Question

The entropy of a crystalline substance at absolute zero on the basis of third law of thermodynamics should be taken as:
A. 100
B. 50
C. Zero
D. Different for different substances

Answer 1

Hint: This question is based on the concept of the third law of thermodynamics, which was proposed by Nernst. The statement of the third law of thermodynamics must be known, it is related to the entropy of a pure substance.

Complete step by step answer:
Now, first let us know about the third law of thermodynamics. It states that the entropy of a system will approach zero as the value of temperature approaches zero.
It also says that the system would be in a state having minimum thermal energy, when the temperature has approached zero.
The above mentioned statement is only applicable in the case of perfect crystal.
As we know, in the third law entropy is relatable with the number of microstates.
According to the formula i.e.S-S$_0$ = k$_{B}$ ln$\omega$, here $\omega$ refers to the total microstates, S is the entropy, S$_0$ initial entropy, and k$_{B}$ is the Boltzmann constant.
Thus, for the perfect crystal we have one ground state, it means one microstate i.e. $\omega$ = 1
So, ln 1 = 0
Then, the value of entropy is equal to the zero.
Therefore, the entropy of a crystalline substance at absolute zero on the basis of the third law of thermodynamics should be taken as zero.
Hence, the correct option is (C).

Note: Don’t get confused while talking about the microstates. Just remember that for perfect crystalline, there is only one microstate possible. It is determined by the number of possible ground states in all other cases.