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

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,

$\frac{P'}{P} = \frac{T'}{T}$

$T' = \frac{P'}{P}\times T$

$T' = \frac{438}{130}\times 320 = 1078 K$

Now, change in entropy is given by:

$m(s' - s) = \int_{T}^{T'}\frac{1}{T}mC_{v}dT$

$(s' - s) = 0.718[ln\frac{T'}{T}]_{320}^{1078}$

$(s' - s) = ln\frac{1078}{320} = 0.8716$

Therefore, change in entropy is 0.8712 kJ/kgK