# 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?

##### 1 Answer

You mean,

At low temperatures,

The **molar entropy** at

#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

#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 *Physical Chemistry*, McQuarrie):

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

We will find that this form makes

#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(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")#