Third Law Of Thermodynamics

Third Law Of ThermodynamicsThe measure of specific realizations which realizes a thermodynamic system in a defined state as specified by macroscopic observations is known as entropy. Entropy can be commonly understood as a disorder. The entropy of a system increases with a rise in temperature, this happens because the molecular motion of the particles will increase with the rise in temperature. Conversely, with a decrease in temperature entropy will decrease. In 1906, Nernst made a significant observation related to the entropy of a perfect crystal at zero Kelvin and stated the third law of thermodynamics.

Statement of the Third Law of Thermodynamics

The third law of thermodynamics states that ” The entropy of a perfect crystal at absolute zero temperature is exactly equal to zero.”

At absolute zero that is zero Kelvin, the system is said to possess minimum energy. This statement of the third law of thermodynamics holds true only when the perfect crystal has minimum energy state. If the system has minimum disturbances it means that the temperature has approached zero Kelvin where molecular motion seizes to exist. Standard molar entropy is defined as the entropy per mole of a substance at standard conditions and specified temperature.

Application Of Third Law Of Thermodynamics

The most significant use of the third law of thermodynamics is that it helps in calculating the absolute entropy of a substance at room temperature (or any temperature T). These determinations are based on the heat capacity measurements of the substance. For any solid, let S0 be the entropy at 0 K and S be the entropy at T K, then

ΔS = S – S0 = \( \int^T_0 \frac {C_p dT}{T}\)

According to the third law of thermodynamics, we know that S0= 0 at 0 K,

S = \( \int^T_0 \frac{C_p}{T}\) dT

We obtain the value of this integral by plotting the graph of Cp/ T versus T and then finding the area of this curve from 0 to T. The simplified expression for the absolute entropy of a solid at temperature T is as follows:

S = \( \int^T_0 \frac{C_p}{T}\) dT =\( \int^T_0 C_p\) d lnT

= Cp ln T = 2.303 Cp log T

Here Cp is the heat capacity of the substance at a constant pressure and this value is assumed to be constant in the range of 0 to T K.

We have seen a brief description of the third law of thermodynamics, its importance, and applications. For any further query on this topic call the mentor support team at Byju’s-The learning App.


Practise This Question

Calculate the entropy change accompanying the following change of state

5 mol of O2(27C,1atm)  5mol of O2(117C,5atm)

CP for O2 = 6.95 cal deg1 mol1