Answer
$θ_{D}$ = 375.14 K
Work Step by Step
Given:
Constant A in Equation 19.2 is $\frac{12π^{4}R}{5 θ_{D}^{3}}$
R- gas constant
$θ_{D}^{3}$ - Debye temperature
Required:
$θ_{D}^{3}$ of aluminum with $c_{v} = 4.60 \frac{J}{kg.K} $ at 15 K
Solution:
$A = \frac{C_{v}}{T^{3}} = \frac{(4.60 \frac{J}{kg.K})(\frac{1 kg}{1000 g})(26.98 \frac{g}{mol}) }{(15 K^{3})} = 3.68 \times 10^{-5} \frac{J}{mol.K^{4}}$
$A = 3.68 \times 10^{-5} \frac{J}{mol.K^{4}} = \frac{12π^{4}R}{5 θ_{D}^{3}}$
Computing for $θ_{D}$,
$θ_{D} = (\frac{(12)(π)^{4}R}{5A})^{1/3} = (\frac{12π^{4}(8.31 \frac{J}{mol.K})}{(5)(3.68 \times 10^{-5} \frac{J}{mol.K^{4}})})^{1/3}$ = 375.14 K
