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Question: For a perfectly crystalline solid \({C_{p.m}} = a{T^3}\), where \(a\) is constant. If \({C_{p.m}}\) ...

For a perfectly crystalline solid Cp.m=aT3{C_{p.m}} = a{T^3}, where aa is constant. If Cp.m{C_{p.m}} is 0.42J/K−mol0.42J/K - mol at 10K10K, molar entropy at 10K10K is
(A) 0.42J/K−mol0.42J/K - mol
(B) 0.14J/K−mol0.14J/K - mol
(C) 4.2J/K−mol4.2J/K - mol
(D) Zero

Explanation

Solution

As we know that crystalline solids are the solids whose constituent atoms, molecules or ions are arranged in a highly ordered manner . Entropy of a system is the measure of randomness. In the given question we have given the expression of molar heat capacity. We will have the given values and find the value of aa and then proceed further.

Complete step-by-step answer: As in the question we have been given the value of Cp.m{C_{p.m}} =0.42J/K−mol = 0.42J/K - mol at 10K10K and we know that Cp.m=aT3{C_{p.m}} = a{T^3}. So we will find the value of aa.
Cp.m=aT3 ⇒0.42=a(10)3 ⇒a=0.42×10−3  {C_{p.m}} = a{T^3} \\\ \Rightarrow 0.42 = a{(10)^3} \\\ \Rightarrow a = 0.42 \times {10^{ - 3}} \\\
Now we know that the expression of finding the molar entropy of a crystalline solid is :
Sm=∫010Cp.mTdT ⇒Sm=∫010aT3TdT ⇒Sm=∫010aT2dT ⇒Sm=a3[103−0]=0.423 ⇒Sm=0.14J/K−mol  {S_m} = \int\limits_0^{10} {\dfrac{{{C_{p.m}}}}{T}} dT \\\ \Rightarrow {S_m} = \int\limits_0^{10} {\dfrac{{a{T^3}}}{T}} dT \\\ \Rightarrow {S_m} = \int\limits_0^{10} {a{T^2}} dT \\\ \Rightarrow {S_m} = \dfrac{a}{3}\left[ {{{10}^3} - 0} \right] = \dfrac{{0.42}}{3} \\\ \Rightarrow {S_m} = 0.14J/K - mol \\\

From the above explanation and calculation it is clear to us that the correct answer of the given question is option : (B) 0.14J/K−mol0.14J/K - mol .

Additional information: Molar heat capacity is the heat that is to be given to one mole of a substance to increase the temperature of the substance by one unit. Entropy is an extensive property. Gases have the highest entropy .
Crystalline solids are anisotropic in nature. Examples of crystalline solids are diamond and quartz. Crystalline solids have sharp melting points as compared to amorphous solids because the constituent particles like atoms, molecules and ions are at the same distance from the neighbouring constituents. There is a regularity in the crystalline lattice.

Note: Always remember that the entropy is the measure of randomness. The formula of finding the molar entropy is Sm=∫T22T1Cp.mTdT{S_m} = \int\limits_{{T_{22}}}^{{T_1}} {\dfrac{{{C_{p.m}}}}{T}} dT . Here Cp.m{C_{p.m}} is the molar heat capacity. Always avoid silly mistakes and calculation mistakes while solving the numerical problem.