Answer :
Answer:
The change of entropy is 1.229 kJ/(kg K)
Explanation:
Data
[tex] T_1 = 300 K [/tex]
[tex] T_2 = 900 K[/tex]
[tex] p_1= 400 kPa[/tex]
[tex] p_2= 300 kPa[/tex]
[tex] R= 0.287 kJ/(kg K)[/tex] (Individual Gas Constant for air)
For variable specific heats
[tex]s(T_2, p_2) - s(T_1, p_1) = s^0(T_2) - s^0(T_1) - R \, ln \frac{p_2}{p_1}[/tex]
where [tex] s^0(T) [/tex] is evaluated from table attached
[tex] s^0(900 K) = 2.84856 kJ/(kg K)[/tex]
[tex] s^0(300 K) = 1.70203 kJ/(kg K)[/tex]
Replacing in equation
[tex]s(900 K, 300 kPa) - s(300 K, 400 kPa) = 2.84856 kJ/(kg K) - 1.70203 kJ/(kg K) - 0.287 kJ/(kg K) \, ln \frac{300 kPa}{400 kPa}[/tex]
[tex]s(900 K, 300 kPa) - s(300 K, 400 kPa) = 1.229 kJ/(kg K)[/tex]
