# Specific heat capacity at constant pressure (C_P) of certain solid at 4.2K is 0.43j/kgmol. What is its molar entropy at that temperature?

Jul 22, 2018

You mean, $\text{J/mol"cdot"K}$...?

At low temperatures, $\overline{S} \approx {\overline{C}}_{P} / 3$.

The molar entropy at $\text{4.2 K}$ is defined relative to $\text{0 K}$ being barS("0 K") = "0 J/mol"cdot"K", and is given at constant atmospheric pressure by...

$\overline{S} \left(T\right) = {\overbrace{\overline{S} \left({T}_{1}\right)}}^{S \left(\text{0 K}\right) = 0} + {\int}_{{T}_{1}}^{{T}_{2}} {\left(\frac{\partial \overline{S}}{\partial T}\right)}_{P} \mathrm{dT}$

Also,

$\frac{1}{T} {\left(\frac{\partial \overline{H}}{\partial T}\right)}_{P} = {\overline{C}}_{P} / T = {\left(\frac{\partial \overline{S}}{\partial T}\right)}_{P}$

where $H$ is enthalpy and ${\overline{C}}_{P}$ is the molar specific heat capacity at constant pressure.

Therefore,

$\overline{S} \left(\text{4.2 K") = int_(0)^("4.2 K}\right) {\overline{C}}_{P} / T \mathrm{dT}$

However, we cannot take $\ln \left(0\right)$:

$\overline{S} \left(\text{4.2 K}\right) = {\overline{C}}_{P} \ln | {T}_{2} / {T}_{1} |$

= ("0.43 J/mol"cdot"K")ln(4.2/0) = ul"UNDEFINED"

Instead, we assume the solid follows the Debye model at low temperatures, so that ${\overline{C}}_{P}$ is given in terms of the Debye temperature ${\Theta}_{D}$ (Physical Chemistry, McQuarrie):

${\overline{C}}_{P} = \frac{12 {\pi}^{4}}{5} R {\left(\frac{T}{\Theta} _ D\right)}^{3}$

We will find that this form makes $\overline{S}$ able to be evaluated. Hence,

color(blue)(barS("4.2 K")) = int_(0)^("4.2 K") barC_P/TdT

$= {\int}_{0}^{\text{4.2 K}} \frac{12 {\pi}^{4}}{5} R {\left(\frac{1}{\Theta} _ D\right)}^{3} {T}^{2} \mathrm{dT}$

= {:1/3[(12pi^4)/5 R (T/Theta_D)^3]|:}_(0)^("4.2 K")

We should notice that now, $\overline{S} \approx {\overline{C}}_{P} / 3$ due to the definition of $\overline{S}$ at $\text{0 K}$, i.e. that

$\overline{S} \left({T}_{\text{low") = {:1/3[(12pi^4)/5 R (T/Theta_D)^3]|:}_(0)^(T_"low") ~~ 1/3barC_P(T_"low}}\right)$.

Therefore,

= 1/3(barC_P("4.2 K") - cancel(barC_P("0 K"))^(0))

$= \frac{1}{3} \left(\text{0.43 J/mol"cdot"K}\right)$

$= \textcolor{b l u e}{\text{0.143 J/mol"cdot"K}}$