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.

Question

Justin David

Chemistry

31 July, 14:20

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.

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Kaylee Meza · Answer 1

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