Question

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?

Answer 1

You mean, `"J/mol"cdot"K"`...?

At low temperatures, `barS ~~ barC_P/3`.

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

`barS(T) = overbrace(barS(T_1))^(S("0 K") = 0) + int_(T_1)^(T_2) ((delbarS)/(delT))_PdT`

Also,

`1/T((delbarH)/(delT))_P = barC_P/T = ((delbarS)/(delT))_P`

where `H` is enthalpy and `barC_P` is the molar specific heat capacity at constant pressure.

Therefore,

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

However, we cannot take `ln(0)`:

`barS("4.2 K") = barC_Pln|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 `barC_P` is given in terms of the Debye temperature `Theta_D` (Physical Chemistry, McQuarrie):

`barC_P = (12pi^4)/5R(T/Theta_D)^3`

We will find that this form makes `barS` able to be evaluated. Hence,

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

`= int_(0)^("4.2 K") (12pi^4)/5R(1/Theta_D)^3T^2dT`

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

We should notice that now, `barS ~~ barC_P/3` due to the definition of `barS` at `"0 K"`, i.e. that

`barS(T_"low") = {:1/3[(12pi^4)/5 R (T/Theta_D)^3]|:}_(0)^(T_"low") ~~ 1/3barC_P(T_"low")`.

Therefore,

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

`= 1/3 ("0.43 J/mol"cdot"K")`

`= color(blue)("0.143 J/mol"cdot"K")`