Answer:
4.7 kJ/kmol-K
Explanation:
Using the Debye model the specific heat capacity in kJ/kmol-K
c = 12π⁴Nk(T/θ)³/5
where N = avogadro's number = 6.02 × 10²³ mol⁻¹, k = 1.38 × 10⁻²³ JK⁻¹, T = room temperature = 298 K and θ = Debye temperature = 2219 K
Substituting these values into c we have
c = 12π⁴Nk(T/θ)³/5
= 12π⁴(6.02 × 10²³ mol⁻¹)(1.38 × 10⁻²³ JK⁻¹)(298 K/2219 K)³/5
= 9710.83(298 K/2219 K)³/5
= 1942.17(0.1343)³
= 4.704 J/mol-K
= 4.704 × 10⁻³ kJ/10⁻³ kmol-K
= 4.704 kJ/kmol-K
≅ 4.7 kJ/kmol-K
So, the specific heat of diamond in kJ/kmol-K is 4.7 kJ/kmol-K
Answer:
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