Question

Dagastinium, on the other hand, is a rare material that has a temperature dependent constant pressure heat capacity of cP = 0.1 x T J/(mol.K) in the range from 273 to 303 K. What is the entropy change of 2 moles of Dagastinium between 273 and 303 K?

Answer 1

Answer : The change in entropy is 6 J/K

Explanation :

To calculate the change in entropy we use the formula:

`\Delta S=\int (dQ)/(T)`

and,

`Q=nC_pdT`

`\Delta S=n\int\limits^(T_f)_(T_i){(C_(p)dT)/(T)`

where,

`\Delta S` = change in entropy

n = number of moles = 2 moles

`T_f` = final temperature = 303 K

`T_i` = initial temperature = 273 K

`C_(p)` = heat capacity at constant pressure =
`0.1* T(J/K.mol)`

Now put all the given values in the above formula, we get:

`\Delta S=2\int\limits^(303)_(273){((0.1* TdT)/(T)`

`\Delta S=2* 0.1\int\limits^(303)_(273)dT`

`\Delta S=2* 0.1* [T]^(303)_(273)`

`\Delta S=2* 0.1* (T_f-T_i)`

`\Delta S=2* 0.1* (303-273)`

`\Delta S=6J/K`

Therefore, the change in entropy is 6 J/K