Answer:
a. Heat Capacity = 1.756J/mol-K
b. Heat Capacity = 24.942J/mol-k
Explanation:
Given
Constant volume Cv = 0.81J/mol-k
T1 = 34K
Td = Debye temperature = 306 K. Estimate the heat capacity (in J/mol-K) a. 44 K
First, The value of the temperature-independent constant.
Using Cv = AT³
Make A the subject of formula
A = Cv/T³
Substitute each values
A = 0.81/34³
A = 0.000020608589456543
A = 2.061 * 10^-5J/mol-k
The heat capacity changes with the temperature; below is the relationship between heat capacity and the temperature
Cv = AT³
So, The heat capacity when T = 44k is then calculated as
Cv = 2.061 * 10^-5 * 44³
Cv = 1.755522084266232
Cv = 1.756J/mol-K
(b) at 477 K.
Because the temperature is larger than the Debye temperature, the specific heat is calculated using as:
Cv = 3R
Where R = universal gas constant
R = 8.314J/mol-k
Cv = 3 * 8.314
Cv = 24.942J/mol-k
Answer:
a) 1.75 J/mol-k
b) 24.94 J/mol-k
Explanation:
We are given:
Cv = 0.81J/mol-k
Debye temperature = 306k
Heat capacity varies with temperature and the relationship between them is given as:
Cv = AT³
To find temperature independent constant A, we have:
[tex] A = \frac {C_v}{T^3} [/tex]
[tex] A = \frac{0.81J/mol-k}{34^3} [/tex]
[tex] A = 2.06*10^-^5 [/tex]
a) for temperature at 44k
Cv = AT³
[tex] C_v = 2.06*10^-^5 * 44^3 [/tex]
Cv = 1.75 J/mol-k
b) for T at 477k
Since the temperature 447k is higher than Debye's temperature, the specific heat will be given as:
Cv = 3R
Where R = universal gas constant = 8.314 J/mol-k
Therefore,
Cv = 3 * 8.314 J/mol-k
Cv = 24.94 J/mol-k
