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:

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