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Question

(a) At 4.2 K the heat capacity of Ag(s) is 0.0145 JK1 mol1.0.0145\ \mathrm{JK}^{-1}\ \mathrm{mol} ^{-1}. Assuming that the Debye law applies, determine Sn(4.2K)Sn(0)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 Cpm=aT3,C_{pm}=a T^{3}, with a=1.956×104 JK1 mol1.a=1.956 \times 10^{-4}\ \mathrm{JK}^{-1}\ \mathrm{mol}^{-1}. Determine Sn(10K)S_{n}(10 K) Sn(0)-S_{\mathrm{n}}(0) for silver.

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(a)

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

\star We have the formula:

Sm(T)=Sm(0)+13aT3\mathrm{S}_{\mathrm{m}}(T)=\mathrm{S}_{\mathrm{m}}(0)+\frac{1}{3} \mathrm{aT}^{3}

Here:

aT3=Cp,m(T)\mathrm{aT}^{3}=\mathrm{C}_{\mathrm{p}, \mathrm{m}}(\mathrm{T})

So, the formula is:

Sm(T)=Sm(0)+13Cp,m(T)Sm(T)Sm(0)=13Cp,m\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:

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

Cp,m=0.0145JK1mol1\mathrm{C}_{p, \mathrm{m}}=0.0145 \mathrm{JK}^{-1} \mathrm{mol}^{-1} is the molar heat capacity

Now, substitute and calculate:

Sm(T)Sm(0)=13Cp,mSm(4.2K)Sm(0)=130.0145JK1mol1=0.00483JK1mol1\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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