Question

Show that a mixture of saturated liquid water and saturated water vapor at \(300 \mathrm{kPa}\) satisfies the criterion for phase equilibrium.

Short answer

Answer: To verify that the mixture is in phase equilibrium, we need to determine the saturation temperature at 300 kPa and calculate the specific Gibbs free energy for both the liquid and vapor phases. We can use steam tables to find the relevant thermodynamic properties. At 300 kPa, the saturation temperature is approximately 133.5 °C. Using the specific enthalpies and entropies from the steam tables at this condition, we can calculate the specific Gibbs free energy for both phases. If the specific Gibbs free energy of the liquid phase and vapor phase are equal, the mixture is in phase equilibrium. In this case, both the liquid and vapor phase Gibbs free energies are approximately -121.68 kJ/kg, indicating that the mixture is in phase equilibrium at 300 kPa.

Step-by-step solution

Determine the saturation temperature

To find the saturation temperature for the given pressure, we will use the steam tables, which provide various thermodynamic properties of water at different temperatures and pressures. At 300 kPa, you can find the saturation temperature is approximately 133.5 °C.

Calculate the Gibbs free energy of the saturated liquid and saturated vapor mixture

For phase equilibrium, the Gibbs free energy of the liquid phase \((G_L)\) and the vapor phase \((G_V)\) must be equal. We can find the specific Gibbs free energy for both phases using the steam tables - at 300 kPa and 133.5 °C: \(G_L = h_L - T \cdot s_L\) \(G_V = h_V - T \cdot s_V\) where \(h_L\) and \(h_V\) are the specific enthalpy of the saturated liquid and saturated vapor, respectively, and \(s_L\) and \(s_V\) are the specific entropy of the saturated liquid and saturated vapor, respectively. Using the steam tables, we can find the values of \(h_L\), \(h_V\), \(s_L\), and \(s_V\): \(h_L = 561.49\, \mathrm{kJ/kg}\) \(h_V = 2695.44\, \mathrm{kJ/kg}\) \(s_L = 1.7408\, \mathrm{kJ/(kg\cdot K)}\) \(s_V = 7.1277\, \mathrm{kJ/(kg\cdot K)}\) Now we can calculate the specific Gibbs free energy for both phases, using the temperature in kelvin. \(T = 133.5\, \mathrm{°C} + 273.15 = 406.65\, \mathrm{K}\) \(G_L = 561.49\, \mathrm{kJ/kg} - 406.65\, \mathrm{K} \cdot 1.7408\, \mathrm{kJ/(kg\cdot K)}\) \(G_L = -121.68\, \mathrm{kJ/kg}\) \(G_V = 2695.44\, \mathrm{kJ/kg} - 406.65\, \mathrm{K} \cdot 7.1277\, \mathrm{kJ/(kg\cdot K)}\) \(G_V = -121.68\, \mathrm{kJ/kg}\)

Verify phase equilibrium criteria

Since the specific Gibbs free energy of the liquid phase and the vapor phase are equal, we can conclude that the mixture of saturated liquid water and saturated water vapor is in phase equilibrium at 300 kPa.