Question

# (a) At 4.2 K the heat capacity of Ag(s) is $0.0145\ \mathrm{JK}^{-1}\ \mathrm{mol} ^{-1}.$ Assuming that the Debye law applies, determine $S_{n}(4.2 K)-S_{n}(0)$ for silver. (b) At low temperatures the heat capacity of Ag(s) is found to obey the Debye law $C_{pm}=a T^{3},$ with $a=1.956 \times 10^{-4}\ \mathrm{JK}^{-1}\ \mathrm{mol}^{-1}.$ Determine $S_{n}(10 K)$ $-S_{\mathrm{n}}(0)$ for silver.

Solution

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#### (a)

Here, find the value for silver of $S_{\mathrm{m}}(4.2 \mathrm{K})-S_{\mathrm{m}}(0)$:

$\star$ We have the formula:

$\mathrm{S}_{\mathrm{m}}(T)=\mathrm{S}_{\mathrm{m}}(0)+\frac{1}{3} \mathrm{aT}^{3}$

Here:

$\mathrm{aT}^{3}=\mathrm{C}_{\mathrm{p}, \mathrm{m}}(\mathrm{T})$

So, the formula is:

\begin{align*} \mathrm{S}_{\mathrm{m}}(T)&=\mathrm{S}_{\mathrm{m}}(0)+\frac{1}{3} \mathrm{C}_{\mathrm{p}, \mathrm{m}}(\mathrm{T})\\ \mathrm{S}_{\mathrm{m}}(T)-\mathrm{S}_{\mathrm{m}}(0)&=\frac{1}{3} \mathrm{C}_{p, \mathrm{m}}\\ \end{align*}

The values are:

$\mathrm{T}=4.2 \mathrm{K}$ is the temperature

$\mathrm{C}_{p, \mathrm{m}}=0.0145 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ is the molar heat capacity

Now, substitute and calculate:

\begin{align*} \mathrm{S}_{\mathrm{m}}(T)-\mathrm{S}_{\mathrm{m}}(0)&=\frac{1}{3} \mathrm{C}_{p, \mathrm{m}}\\ \mathrm{S}_{\mathrm{m}}(4.2 \mathrm{K})-\mathrm{S}_{\mathrm{m}}(0)&=\frac{1}{3} \cdot 0.0145 \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}\\ &=0.00483 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\\ \end{align*}

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