Q85P - University Physics with Modern Physics

Question

 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?

Step-by-Step Solution

Answer

    (a) The heat required to raise the temperature of the $1.50 \mathrm{mol} \mathrm{of} \mathrm{rock} \mathrm{salt} \mathrm{from} 10.0 \mathrm{K} \mathrm{to} 40.0 \mathrm{K} \mathrm{is} 83.6 \mathrm{J}.$

    (b) The given range has an average molar heat capacity of $1.86\mathrm{J}/\mathrm{mol}\cdot \mathrm{K}$.

    (c)  At $40.0 \mathrm{K}$, The true molar heat capacity is $5.60 \mathrm{J}/\mathrm{mol}\cdot \mathrm{K}$.

Step 1:Given Data

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

• $\mathrm{q}=281\mathrm{K}$
  

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

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

•  Final Temperature: $40.0 \mathrm{K}$   

Step 2: Concept of Debye’s T3 law.

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.

Step 3: (a) Determination of heat required to raise the temperature of the 1.50 mol of rock salt from 10.0K to 40.0K.

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.

Step 4: (b) Determination of average molar heat capacity in the given range.

The total specific heat relation is gives as,

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

Average specific is therefore given as,

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

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

Step 5: (c) Determination of true molar heat capacity at 40.0K.

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 $40.0 {\mathrm{K}}^3 \mathrm{is} 5.60 \mathrm{J}/\mathrm{mol}.\mathrm{K}$