At very low temperatures the molar heat capacity of rock salt varies with temperature according to Debye’s T3 law:

$\mathbf{C}\mathbf{=}\mathbf{K}\frac{{\mathbf{T}}^{\mathbf{3}}}{{\mathbf{\theta }}^{\mathbf{3}}}$

where $\mathbf{k}\mathbf{=}\mathbf{}\mathbf{1940}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{mol}\mathbf{}\mathbf{x}\mathbf{}\mathbf{K}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\theta }\mathbf{=}\mathbf{281}\mathbf{K}$. (a) How much heat is required to raise the temperature of 1.50 mol of rock salt from 10.0 K to 40.0 K? (Hint: Use Eq. (17.18) in the form dQ = nCdT and integrate.) (b) What is the average molar heat capacity in this range? (c) What is the true molar heat capacity at 40.0 K?

• $\mathrm{k}=1940\mathrm{J}/\mathrm{mol}\times \mathrm{K}$
  

• localid="1668058665075" $\mathrm{q}=281\mathrm{K}$
  

• Amount of rock salt:$1.50\hspace{0.33em}\mathrm{mol}$

• Initial temperature:$10.0\mathrm{K}$

• Final Temperature: $40.0\mathrm{K}$

The Debye’s law in solid state physics states that for any solid its specific heat capacity is directly proportional to the third power of thermodynamic temperature.

Mathematically,

${\mathbf{C}}_{\mathbf{v}}\mathbf{=}\frac{\mathbf{12}{\mathbf{\pi }}^{\mathbf{4}}\mathbf{}{\mathbf{Nk}}_{\mathbf{B}}\mathbf{}{\mathbf{T}}^{\mathbf{3}}}{\mathbf{5}{\mathbf{T}}_{\mathbf{D}}\mathbf{3}}$

Here, N is the Avogadro number, KB is the Boltzmann constant, and TD is the Debye temperature.

From the given data,

$d\mathrm{Q}=\mathrm{nCdT}$

Integrate both sides and set the temperature limits,

$$\begin{gathered} \mathrm{Q}=\mathrm{n}{\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2}\mathrm{Cdt} \\ =\mathrm{n}{\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2}\mathrm{K}\left(\frac{{\mathrm{T}}^3}{{\mathrm{\theta }}^3}\right)\mathrm{dT} \\ =\frac{\mathrm{nk}}{{\mathrm{\theta }}^3}{\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2}{\mathrm{T}}^3\mathrm{dT} \\ =\left(\frac{\mathrm{nk}}{{\mathrm{\theta }}^3}\right)\frac{1}{4}{\left[{\mathrm{T}}^4\right]}_{{\mathrm{T}}_1}^{{\mathrm{T}}_2} \end{gathered}$$

Solve the above equation and substitute all the given values,

$$\begin{gathered} \mathrm{Q}=\frac{\mathrm{nk}}{4{\mathrm{\theta }}^3}\left({{\mathrm{T}}_2}^4-{{\mathrm{T}}_1}^4\right) \\ =\frac{\left(1.50\mathrm{mol}\right)\left(1940\mathrm{J}/\mathrm{mol}.\mathrm{K}\right)}{4{\left(281\mathrm{K}\right)}^3}{\left(40.0\mathrm{K}\right)}^4-{\left(10.0\mathrm{K}\right)}^4 \\ =83.6\mathrm{J} \end{gathered}$$

Thus the heat required to raise the temperature of the 1.50 mol of rock salt from 10.0K to 40.0K is 83.6 J.

The total specific heat relation is gives as,

$C=\frac{\Delta Q}{\Delta T}$

Average specific is therefore given as,

$$\begin{gathered} C_{av}=\frac{1}{n}\frac{\Delta Q}{\Delta T} \\ =\frac{1}{1.50mol}\left(\frac{83.6J}{40.0K-10.0K}\right) \\ =1.86J/mol.K \end{gathered}$$

Thus, the average specific in the temperature range of 10.0 K to 40.0 K is

The expression for true molar heat is given as

$C=k\frac{T^3}{q^3}$

Substitute the values,

$$\begin{gathered} \mathrm{C}=\left(1940\mathrm{J}/\mathrm{mol}.\mathrm{K}\right){\left(\frac{40.0\mathrm{K}}{281\mathrm{K}}\right)}^3 \\ =5.60\mathrm{J}/\mathrm{mol}.\mathrm{K} \end{gathered}$$

Thus the true specific heat at $40.0{\mathrm{K}}^3\mathrm{is}5.60\mathrm{J}/\mathrm{mol}.\mathrm{K}$