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The third law of thermodynamics states that the entropy of a crystal in its pure state at temperature of zero kelvin or absolute zero temperature is equivalent to zero. It studies different forms of energy and the relationship between them.
- The third law of thermodynamics is related to the behaviour of systems as the temperature is equivalent to absolute zero.
- The law was proposed by Walter Nernst.
- It is also referred to as Nernst Law.
- This absolute zero is a constant value that can’t be supported by any alternate factors, like applied magnetic field or pressure.
- In other words, the third law of thermodynamics describes entropy.
- It describes the behaviour of entropy and the properties of a system.
- At absolute zero temperature, the system has the minimum possible state of energy.
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Key Terms: Third Law of Thermodynamics, Entropy, Absolute Zero, Temperature, Energy, Kelvin, Heat Exchange, Molecules, Ground State
Third Law of Thermodynamics
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The third law of thermodynamics states that once the system reaches absolute zero temperature, its entropy reaches a constant value. The law also helps in explaining the behaviour of solids at low temperatures.
- It means with both temperatures at absolute zero, it creates a minimum level of randomness in a system.
- If the combination of two thermodynamic systems forms an isolated system, it will create an energy exchange between them.
- The third law of thermodynamics primarily determines modern industries' work patterns and machinery design concepts.
- It can be applied to anything and everything in this universe because the law comprehensively relates to the universe.
- According to the law, everything falls under one system and depends entirely on the temperature.
Third Law of Thermodynamics
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Absolute Zero
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Absolute zero is the minimum point of temperature at which there is nearly no particle in motion, and no exchange of heat takes place. The lowest temperature equals -273.15 degrees Celsius, -459.67 degrees Fahrenheit and 0 degrees Kelvin. The developments that can be noticed in a closed system at absolute zero temperature are as follows:
- There is no heat in the system
- The energy points of all the system's atoms and molecules are at the minimum level.
Hence, there is only a single accessible microstate for a system at absolute zero temperature, and it is known as the ground state. The third law of thermodynamics states the entropy of the system mentioned above is precisely zero.
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Entropy
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Entropy is defined as the measure of system thermal energy per unit temperature. It also measures the molecular disorder in a system. The term entropy is denoted by the symbol 'S'.
- It is relative to the number of microstates (a definite microscopic state possessed by a system) that a system can access.
- This means the more microstates the closed system has, the higher its entropy will be.
- The microstate wherein the system has the lowest energy is termed the system's ground state.
- Any variation in entropy associated with a process from its initiation to its end is denoted by ΔS.
- Nature plays a vital role in creating disorder compared to order, like ageing, rusting, decaying, etc.
- While work is done, the usable energy is converted to unusable energy.
- The more the energy is dispersed, the greater would be the entropy.
Entropy
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Trends & Physical Properties of Entropy
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The physical properties of entropy are as follows:
- Melting and evaporation of the system lead to higher entropy.
- Mixing of solids and liquids results in higher entropy.
- Dissolving gas in water lowers the entropy.
- Hard and brittle materials have lower entropy than pliable solids like metals.
- Chemical complexity results in higher entropy.
- An isothermal process states when the heat change (Q) is divided by the absolute temperature (T), it results in a change in entropy (ΔS).
- The SI units representing entropy are J/K (joules/degrees Kelvin).
ΔS = Q/T
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Third Law of Thermodynamics - Mathematical Explanation
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In statistical mechanics the equation representing the third law of thermodynamics is as follows:
S - S0 = kB lnΩ
Here,
- S represents the entropy of the system
- S0 represents the initial entropy
- kB represents the Boltzmann constant
- Ω specifies the total number of microstates consistent with the macroscopic configuration of the system.
In the case of a perfect crystal with only 1 unique ground state, Ω = 1. Hence, the revised equation can be:
S – S0 = kB ln (1) = 0 [because ln(1) = 0]
- While zero is selected as the system’s initial entropy, its value of ‘S’ can be obtained as follows:
S - 0 = 0 ⇒ S = 0
Therefore, a perfect crystal entropy at absolute zero temperature is zero.
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Application of the Third Law of Thermodynamics
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One of the prime applications of the third law of thermodynamics is that it supports calculating the absolute entropy at any particular temperature (T). Such conclusions are derived from the measurements of the heat capacity of the substance.
- In the case of solids, if S0 is the entropy at 0 degree K and S is the entropy at T degree K, then:
ΔS = S - So = ∫0T Cp dT/T
- Based on the third law of thermodynamics, S0 = 0 at 0 K, thus,
S = ∫0T Cp / T . dT
- This integral value can be found by plotting the graph of Cp / T versus T followed by determining the area of this curve starting from 0 to T.
- However, a simpler representation for the absolute entropy of a solid at temperature T is stated below:
S = ∫0T Cp / T . dT
S = ∫0T Cp d ln T
Cp In T = 2.303 Cp log T
- Here, Cp is represented as the heat capacity of the substance at constant pressure where the value is considered as constant in the 0 to T K range.
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Third Law of Thermodynamics Example
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The temperature of any gaseous form of water, such as steam or water vapour, is high, and the molecules inside these gases move randomly. Thus, it results in higher entropy.
- However, once the steam or water vapour cools down below 100 degrees Celsius, it will change into water.
- This will restrict the movement of the molecules and result in lower entropy.
- Another example of the third law of thermodynamics includes the measurement of the level of success and happiness in a family.
- Entropy is responsible for the measurement of success and happiness in life.
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Contradiction with the Other Laws of Thermodynamics
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Contrary to the third law of thermodynamics that determines absolute zero as a state, the second law of thermodynamics rules out the fact that temperature can actually become zero.
- As per the second law, it has been established that the spontaneous movement of heat from a colder medium to a hotter medium is not possible.
- In the process to reach absolute zero, the system tends to capture heat energy from the external environment.
- Thus, absolute zero temperature can never be achieved.
According to the first law of thermodynamics, the possibility of creation and destruction of energy is completely ruled out. In such a situation the heat energy comes from an external source outside the system. Therefore, once again the possibility of the system reaching absolute zero is eliminated.
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Things to Remember
- Third Law of Thermodynamics states that the entropy of pure crystal is zero at absolute zero temperature.
- The entropy of mixing with non-reacting substances in a system is invariably higher.
- Entropy is regarded as an extensive property.
- The process continues naturally in the direction of higher randomness or disorder.
- The application of the third law is helpful in calculating the properties of thermodynamics.
- The third law of thermodynamics determines the behaviour of solids at low temperatures.
- The law also supports examining chemical and phase equilibrium.
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Sample Questions
Ques. When does the third law of thermodynamics apply? (2 marks)
Ans. According to the third law of thermodynamics, “the entropy of a crystal in its pure state and at a zero Kelvin or absolute zero temperature is equivalent to zero. At this temperature, the system is considered to maintain the lowest energy. This statement of the third law of thermodynamics is applicable only if the crystal in its pure state has the lowest state of energy.
Ques. Predict in which of the following, entropy increases/decreases: (5 marks)
- A liquid crystallizes into a solid.
- The temperature of a crystalline solid is raised from 0 K to 115 K.
- 2NaHCO3 (s) → Na2CO3 (s) + CO2 (g) + H2O (g)
- H2 (g) → 2H (g)
Ans. The answers are as follows:
- Entropy decreases after freezing because the molecules attain an ordered state.
- The constituent particles are static at 0 K and entropy is minimum. If the temperature is raised to 115 K, these begin to move and oscillate about their equilibrium positions in the lattice and the system becomes more disordered, therefore entropy increases.
- Reactant, NaHCO3 is solid with lower entropy.
- The products are made up of one solid and two gases. Therefore, a higher entropy condition is expressed by the products.
- In this case, one molecule provides two atoms i.e. the number of particles increases resulting in a state of greater disorder.
- Two molecules of H atoms have higher entropy than one molecule of hydrogen atoms.
Ques. For the reaction NH4Cl(S)→NH3(g)+HCl(g) at 250 C, enthalpy change Δ=+177kJ mol−1 and entropy change ΔS = +285JK−1 mol−1. Calculate free energy change ΔG at 250C and product whether the reaction is spontaneous or not. (3 marks)
Ans. Given, ΔH=177 kJ/mol, ΔS=285 JK−1mol−1, T=298 K
Using the relation, ΔG=?H−T?S
- ΔG=(177×103 J/mol)−(298 K)(285 JK−1mol−1)
- ΔG=92070 J/mole
- ΔG=92.07 kJ/mole
- ΔG>0
Since ΔG is positive, therefore, the reaction is nonspontaneous.
Ques. For a reversible spontaneous reaction Δs is: (1 mark)
- ΔE/T
- PΔV/T
- q/T
- RT logK
Ans. The answer is 3. q /T
Explanation: Ds = dq.rev/T
- Δs = dq.rev/T
- dq.rev/T (for rev. proves)
- Δs = q/T
Ques. What is the impact of the Third Law of Thermodynamics on modern science and technology? (2 marks)
Ans. The third law of thermodynamics expresses a vital point in relation to the entropy of the universe which contradicts the theory of the expansion of the universe. In a universe that expands at a higher rate, the entropy will be greater with a lower temperature. Finally, on reaching zero Kelvin temperature, the entropy of the universe would subsequently be zero and it is fundamentally not possible.
Ques. What is the role of the Third Law of Thermodynamics in refrigeration? (3 marks)
Ans. According to the third law, there cannot be any environmental parameter that can be influenced to modify the entropy of a system at absolute zero. Refrigeration involves a cycle of alternating processes.
- It starts with maintaining a constant temperature and changing other parameters to generate heat from a system that would eventually result in lowering the entropy.
- The following step includes insulating the system and reversing the change in the parameter.
- The result is a lowering of the temperature.
- Thus, the third law of thermodynamics compatible with the alternate isothermal and adiabatic processes works and that helps in better refrigeration.
Ques. For the reaction 2Cl(g)→Cl2(g) what are the signs of ΔH and ΔS? (2 marks)
Ans. ΔH and ΔS are negative. The given reaction represents the forming of chlorine molecules from chlorine atoms. Bond formation occurs in this situation. Therefore, energy is being released. Hence, ΔH is negative. Moreover, the randomness of two moles of atoms is greater than one mole of atom. With lower spontaneity, ΔS is negative for the above reaction.
Ques. What are the difference between endothermic and exothermic reaction? (3 marks)
Ans. The difference between endothermic and exothermic reaction are as follows:
| Endothermic Reaction | Exothermic Reaction |
|---|---|
| It is a form of reaction that absorb energy from the system. | It is a form of reaction that release energy from the system. |
| In this energy is absorbed in the form of heat. | In this energy is absorbed in the form of heat, electricity, light or sound. |
| For example: photosynthesis | For example: rusting of iron |
Ques. Why is heat energy released or absorbed in a chemical reaction? (2 marks)
Ans. In a chemical reaction, chemical bonds are either created or destroyed. When bonds are formed, heat energy is released, and similarly, when chemical bonds are broken, then heat energy is absorbed. Since molecules want to remain together during chemical bonds' formation or breakage, heat energy is released or absorbed in a chemical reaction.
Ques. State the first and second laws of thermodynamics? (2 marks)
Ans. According to the first law of thermodynamics, energy can neither be created nor destroyed but transferred from one form to another. According to the second law, a system's thermodynamics entropy continuously increases in a chemical reaction.
Ques. What are the physical properties of entropy? (3 marks)
Ans. The physical properties of entropy are as follows:
- The entropy of a substance decreases when gas is dissolved in water.
- The entropy of a substance increases when a solid is mixed with liquid.
- The entropy of a substance increases when its melting point increases.
- The entropy of a substance increases with increases in the complexity of a chemical reaction.
Ques. Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions at 298 K. Given ΔrH0 = - 286 kJ mol−1? (3 marks)
Ans. H2 (g) +21O2 (g) ⟶H2O (l) ΔfH0 = −276KJ/mol
- From the above equation,
- At 298K, when 1 mole of H2O(l) is formed, 276KJ of heat is released. The same amount of heat is absorbed by the surroundings.
- qsurr. =+ 276KJ/mol; T=298K
- As we know that,
- ΔSsurr.= qsurr. / T
- ΔSsurr. = 276 / 298 = 0.92 KJ / mol
Hence the entropy change in surroundings will be 0.92 KJ/mol.
Ques. Explain the Zeroth law of thermodynamics? (2 marks)
Ans. Zeroth Law of Thermodynamics states that when body A is in thermal equilibrium with two separate bodies, B and C, respectively, then body B and C are in thermal equilibrium with each other. The law was proposed by Ralph L Fowler and is based on temperature measurement. Temperature is an important parameter as it will determine whether heat exchange between object take place or not.
Ques. Given the following entropy values (in J/K−mol) at 298K and 1atm H2(g)=120.6, Cl2(g)=220.0, HCl(g)=176.7. The entropy change (in J/K−mol) for the reaction H2 (g) + Cl2 (g) → 2HCl (g), is: (3 marks)
Ans. ΔSHCl(g) = 176.7J/mol
- ΔSH2(g) =120.6J/Kmol
- ΔSCl2(g) = 220.0J/Kmol
- H2(g) + Cl2(g) → 2HCl(g)
- ΔSreaction= 2×ΔSHCl(g) −(ΔSH2(g)+ΔSCl2(g))
- 2×176.7−(120.6+220)
- 353.2 − 340.6
- 12.6 J/K−mol
Ques. What is the chemical reaction involved in the third law of thermodynamics? (1 mark)
Ans. The chemical reaction involved in the third law of thermodynamics is as follows:
S - S0 = kB lnΩ
where, S represents the entropy of the system
- S0 represents the initial entropy
- kB represents the Boltzmann constant
- Ω specifies the total number of microstates consistent with the macroscopic configuration of the system.
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