Question

For some material, the heat capacity at constant volume Cv at 34 K is 0.81 J/mol-K, and the Debye temperature is 306 K. Estimate the heat capacity (in J/mol-K) (a) at 44 K, and (b) at 477 K.

Answer 1

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:

$A = \frac {C_v}{T^3}$

$A = \frac{0.81J/mol-k}{34^3}$

$A = 2.06*10^-^5$

a) for temperature at 44k

Cv = AT³

$C_v = 2.06*10^-^5 * 44^3$

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

Answer 2

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