Question

The specific heat \(C_p(T)\) of one mole of a material as a function of temperature \(T\) is given as \(C_p(T) = AT + BT^3\), where \(A = 0.695\) mJ.mol\(^{-1}\).K\(^{-2}\) and \(B = 0.045\) mJ.mol\(^{-1}\).K\(^{-4}\). When \(T\) is changed from 1 K to 10 K at constant pressure, the change in entropy \(\Delta S\) in mJ.mol\(^{-1}\).K\(^{-1}\) (rounded off to one decimal place) is

Hint:

Hint:
Divide \(C_p\) by \(T\) first, then integrate the resulting polynomial from 1 K to 10 K.

Answer 1

Correct Answer: 21.2

Step 1: Split the heat capacity by its physical origin.
The term $AT$ in $C_p(T)$ behaves like the electronic heat capacity of a metal (linear in $T$), while $BT^3$ behaves like the Debye lattice heat capacity (cubic in $T$). Treating them as two separate contributions to the entropy works as a check, since each piece integrates to a simple closed form on its own.

Step 2: Entropy from the electronic term.
\[ \Delta S_{el} = \int_1^{10} \frac{AT}{T}\,dT = A\int_1^{10} dT = A(10-1) = 0.695 \times 9 = 6.255 \text{ mJ.mol}^{-1}.\text{K}^{-1} \]

Step 3: Entropy from the lattice term.
\[ \Delta S_{ph} = \int_1^{10} \frac{BT^3}{T}\,dT = B\int_1^{10} T^2\,dT = B\left[\frac{T^3}{3}\right]_1^{10} = 0.045 \times \frac{999}{3} = 0.045 \times 333 = 14.985 \text{ mJ.mol}^{-1}.\text{K}^{-1} \]

Step 4: Add the two contributions.
\[ \Delta S = \Delta S_{el} + \Delta S_{ph} = 6.255 + 14.985 = 21.24 \text{ mJ.mol}^{-1}.\text{K}^{-1} \]

Final Answer:
Both the electronic piece and the lattice piece add up to the same total the direct integration gives. \[ \boxed{\Delta S = 21.2 \text{ mJ.mol}^{-1}.\text{K}^{-1}} \]