Question

At very low temperatures, the molar specific heat ${\mathit{C}}_{\mathbf{V}}$of many solids is approximately ${\mathit{C}}_{\mathbf{v}}\mathbf{=}\mathit{A}{\mathit{T}}^{\mathbf{3}}$, where depends on the particular substance. For aluminum,$\mathit{A}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{15}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{.}{\mathit{K}}^{\mathbf{4}}$. Find the entropy change for 4.00 mlof aluminum when its temperature is raised from 5.00 kto 10.0 k.

Short answer

Entropy change of aluminum is$0.0368J/K$ .

Step-by-step solution

The given data

a) Molar specific heat at constant volume, $C_v=A{T}^3$where$A=3.15\times {10}^{-5}J/mol.K^4$

b) Number of moles present in aluminum,$n=4.00mol$

c) Temperature raise of aluminum, from$T_i=5.00K$ to$T_f=10.0K$

Understanding the concept of entropy change

Entropy change is a phenomenon that quantifies how disorder or randomness has changed in a thermodynamic system. We can use the formula for entropy change in terms of specific heat at constant volume. Then by inserting specific heat at constant volume and the given values in the formula and taking the integral, we can get the entropy change of aluminum.

Formula:

The entropy change of a gas at constant volume, $∆S=\int \frac{n{C}_{v}dt}{T}$ …(i)

Calculation of the entropy change of aluminum

Substituting the given data in equation (i), the entropy change of aluminum can be given as follows:

$$\begin{gathered} ∆S={\int }_{5.00k}^{10.00k}\frac{n\left(A{T}^3\right)dt}{T} \\ ={\int }_{5.00k}^{10.00k}n\left(A{T}^3\right)dt \\ ={\left[\frac{nA{T}^3}{3}\right]}_{5.00k}^{10.00k} \\ =\frac{4.00mol\times 3.15\times {10}^{-5}J/mol.K^4\left({10}^3-5^3\right)K^3}{3} \\ =0.03675J/K \\ \approx 0.0368J/K \end{gathered}$$

Hence, the value of the entropy change is$0.0368J/K$