Answer:
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,
[tex]\frac{P'}{P} = \frac{T'}{T}[/tex]
[tex]T' = \frac{P'}{P}\times T[/tex]
[tex]T' = \frac{438}{130}\times 320 = 1078 K[/tex]
Now, change in entropy is given by:
[tex]m(s' - s) = \int_{T}^{T'}\frac{1}{T}mC_{v}dT[/tex]
[tex](s' - s) = 0.718[ln\frac{T'}{T}]_{320}^{1078}[/tex]
[tex](s' - s) = ln\frac{1078}{320} = 0.8716[/tex]
Therefore, change in entropy is 0.8712 kJ/kgK
