant, the 'pressure exerted by the gaseous molecules' will be a constant

■ This pressure at equilibrium is called equilibrium vapor pressure or saturated vapor pressure

9. We attached the words ‘equilibrium/saturated’ to vapour pressure

• So, is there an ‘ordinary vapour pressure’ also ?

• The answer is: Yes

• This can be explained in 4 steps:

(i) In step (6), we saw that:

    ♦ As time passes, more and more gaseous molecules are being added

          ✰ This is a situation which continues till equilibrium

(ii) So before reaching equilibrium we can measure the vapour pressure at any instant

    ♦ Let the vapour pressure measured at time t₁ be p_vapour(1)

    ♦ Let the vapour pressure measured at time t₂ be p_vapour(2)

(iii) p_vapour(1) Will be different from p_vapour(2) 

• This is because:

    ♦ The number of gaseous molecules at time t₁

    ♦ will be different from 

    ♦ The number of gaseous molecules at time t₂

(iv) So we can write:

• The vapour pressures measured before attaining equilibrium are reported simply as: Vapour pressure

• The vapour pressure measured at equilibrium is reported as: Equilibrium vapour pressure or Saturated vapour pressure

10. Whenever we report a vapour pressure (equilibrium or ordinary), we must mention the temperature also

• This is because:

    ♦ Vapour pressure is temperature dependent

    ♦ When the temperature is high, there will be a greater number of gaseous molecules

          ✰ This will create a greater vapour pressure

    ♦ When the temperature is low, there will be a lesser number of gaseous molecules

          ✰ This will create only a lower vapour pressure

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• Based on the above discussion, we can now write some basics about boiling

• The basics can be written in 5 steps:

1. Consider the vessel in fig.5.29(b) above

• We see that, the ‘molecules which break away from the liquid body’ are not free to escape because, there is a lid 

2. Consider the vessel in fig.5.29(a) above

• We see that, the ‘molecules which break away from the liquid body’ are free to escape because, there is no lid

3. But in fact, in fig.a, there is an invisible lid. This can be explained in 5 steps:

(i) We know that, atmospheric pressure acts every where

(ii) Pressure is: Force per unit area

• So atmospheric pressure is: Force exerted by atmosphere on unit area

(iii) The atmosphere can be considered as a mixture of the molecules of N₂, O₂, CO₂ etc.,

• These molecules are pulled down by gravity

• So these molecules apply a down ward weight on substances below them

• When we divide this weight by area, we get: Weight per unit area

    ♦ This 'weight per unit area' is the atmospheric pressure

(iv) Consider the molecules (both gaseous and liquid) in the vessel in fig.5.29(a)

• Each of those molecules will experience the atmospheric pressure

(v) That means, all the molecules in the vessel in fig.a are pushed down by the atmospheric pressure

• This, in effect, is a lid

4. If the gaseous molecules in fig.5.29(a), can over come the ‘pushing down by the atmosphere’, they can escape

• So let us consider the conditions by which the molecules can over come the ‘pushing down by the atmosphere’

• It can be written in 5 steps:

(i) Let us heat the liquid from the bottom of the vessel

• The temperature of the liquid will begin to increase

(ii) More and more liquid molecules will attain the required energy to break free from the liquid body

• So more and more molecules will turn into gaseous state

• So more and more molecules will get collected between the lid (here, the lid is atmospheric pressure) and the surface of the liquid

(iii) That means, number of gaseous molecules in that region increases continuously

• When the number of gaseous molecules increases, the ‘vapor pressure’ that we saw earlier increases

• Due to the continuous heating, the vapor soon becomes equal to the atmospheric pressure

(iv) With a little more heating, the vapor pressure becomes greater than atmospheric pressure

• The ‘vapor pressure is greater’ means:

    ♦ The force per unit area exerted by the vapor

    ♦ is greater than

    ♦ The force per unit area exerted by the atmosphere

• So the vapor pushes away the atmosphere and escapes from the vessel

(v) As the heating is continued, the vapor pressure also increases

• More and more liquid molecules first break away from the liquid body

• After that, they push away the atmosphere and escape from the vessel

5. At this stage, we see bubbles coming from the bottom of the vessel

• Bubbles originate at the bottom of the vessel because, heating is done at the bottom

• The liquid molecules at the bottom attains the required energy to break away

• They push the other molecules in all directions

• So a bubble is made up of ‘high energy liquid molecules’

• Since these ‘high energy molecules’ push the other molecules in all directions, bubbles have a near spherical shape

• When bubbles begin to appear, we say that: Boiling has begun

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• Next we will see the influence of atmospheric pressure on boiling. It can be written in steps:

1. In our earlier classes, we have learnt that:

Boiling point is the temperature at which a liquid starts to boil at standard atmospheric pressure

2. From the above definition, it is clear that:

The temperature should be measured 'when the atmospheric pressure is at a standard value'

• In other words:

If we measure the temperature 'when the atmospheric pressure is not at a standard value', we cannot report it as an acceptable boiling point

3. For example, at mountain tops, the atmospheric pressure will be low

• The air molecules of the atmosphere will not be ‘pushing down’ with the same force as at sea level

• The vapour pressure will quickly become equal to the surrounding atmospheric pressure

• A lower temperature will be sufficient to build up the required vapour pressure

    ♦ So we will find that, the liquid boils at a lower temperature

    ♦ We cannot report this low temperature as the boiling point

4. In order to avoid any doubts, we can report the pressure value also. Thus we have:

The temperature at which the vapor pressure of a liquid is equal to the external pressure is called boiling temperature at that pressure

5. If the external pressure is 1 atm, we do not write the pressure

• Instead, we use the word ‘normal’

• This can be explained in 3 steps:

(i) Suppose that, the ‘temperature at which a liquid boils’ is T₁ K

(ii) Also suppose that, when this T₁ K is measured, the external pressure is 1 atm

(iii) We can report the result in two ways:

    ♦ The boiling point of the liquid is T₁ K at 1 atm pressure

    ♦ The normal boiling point of the liquid is T₁ K

          ✰ When the word ‘normal’ is used, a ‘pressure of 1 atm’ is implied

6. Similarly, if the external pressure is 1 bar, we do not write the pressure

• Instead, we use the word ‘standard’

• This can be explained in steps:

(i) Suppose that, the ‘temperature at which a liquid boils’ is T₂ K

(ii) Also suppose that, when this T₂ K is measured, the external pressure is 1 bar

(iii) We can report the result in two ways:

    ♦ The boiling point of the liquid is T₂ K at 1 bar pressure

    ♦ The standard boiling point of the liquid is T₂ K

          ✰ When the word ‘standard’ is used, a ‘pressure of 1 bar’ is implied

7. Let us see if there is any difference between the two methods mentioned in (5) and(6). We will take water as an example

• The normal boiling point of water is 100 ^oC

• The standard boiling point of water is 99.6 ^oC

• So indeed there is a difference

• The explanation can be given in 4 steps:

(i) 99.6 is less than 100

(ii) This implies that:

    ♦ If the external pressure is 1 bar,

    ♦ the molecules can escape a bit more easier than

    ♦ If the external pressure is 1 atm

(iii) This implies that:

    ♦ A pressure of 1 bar

    ♦ is less than

    ♦ A pressure of 1 atm

(iv) This is indeed true. Converting both units into M pa, we get:

    ♦ 1 bar = 0.1 M pa

    ♦ 1 atm =0.101 M pa

• It is clear that:

    ♦ If we take 1 atm, we get a pressure of 0.101 M pa

    ♦ If we take 1 bar, we get a pressure of only 0.1 M pa

• So 1 bar is slightly less than 1 atm

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• Based on the above discussion, we can now explain why it is difficult to cook food on mountain tops

• The explanation can be written in 3 steps:

1. At mountain tops, water boils at low temperatures

• We know the reason:

A low temperature is sufficient to give the water molecules the required energy to break off

2. So compare the two items:

(i) Molecules of some boiling water at the mountain top

(ii) Molecules of some boiling water at the sea level

• obviously, (i) will be having lesser energy

3. Now compare the two items:

(i) Raw food material put into the water mentioned in 2(i)

(ii) Raw food material put into the water mentioned in 2(ii)

• The raw food mentioned in (i) has only a 'lesser energy available to absorb' when compared to (ii)

• So we will need to cook it for a longer time in order to make it edible

• That is why we say that, it is difficult to cook food at mountain tops

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• Similarly, we can explain why it is easier to cook food in a pressure cooker

• The explanation can be written in 4 steps:

1. Inside the pressure cooker, the pressure is high

• That means, a high pressure is pressing down on the water molecules

2. Water in a pressure cooker will boil only at a high temperature

• We know the reason:

A high temperature is necessary to give the water molecules the required energy to break off

3. So compare the two items:

(i) Molecules of some boiling water in a pressure cooker

(ii) Molecules of some boiling water in an ordinary vessel

• obviously, (i) will be having greater energy

4. Now compare the two items:

(i) Raw food material put into the water mentioned in 3(i)

(ii) Raw food material put into the water mentioned in 3(ii)

• The raw food mentioned in (i) has a 'greater energy available to absorb' when compared to (ii)

• So we will need to cook it only for a shorter time in order to make it edible

• That is why we say that, it is easier to cook food in a pressure cooker

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• Next we will see how the ‘relation between boiling point and external pressure’ can be shown graphically

• We will use water as an example. It can be written in 7 steps:

1. Apply a pressure of p₁ on a sample of water

• Heat that water until it boils

• Note down the temperature T₁ at which it boils

2. Apply a pressure of p₂ on that sample

• Heat it until it boils

• Note down the temperature T₂ at which it boils

3. Repeat the steps several times with different pressure values. We get the points: (p₁,T₁), (p₂,T₂), (p₃, T₃) . . . so on . . .

4. Plot temperature along the x-axis

• Plot pressure along the y-axis

• We will get the green curve shown in fig.5.30(a) below:

Fig.5.30

5. Now, 1 atm pressure is equivalent to 760 mm mercury

• So draw a horizontal white dashed line through 760 mm mercury

• This horizontal dashed line will intersect the green curve at a point

6. Through this point of intersection, draw a vertical white dashed line

• This vertical dashed line will meet the x-axis at the normal boiling point

• For water, it will be 373 k

7. Graphs of other liquids can be drawn in this way

• Fig.b shows the graphs of diethyl ether, carbon tetrachloride etc.,

• Once their graphs are plotted, we can easily find the normal boiling points by drawing vertical dashed lines

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• In the next section, we will see surface tension and viscocity

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