Question

0.2 kg of air is heated in a constant volume process from 20 to 100(degrees celsius). The specific heat at constant volume is 0.7186J/ Kg K. Calculate the change in entropy of this process.

Answer 1

`\delta \ S =0.034679 \ J/K`

Step-by-step explanation:

The formula for calculating the change in entropy of a process can be expressed as:

`\delta \ S = \int\limits \ (d \theta )/(T)`

`\delta \ S = mc \int\limits^(T_2)_(T_1) \ \frac {d T}{T}}`

`\delta \ S = mc \ In (T) ^(T_2)_(T_1)`

`\delta \ S = mc \ [In \ (T_2) - In \ ({T_1})]`

`\delta \ S = mc \ In \ ((T_2)/(T_1))`

Given that:

mass m = 0.20 kg

specific heat constant c = 0.7186J/ Kg K

`T_2` = 100° C = ( 100 + 273.15) = 373.15 K

`T_1` = 20° C = ( 20 + 273.15) = 293.15 K

Replacing our values into above equation; we have :

`\delta \ S = (0.20)(0.7186) \ In \ ((373.15)/(293.15))`

`\delta \ S =0.034679 \ J/K`