Question

For a solid at $TK{S}_T-S_0=\Delta S={\int }_O^T\frac{CpdT}{T}=C_{P}lnT=2.303C_{P}logT$

• A. True
• B. False

Answer 1

The correct option is A

True

Yes this statement is true

According to the third law of Thermodynamics, "At absolute zero, the entropy of a perfectly crystalline substance is taken as zero”

(i) For a solid at $TK$

$S_T-S_0=\Delta S={\int }_O^T\frac{CpdT}{T}$ Cp In T = 2.303 Cp log T

Where $S_T$ and $S_0$ are the entropies at $TK$ and $0K$ respectively and $S_0=0$ (according to third law of thermodynamics).

The value of integral can be calculated from the graph of $\frac{C_P}{T}$ against temperature $\left(T\right)$. The area under the curve between $0K$ and $TK$ gives the value of integral and, thus, the value of $S$ at temperature $T$.

(ii) For liquids and gases, the absolute entropy at a given temperature $T$ is given by the expression
$S={\int }_O^{T_f}\frac{Cp\left(s\right)dT}{T}+\frac{\Delta H_f}{T_f}+{\int }_{T_f}^{T_b}\frac{Cp\left(1\right)dT}{T}+\frac{\Delta H_{vap}}{T_b}+\frac{T}{T_b}\frac{Cp\left(g\right)dT}{T}$