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?

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

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

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")#