Third Law Of Thermodynamics - Practice Questions & MCQ

The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero. This is called the third law of thermodynamics.

• At any pressure, the entropy of every crystalline solid in thermodynamic equilibrium at absolute zero is zero.

• It is impossible to reduce the temperature of any system to absolute zero by any process. 

• As the absolute temperature approaches zero, increment in entropy for isothermal process in crystalline solids approaches zero, 

i.e. S=0 at T=0

Or $\lim _{T \rightarrow 0}, S \rightarrow 0$

If molar heat capacities of a substance (Cp) are measured at different temperature and a graph between Cp/T vs T is drawn and it shows this type of behavior.

Now let $S_M^o$ be the entropy of the substance  at zero Kelvin and S_M is its molar entropy at Kelvin then 

 $\begin{aligned} & \Delta \mathrm{S}=\mathrm{S}_{\mathrm{M}}-\mathrm{S}_{\mathrm{M}}^{\mathrm{o}} \\ & \therefore \Delta \mathrm{S}=\mathrm{S}_{\mathrm{M}} \\ & (\because\left.\mathrm{S}_{\mathrm{M}}^{\mathrm{o}}=0, \text { according to 3rd Law of thermodynamics }\right) \\ & \Delta \mathrm{S}=\frac{\mathrm{q}}{\mathrm{T}} \\ & \Delta \mathrm{S}=\int_0^{\mathrm{T}} \frac{\mathrm{C}_{\mathrm{p}} \cdot \mathrm{dT}}{\mathrm{T}}, \quad \because \mathrm{q}=1 \times \mathrm{C}_{\mathrm{p}} \cdot \mathrm{dT} \\ & \therefore \Delta \mathrm{S}=\mathrm{S}_{\mathrm{M}}=\int_0^{\mathrm{T}} \frac{\mathrm{C}_{\mathrm{p}} \cdot \mathrm{dT}}{\mathrm{T}}\end{aligned}$

The area under the curve or graph of Cp/T vs T determined from zero Kelvin to any desired temperature would be molar entropy change going from zero to desired T.