Question

Air in a rigid tank is initially at 320 K and 130 kPa. Heat is added until the final pressure is 438 kPa. What is the change in entropy of the air? Do NOT assume constant specific heats

Answer 1

Change in entropy is 0.8712 kJ/kgK

Given:

Initial Temperature, T = 320 K

Initial Pressure, P = 130 kPa

Final Pressure, P' = 438 kPa

Solution:

Here, a rigid tank is considered, therefore, the volume of the tank is constant and for a process at constant volume:

Pressur, P ∝ Temperature, T

Therefore,

`(P')/(P) = (T')/(T)`

`T' = (P')/(P)* T`

`T' = (438)/(130)* 320 = 1078 K`

Now, change in entropy is given by:

`m(s' - s) = \int_(T)^(T')(1)/(T)mC_(v)dT`

`(s' - s) = 0.718[ln(T')/(T)]_(320)^(1078)`

`(s' - s) = ln(1078)/(320) = 0.8716`

Therefore, change in entropy is 0.8712 kJ/kgK