Question

A very simple assumption for the specific heat of a crystalline solid is that each vibrational mode of the solid acts independently and is fully excited and thus cv=3NAkB=24.9 kJ/(kmol⋅K). This is called the law of Dulong and Petit. Calculate the Debye specific heat (in units of kJ/(kmol⋅K) of diamond at room temperature, 298 K. Use a Debye temperature of 2219 K.

Answer 1

Answer:

The Specific Heat of Diamond in kJ/kmol-K is 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 & θ = Debye Temperature = 2219 K  

Substituting these values into c we get:

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