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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/(kmolā‹…K). This is called the law of Dulong and Petit. Calculate the Debye specific heat (in units of kJ/(kmolā‹…K) of diamond at room temperature, 298 K. Use a Debye temperature of 2219 K.ā€‹

Respuesta :

oladayoademosu

Answer:

4.7 kJ/kmol-K

Explanation:

Using the Debye model the specific heat capacity in kJ/kmol-K

c = 12Ļ€ā“Nk(T/Īø)Ā³/5

where N = avogadro's number = 6.02 Ɨ 10Ā²Ā³ molā»Ā¹, k = 1.38 Ɨ 10ā»Ā²Ā³ JKā»Ā¹, T = room temperature = 298 K and Īø = Debye temperature = 2219 K Ā 

Substituting these values into c we have

c = 12Ļ€ā“Nk(T/Īø)Ā³/5 Ā 

= 12Ļ€ā“(6.02 Ɨ 10Ā²Ā³ molā»Ā¹)(1.38 Ɨ 10ā»Ā²Ā³ JKā»Ā¹)(298 K/2219 K)Ā³/5

= 9710.83(298 K/2219 K)Ā³/5

= 1942.17(0.1343)Ā³

= 4.704 J/mol-K

= 4.704 Ɨ 10ā»Ā³ kJ/10ā»Ā³ kmol-K

= 4.704 kJ/kmol-K

ā‰… 4.7 kJ/kmol-K

So, the specific heat of diamond in kJ/kmol-K is 4.7 kJ/kmol-K

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