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At absolute zero, the entropy of a pure crystal is zero. This statement is:
a) First law of thermodynamics
b) Second law of thermodynamics
c) Third law of thermodynamics
d) Zeroth law of thermodynamics

Answer
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Hint:The mentioned statement is related to the laws of thermodynamics. It helps in defining the physical quantities like temperature, energy, and entropy that characterize thermodynamic systems at equilibrium. The second and third law mostly relates to the term entropy, and temperature.

Complete answer:
>Firstly, let us discuss the first law of thermodynamics. It is considered to be the law of conservation of energy.
>The second law of thermodynamics states that the total entropy of an isolated system remains constant for the reversible process.
>Now, if we talk about the third law of thermodynamics, it states that the entropy of a system approaches a constant value as the temperature approaches zero. The system must be in a state with the minimum thermal energy at the zero temperature.
>Then, the entropy of a pure crystalline substance at absolute zero temperature is zero.
>Further talking about the zeroth law of thermodynamics, it states if two thermodynamic systems are in thermal equilibrium with a third one, then they are considered to be in thermal equilibrium with each other.
>So, we can conclude that this statement is the third law of thermodynamics.

Hence the correct answer is option ‘c’.

Additional information: Applying the third law of thermodynamics, the system such as glass, it might have more than one minimum energy state, the entropy of a system approaches a constant value. The constant value is known as the residual entropy of the system.

Note::The relation between microstates and entropy, as per the statistical mechanics is given below-
\[S=k\ln W\]
where, S=Entropy; k=Boltzmann constant and W=Microstates.
>Microstates signify the particular energies and configurations of the molecules and atoms and for perfect crystal, it is has one ground state i.e. W=1.
>Therefore, the above-said equation becomes, S=k ln(1) = 0.
>In this way, the entropy of a perfect crystal becomes zero at 0 K as per statistical mechanics also.