Chapter 6.13 - Entropy

In the previous section, we saw some basics about spontaneity. In this section, we will see entropy and the second law of thermodynamics

• We saw that, being exothermic/endothermic is not the only criterion for spontaneity
• So to investigate further, we consider a system which cannot be exothermic or endothermic
    ♦ For that, we consider a system in which there is no heat transfer
    ♦ Such a system is called an isolated system
• It can be written in 18 steps

1. Consider the closed container in fig.6.21(a) below. It is an isolated container
• It is ‘isolated’ because, there is no flow of energy into or out of the container

Fig.6.21

2. On the left side of the container, there are some cold molecules of an ideal gas
    ♦ They are represented in blue color
• The fact that ‘they are cold’ is emphasized by the 'short arrows' indicating 'small velocities'
3. On the right side, there some hot molecules of the same ideal gas
    ♦ They are represented in red color
• The fact that ‘they are hot’ is emphasized by the 'long arrows' indicating 'large velocities'
4. The two sides are separated by a temporary partition shown in green color
• So the left side of the container will be cold and the right side will be hot
5. Let us remove the temporary partition at the middle
• The cold and hot molecules will begin to collide with each other
    ♦ The red molecules will lose some energy
    ♦ The blue molecules will gain some energy
• As a result, all the molecules will attain an intermediate velocity
• Thus, all the molecules will attain an intermediate temperature
• The fact that ‘they are at intermediate temperature’ is emphasized by the 'intermediate arrows' in fig.b, indicating 'intermediate velocities'
6. Since there is no net loss or gain, we can say: the energy is conserved
• But the energy is now distributed evenly
7. Now, two questions arise in our minds:
• After removal of the partition and intermixing,
    ♦ Why don't the blue molecules become cold and return to the left side?
    ♦ Why don't the red molecules become hot and return to the right side?
• If they return to their original sides, it would look like as in fig.c
◼ So the question is this:
Why don't we see the arrangement in fig.c?
• We can find the answer with the help of simple statistics
• The following steps from (8) to (11) will give the answer
8. Once the partition is removed, there are many possible arrangements
• The arrangement in fig.c is only one among the many possible arrangements
9. In fact, the intermixed arrangement in fig.b is also only one among the many possible arrangements
    ♦ A blue molecule at top left in fig.b, may move to bottom right
    ♦ A red molecule at bottom left in fig.b, may move to top right
    ♦ so on
• Such movements will give different intermixed arrangements
10. But the ordered arrangement in fig.c do not have many options
• This is because:
    ♦ The blue molecules have to stay on the left side
    ♦ The red molecules have to stay on the right side
11. So it is clear that:
    ♦ The number of possible intermixed arrangements as in fig.b
    ♦ is far greater than
    ♦ The number of ordered arrangements as in fig.c
12. We know that, even one mole of a gas will contain 6.023  × 10²³ molecules
• Remember that one million is only 1 × 10⁵
• So in a practical situation,
    ♦ The probability of finding intermixed arrangements as in fig.b is nearly 1 (100%)
        ✰ Like the decimal: 0.9999
    ♦ The probability of finding an ordered arrangement as in fig c is very close to zero
        ✰ Like a fraction with one in the numerator and 10²³ in the denominator
        ✰ An arrangement with such a low probability will not occur
• So we never see cold molecules on one side and hot molecules on the other side
13. If heat is to flow from cold to hot, it means that:
• The cold portions get colder and colder and settle down at a portion of the container
• The hot portion get warmer and warmer and occupy the remaining portion of the container
◼ Such an orderly arrangement have only a very low probability
• That is why we never see spontaneous flow of heat from a cold body to a hot body
14. Now consider a cold body and a hot body of same material and size
• The molecules in the cold body will be more or less stationary
    ♦ They do not move much
• The molecules in the hot body will be moving greater distances
15. We can compare the ‘number of possible arrangements’
• The ‘number of possible arrangements’ for the molecules in the hot body is larger
    ♦ Because, the hot molecules have greater mobility
• The ‘number of possible arrangements’ for the molecules in the cold body is smaller
    ♦ Because, the cold molecules have lesser mobility
16. ‘Number of possible arrangements’ is called entropy
• It’s symbol is ‘S’
◼ We can say:
    ♦ A hot body
    ♦ Has greater entropy
    ♦ Than a cold body
17. Besides temperature, the 'natural arrangement' of molecules also has a major role in deciding entropy of a body
• We know that:
    ♦ Molecules in a gas, naturally have very high mobility
    ♦ Molecules in a liquid, naturally have intermediate mobility
    ♦ Molecules in a solid, naturally have very low mobility
• So it is clear that
    ♦ Gases have the highest entropy
    ♦ Liquids have intermediate entropy
    ♦ Solids have the least entropy
◼ We can write: S_Gas > S_Liquid >S_Solid
18. Comparison between a neat room and a messy room is often used to explain entropy. It can be written in 6 steps:
(i) Consider a neat room
• We call it neat because, every object in that room is in it’s proper place
   ♦ Books are stacked neatly in the book shelf
   ♦ The table-lamp is at the left side of the table
   ♦ Monitor, keyboard and mouse are kept together as a system
   ♦ so on . . .
(ii) Consider the messy room
• We call it messy because, objects are not in their proper places
   ♦ Some books are on the shelf, some on the table, some on the floor etc.,
   ♦ The table-lamp is placed at a corner of the room
   ♦ Monitor is on the table
   ♦ But keyboard is on the floor
   ♦ Mouse is below the bed
    ♦ so on . . .
(iii) Now let us consider the possible arrangements in the neat room
• The neat room can have many different arrangements and still be neat
   ♦ In the book shelf
         ✰ The physics text book can be stacked below the chemistry text book
         ✰ Or the chemistry text book can be stacked below the physics text book
         ✰ In either arrangement, the room will be neat
   ♦ The table-lamp can be either on left or right side of the table according to convenience
         ✰ In either arrangement, the room will be neat
   ♦ The mouse can be either on the left or right side of the monitor according to convenience
         ✰ In either arrangement, the room will be neat
• Thus we see that, there are many possible arrangements in a neat room. The room will still be neat
(iv) Next let us consider the possible arrangements in the messy room
• The messy room can also have many different arrangements and still be messy
   ♦ The books can be any where in the room
         ✰ Above the shelf, below the shelf, north east corner, south west corner etc.,
         ✰ The room will still be messy
   ♦ Like wise, the table-lamp, monitor, keyboard, mouse etc., can also be any where
         ✰ The room will still be messy
(v) Now we can write about the number of possible arrangements
◼  The neat room has many number of possible arrangements by which it can be neat
◼  The messy room has infinite number of possible arrangements by which it can be messy
(vi) Applying entropy:
   ♦ The number of possible arrangements for the messy room
   ♦ is greater than
   ♦ The number of possible arrangements for the neat room
• So we say that:
   ♦ The messy room
   ♦ has greater entropy
   ♦ Than the neat room

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• Now we will see the role of entropy in a chemical/physical process
• During chemical reactions, there will be rearrangements of atoms and ions
   ♦ If the products have a more disordered state than the reactants, we can say:
         ✰ Entropy increases
   ♦ If the products have a more ordered state than the reactants , we can say:
         ✰ Entropy decreases
• The following solved example demonstrates this aspect

Solved example 6.22
Predict in which of the following, entropy increases/decreases:
(i) A liquid crystallizes into a solid.
(ii) Temperature of a crystalline solid is raised from 0 K to 115 K
(iii) 2NaHCO₃(s)  → Na₂CO₃(s) + CO₂(g) + H₂O(g)
(iv) H₂(g) → 2H(g)
Solution:
Part (i)
(i) Let A be the initial state and B the final state
   ♦ Let S_A be the entropy of  the system at state A
   ♦ Let S_B be the entropy of  the system at state B
(ii) Comparing the states:
   ♦ At state A, the system is in liquid state
   ♦ At state B, the system is the crystalline (solid) state
(iii) Comparing entropies:
   ♦ Liquid state
   ♦ has a higher entropy than
   ♦ Solid state
• So S_B will be smaller than S_A
• We can predict that, in this process, the entropy decreases
Part (ii)
(i) Let A be the initial state and B the final state
   ♦ Let S_A be the entropy of  the system at state A
   ♦ Let S_B be the entropy of  the system at state B
(ii) Comparing the states:
   ♦ At state A, the system is in solid state but the temperature is 0 K
   ♦ At state B also, the system is the solid state, but the temperature is higher
(iii) Comparing entropies:
   ♦ At 0 K, all molecules become motion less
   ♦ So the solid at 0 K has lesser disorder
• So S_B will be larger than S_A
• We can predict that, in this process, the entropy increases
Part (iii)
(i) Let A be the initial state and B the final state
   ♦ Let S_A be the entropy of  the system at state A
   ♦ Let S_B be the entropy of  the system at state B
(ii) Comparing the states:
   ♦ At state A, the system is in solid state
   ♦ At state B, the system has one solid and two gases
(iii) Comparing entropies:
   ♦ At state B, there is greater entropy due to the presence of two gases
• So S_B will be larger than S_A
• We can predict that, in this process, the entropy increases
Part (iv)
(i) Let A be the initial state and B the final state
   ♦ Let S_A be the entropy of  the system at state A
   ♦ Let S_B be the entropy of  the system at state B
(ii) Comparing the states:
   ♦ At state A, the system is in gaseous state
   ♦ At state B also, the system is in gaseous state
(iii) Comparing entropies:
   ♦ At state B, there is greater entropy due to the presence of more particles
         ✰ Each H2 molecule is dissociated into two H atoms
         ✰ Larger number of particles increases the 'possible number of arrangements'
• So S_B will be larger than S_A
• We can predict that, in this process, the entropy increases

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For chemical reactions, a more precise method can be used. It can be explained in 8 steps:
1. We have the general form of a balanced equation:
a₁R₁ + a₂R₂ + a₃R₃ + . . . →  b₁P₁ + b₂P₂ + b₃P₃ + . . .   
• R₁, R₂, R₃, . . . are the reactants
    ♦ a₁, a₂, a₃, . . . are the ‘number of mol’ of each of those reactants
• P₁, P₂, P₃, . . . are the products
    ♦ b₁, b₂, b₃, . . . are the ‘number of mol’ of each of those products
2. Let us write the S^⊖ values (standard entropy values). They can be looked up from the data book
• S^⊖ values of the reactants R₁, R₂, R₃, . . .
    ♦ They can be written as: S^⊖[R₁], S^⊖[R₂], S^⊖[R₃], . . .
• S^⊖ values of the products P₁, P₂, P₃, . . .
    ♦ They can be written as: S^⊖[P₁], S^⊖[P₂], S^⊖[P₃], . . .
3. The above S^⊖ values that we obtain from the data book, are related to one mole
• For example, the S^⊖ value of Al₂O₃(s) is 50.92 J K^-1 mol^-1
    ♦ The unit 'J K^-1 mol^-1' indicates that, the values are related to one mol
    ♦ (In the next section, we will see how this unit 'J K^-1 mol^-1' is obtained)
• But in our present case, we have:
    ♦ a₁ mol of reactant R₁
    ♦ a₂ mol of reactant R₂ . . . so on
• So we must multiply the values by the corresponding mol number
• Thus the step (2) can be modified as:
    ♦ S^⊖ values of the reactants: a₁S^⊖[R₁], a₂S^⊖[R₂], a₃S^⊖[R₃], . . .
    ♦ S^⊖ values of the products: b₁S^⊖[P₁], b₂S^⊖[P₂], b₃S^⊖[P₃], . . .
4. Now we find two sums:
(i) Sum for the reactants:
$\mathbf\small{\rm{\sum\limits_{i}{a_i {S^\circleddash}[R_i]}}}$ = a₁S^⊖[R₁] + a₂S^⊖[R₂] + a₃S^⊖[R₃] . . .
(ii) Sum for the products:
$\mathbf\small{\rm{\sum\limits_{i}{b_i {S^\circleddash}[P_i]}}}$ = b₁S^⊖[P₁] + b₂S^⊖[P₂] + b₃S^⊖[P₃] . . .
5. Next we subtract the first sum from the second sum
• The result is: standard entropy change of reaction
    ♦ It is denoted as: ΔS^⊖_r
6. Thus we get:
Eq.6.13: $\mathbf\small{\rm{{\Delta S^\circleddash }_r=\sum\limits_{i}{b_i {S^\circleddash}[P_i]}-\sum\limits_{i}{a_i {S^\circleddash}[R_i]}}}$
7. If 𝚺b_iS^⊖[P_i] is larger than 𝚺a_iS^⊖[R_i], we can say: Products have a greater entropy
• Also, we will get a positive value for ΔS^⊖_r
◼  So we can write:
A positive ΔS^⊖_r indicates an increase in entropy
8. If 𝚺b_iS^⊖[P_i] is smaller than 𝚺a_iS^⊖[R_i], we can say: Products have a lesser entropy
• Also, we will get a negative value for ΔS^⊖_r
◼  So we can write:
A negative ΔS^⊖_r indicates a decrease in entropy

Let us see two solved examples:
Solved example 6.23
Find the entropy change for the following reaction:
4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)
Solution:
1. From the data book, we have:
   ♦ S^⊖ of Fe(s) = 27.28 J K^-1 mol^-1
   ♦ S^⊖ of O₂(g) = 205.14 J K^-1 mol^-1
   ♦ S^⊖ of Fe₂O₃(s) = 87.40 J K^-1 mol^-1
2. Applying the coefficients, we get:
   ♦ S^⊖ of the first reactant = 4 × 27.28
   ♦ S^⊖ of the second reactant = 3 × 205.14
   ♦ S^⊖ of the product = 2 × 87.40
3. Applying Eq.6.13, we get:
ΔS^⊖_r of the reaction = [(2 × 87.40) - (4 × 27.28) - (3 × 205.14)] = -549.74 J K^-1 mol^-1
• This is a negative value. So there is a decrease in entropy

Solved example 6.24
Find the entropy change for the following reaction:
N₂(g) + 3H₂(g) → 2NH₃(g)
Solution:
1. From the data book, we have:
   ♦ S^⊖ of N₂(g) = 191.61 J K^-1 mol^-1
   ♦ S^⊖ of H₂(g) = 130.68 J K^-1 mol^-1
   ♦ S^⊖ of NH₃(s) = 192.45 J K^-1 mol^-1
2. Applying the coefficients, we get:
   ♦ S^⊖ of the first reactant = 1 × 191.61
   ♦ S^⊖ of the second reactant = 3 × 130.68
   ♦ S^⊖ of the product = 2 × 192.45
3. Applying Eq.6.13, we get:
ΔS^⊖_r of the reaction = [(2 × 192.45) - (1 × 191.61) - (3 × 130.68)] = −198.75 J K^-1 mol^-1
• This is a negative value. So there is a decrease in entropy

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• In the next section, we will see the second law of thermodynamics


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Chapter 6 - Thermodynamics

In the previous chapter, we completed a discussion on states of matter. In this chapter we will see thermodynamics.
■ Let us first see an example in which energy is converted from one form to another. It can be written in 5 steps:
1. We know that, when fuels like methane and coal burn, heat is released.
• When fuels burn, the chemical energy stored in the molecules is converted into heat energy.
2. This heat energy can be supplied to a gas inside a cylinder.
3. The heated gas will expand and push the piston of the cylinder.
4. The moving piston will rotate the wheel of an automobile.
5. Thus chemical energy is converted into mechanical energy.
■ Another example is the galvanic cell in which chemical energy is converted into electrical energy.
• There are many situations where we will need to transform energy from one form to another. So some questions arise:
    ♦ How much energy can be obtained from a chemical reaction or process ?
    ♦ What all conditions need to be satisfied to carry out the reaction or process ?
    ♦ How much materials will be required to obtain the desired quantity of energy ?
    ♦ Will it be economical to carry out such a reaction or process ?
• The principles of thermodynamics will help us to find the answers to such questions.

Now we will see some thermodynamic terms. They can be explained in 12 steps:
1. Consider the chemical reaction between two reactants X and Y.
• We will be observing the changes happening to the molecules of X and molecules of Y.
2. Imagine a group consisting of the two items:
    ♦ All the molecules of X.
    ♦ All the molecules of Y.
• Then this group is called the system.
3. Objects and particles other than those mentioned in (2) will constitute the surroundings.
4. System and surroundings together constitute the universe.
• So we can write: Universe = System + Surroundings.
5. The green region in fig.6.1(a) below is the universe.

Fig.6.1

6. In this universe, a container is placed on a table top.
• Inside the container, there are some magenta molecules and some cyan molecules.
    ♦ All those molecules (magenta and cyan) together constitute the system.
7. Usually, the ‘neighborhood of the system’ is affected by the 'reactions or processes' undergone by the system.
• For example, if the system releases some heat, the neighborhood will become hot.
• The ‘neighborhood of the system’ is shown separately in fig.b.
    ♦ This neighborhood is the surroundings.
8. So if we add the molecules into fig.b. we will get fig.c, which is the universe.
9. Movement of matter
• Sometimes matter may move from the system into the surroundings.
    ♦ For example, if the system is water and if that water is heated, some water molecules will escape in the form of steam.
• The reverse can also happen. That is., matter may move from the surroundings into the system.
    ♦ For example, while a reaction is going on, we may add small quantities of one of the reactants.
10. Movement of energy
• Sometimes energy may flow from the surroundings into the system.
    ♦ For example, the gas inside a cylinder may absorb heat from the surroundings.
• The reverse can also happen. That is., energy may flow from the system into the surroundings.
    ♦ For example, if the reaction is exothermic, the heat liberated may flow into the surroundings.
11. To estimate the exact quantities of matter and energy flowing in either direction, we must define a boundary for the system.
    ♦ Every thing inside that boundary will be the system.
    ♦ Every thing outside that boundary will be the surroundings.
12. The boundary can be a thin imaginary layer on the inner surface of the container.
• In fig.6.2(a) below, this layer is shown in green color.

Fig.6.2

• In fig. (b), the container is closed with a lid.
    ♦ The thin layer at the bottom surface of the lid is shown in pink color.
    ♦ It is given another color to indicate that, it can be moved away from the green layer.
• In fig. (c), the container is a cylinder-piston arrangement.
    ♦ The layer at the bottom surface of the piston is given a pink color.
    ♦ It is given a different color to indicate that, it can be moved relative to the green layer.

Different types of systems

• Systems can be classified according to:
    ♦ the movements of matter into or out of the system.
    ♦ the movements of energy into or out of the system.
• The three types of systems are: Open system, Closed system and Isolated system.
• We will now see each of them in detail.

Open system

This can be explained in 3 steps:
1. In this system, there is exchange of matter and energy between the system and the surroundings. This is shown in fig.6.3(a) below:

Fig.6.3

2. The reactants are placed in an open container.
• The imaginary green layer allows the passage of both matter and energy.
• So matter can move in or out through the top.
3. Energy can move in or out through the top as well as the sides.
• For example, heat produced inside is allowed to pass through the green layer.
• It reaches the walls of the container. If the walls are made of materials like copper, heat reaches the outer surface of the container and from there, dissipates into the surroundings.

Closed system

This can be explained in 3 steps:
1. In this system, there is exchange of energy, but no exchange of matter. This is shown in fig.6.3(b) above.
2. The reactants are placed in a closed container.
• Heat produced inside is allowed to pass through the green and pink layers.
• The heat reaches the walls and lid of the container and from there, dissipates into the surroundings.
3. But matter can not move in this way.
• We assume that, the green and pink layers in this closed system do not allow matter to pass through.
• But the layers allow the passage of energy.
 

Isolated system

This can be explained in 2 steps:
1. In this system, neither matter nor energy is exchanged. This is shown in fig.6.3(c) above.
2. We can assume that, the green and pink layers have insulating properties.
• Reaction carried out inside a thermos flask is an example for an isolated system.

State of the system

This can be explained in steps:
1. Consider a system such as the one shown in fig.6.4(a) below:

Fig.6.4

• It will have a certain pressure, volume, temperature and mass.
• Let us see each of these properties in detail.
■ Pressure:
• The molecules in the system will be exerting certain pressure on the walls of the container.
    ♦ This pressure can be measured using a pressure gauge.
    ♦ It is just like measuring the pressure of air inside the wheel of an automobile.
■ Volume:
• The system will be having a certain volume.
    ♦ In fig.6.4(a), it is the volume of the container.
■ Temperature:
The system will be having a certain temperature.
    ♦ This temperature can be easily measured using a thermometer.
• In our previous classes, we have see that, temperature represents the average kinetic energy of the molecules in the system.
■ Mass:
• The system will be having a certain mass.
    ♦ Usually, instead of mass, we use the number of moles, n.
2. If there is no flow of matter and energy into or out of the system, P, V, T and n will remain constant.
• In such a condition, we measure them and record them as: P_A, V_A, T_A and n_A
    ♦ The subscript 'A' indicates the initial state.
• After recording them, we start the reaction or the process.
3. A reaction can be started in different ways. Here we mention two most common methods:
(i) Add desired quantity of another reactant into the system.
(ii) Add heat into the system. For example, some decomposition reactions require heat energy.
4. A process can also be started in different ways. Here we mention two most common methods:
(i) Add heat into the system. A gaseous system will expand by absorbing heat.
(ii) Compress the system using a piston.
5. In any case, when the reaction/process is complete, the system will be having new values for P, V and T. (If any matter was added, n will also change)
• Record the new values as P_B, V_B, T_B and n_B
    ♦ The subscript 'B' indicates the final state.
• So we can write:
    ♦ P_A, V_A, T_A and n_A help us to describe the initial state of the system.
    ♦ P_B, V_B, T_B and n_B help us to describe the final state of the system.
6. P, V, T and n are called state variables.
State variables do not depend on the path. This can be explained using some examples:
Example 1:
• This can be written in 4 steps:
(i) Let V_A and V_B be the initial and final volumes.
(ii) We can heat the system and increase the volume directly from V_A to V_B
    ♦ This is shown in fig.6.4 (b) above.
(iii) We can first cool the system and decrease the volume to a lower value V_A1
    ♦ Then we can heat the system and increase the volume from V_A1 to V_B
    ♦ This is shown in fig. 6.4(c) above.
(iv) In either case, the final volume V_B does not depend on the path.
Example 2:
• This can be written in 4 steps:
(i) Let T_A and T_B be the initial and final temperatures.
(ii) We can heat the system and increase the temperature directly from T_A to T_B
(iii) We can first cool the system and decrease the temperature to a lower value T_A1
    ♦ Then we can heat the system and increase the temperature from T_A1 to T_B
(iv) In either case, the final temperature T_B does not depend on the path.
7. So we have seen variables which do not depend on path. Then a question arises:
■ Are there variables which depend on the path?
• The answer is: Yes.
• Heat is a quantity that depends on the path.
This can be explained using an example. It can be written in 4 steps:
(i) Suppose that, a copper piece is to be heated from 25 °C to 175 °C
(ii) One person may give an excess heat and bring it to 200 °C first. Then he will cool it to 175 °C
• The heat given out during cooling is lost to the surroundings. That heat is not available any more.
• So overall, an excess heat is spent to bring the copper piece to the required 175 °C
(iii) Another person may give the exact amount of heat to bring the copper piece to 175 °C
• So in this case, no heat is lost.
(iv) Thus we see that, heat is path dependent.
■ Another quantity which is path dependent is: Work
We will see it’s details in later sections.

Macro properties and Micro properties

This can be explained in 5 steps:
1. Properties like pressure, volume, temperature, number of moles etc., are called macro properties of the system.
• This is because, they are related to the bulk of the system.
■ We can insert a pressure gauge at any point of the system.
• Whichever be the point of insertion, the reading in the pressure gauge will be the same.
■ We can insert a thermometer at any point in the system.
• Whichever be the point of insertion, the reading in the thermometer will be the same.
■ Volume of a system is the overall volume.
• We will never separate a portion of the system and record it’s volume.
■ n is the overall number of moles.
• We will never separate a portion of the system and record the number of moles in it.
2. We can write:
P, V, T and n are related to the total bulk of the system. So they are called macro properties
• We will see more such macro properties in later sections.
3. Consider the individual molecules of the system. Let us write some properties of those molecules:
■ Each molecule will have it’s own speed and direction.
• In other words, each molecule will have it’s own velocity.
• Some molecules will be moving towards left, some others towards right, some others upwards so on . . .
• Let us number the molecules as 1, 2, 3, . . . , n
    ♦ Then we can write the velocities as: v₁, v₂, v₃, . . . , v_n
■ Since, each molecule has it’s own velocity, each of them will be having it’s own kinetic energy also.
    ♦ We can write the kinetic energies as: KE₁, KE₂, KE₃, . . . , KE_n
■ Since the molecules are always in motion, they will be striking against the walls of the cylinder.
• While striking the walls, each molecule will be exerting a force on the walls.
    ♦ We can write the forces as: F₁, F₂, F₃, . . . , F_n
4. The three properties mentioned in the above step, are related to individual molecules. They are called micro properties of the system.
    ♦ We are considering such properties at the micro level.
5. The macro properties and micro properties are related to each other.
• But for our present discussion in thermodynamics, we do not need the micro properties. We need the macro properties only.
• We can arrive at the required results with out knowing the micro properties.
• In fact, the discoveries in thermodynamics were made long before individual particles like molecules, atoms, electrons, protons etc., were discovered.

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In the next section, we will see internal energy of a system


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