Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The first chamber contains water as a saturated liquid at 300 oC. The second chamber contains a vacuum. The membrane is subsequently pierced and the final pressure is 50 kPa. How much entropy was generated by the steam through this sudden expansion

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C ,

Specific internal energy of saturated liquid at 300°C,  1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa,   = 340.49 kJ/kg

Let the dryness fraction at the second chamber = x

Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is,

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C

From the steam table,

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

ok

explanation: