Second Law of Thermodynamics

Introduction

One of the basic principles ruling physical processes, the Second Law of Thermodynamics deals with the spontaneous direction of energy transfer and the new concept of entropy. It says that the sum of the entropies of an isolated system will always grow larger or remain constant over time, never decreasing, for any natural thermodynamic process. It is the first law to introduce irreversibility in natural processes, while those described by the First Law of Thermodynamics are reversible. The Second Law has many statements, one of which is the Kelvin-Planck statement: "No heat engine can convert all of the heat supplied to it by a single thermal reservoir into work with no other effect." Another is the Clausius statement: "Heat cannot spontaneously flow from a colder body to a hotter one.". These formulations underline the intrinsic limitations of energy transformations and the tendency towards equilibrium and maximum entropy. Another important point that the Second Law supports is that entropy change (ΔS), as the measurement of disorder or randomness, is very important in controlling spontaneity.

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This Story also Contains

1. Introduction
2. Second Law of Thermodynamics
3. Some Solved Examples
4. Summary

Second Law of Thermodynamics

It is not possible to convert heat into work without compensation.

• Work can always be converted into heat but the conversion of heat into work does not take place under all conditions.

• It is impossible to construct a machine that is able to convey heat by a cyclic process from a colder to a hotter body unless work is done on the machine by some outside agency (Clausius statement).

• The heat of the coldest body among those participating In a cyclic process cannot serve as a source of work (Thomson statement).

• It is Impossible by means of the inanimate material agency to derive mechanical work or effort from any portion of matter by cooling it below the temperature of the coldest of the surrounding objects (Kelvin-Planck statement)

• Nature tends to pass from a less probable to a more probable state (Ludwig Boltzmann statement).

• Whenever a spontaneous process takes place it is accompanied by an increase in the total entropy of the universe.

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Some Solved Examples

Example 1: Which of the following processes is iso-entropic?

1)Isobaric process

2) Adiabatic process

3)Isothermal process

4)Isochoric process

Solution

As we have learned,

According to the definition of entropy,

Change in entropy,

$\Delta S=\frac{\Delta Q_{\text {rev }}}{T}$

As the heat change in the adiabatic process is equal to zero. Hence the change in entropy will be zero, which means the entropy will change in the process.

So the adiabatic reversible process will iso-entropic process as well.

Example 2: The entropy (S^o) in (JK^-1mol^-1) of the following substances are :

$\begin{aligned} & \mathrm{CH}_4(\mathrm{~g}): 186.2 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ & \mathrm{O}_2(\mathrm{~g}): 205.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ & \mathrm{CO}_2(\mathrm{~g}): 213.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ & \mathrm{H}_2 \mathrm{O}(\mathrm{l}): 69.9 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\end{aligned}$

The entropy change $\Delta S^0$ for the reaction $\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ is :

1) -312.7

2)-242.8

3)-108.1

4)-37.6

Solution

In the given reaction:

$\begin{aligned} & \mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \\ & \Delta \mathrm{S}=\sum \mathrm{S}^{\circ}(\text { products })-\sum \mathrm{S}^{\circ}(\text { reactants }) \\ & \Delta \mathrm{S}=(213.6+69.9)-\{186.2+2(205)\} \\ & \therefore \Delta \mathrm{S}=-312.7 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\end{aligned}$

Hence, the answer is the option

Example 3: A sample of 18 g of H₂O is slowly heated from 300 K to 400 K. Calculate the change in entropy (in J/K) during heating. (Given: specific heat of water is $4200 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}_{\mathrm{f}}}$)

1)7.81

2) 21.75

3)-7.81

4)36.25

Solution

As we learned,

$\Delta \mathrm{S}=\mathrm{m} \times \mathrm{C}_{\mathrm{P}} \times \operatorname{In} \frac{\mathrm{T}_2}{\mathrm{~T}_1}$

$\Rightarrow \Delta \mathrm{S}=\frac{18}{1000} \times 4200 \times \operatorname{In} \frac{400}{300}$

$\Rightarrow \Delta \mathrm{S}=21.75 \mathrm{~J} / \mathrm{K}$

Hence, the answer is ($\Delta S=21.75 \mathrm{~J} / K)$)

Summary

The Second Law of Thermodynamics states that the total entropy of an isolated system cannot diminish but either increases or remains constant over time; thus, it forms the spontaneity of energy transfer and the irreversibility in processes. It refers to an intrinsic tendency to disorder and equilibrial states; hence, it is not possible to transform heat into work without any other effects, totally. This law is expressed by means of the Kelvin-Planck statement, which forbids the perfect conversion of heat into work, and the Clausius statement, which rejects the spontaneous flow of heat from cold to hot bodies. Introducing the concept of entropy change, ΔS, the Second Law is of help in establishing the spontaneity of a process; positive total entropy changes indicate spontaneity. The implications, in effect, extend to all fields of thermodynamics, physical chemistry, and engineering with regard to how such energy systems are understood and designed.