A rigid container of volume 0.5 m3 contains 1.0 kg of water at 120°C (vf = 0.00106 m3/kg, vg = 0.8908 m3/kg). The state of water is
A mixture of saturated liquid and saturated vapor
Specific volume is \(v=V/m=0.5\ \mathrm{m^3/kg}\). Since \(v_f=0.00106\lt v\lt v_g=0.8908\) at 120°C, water is a saturated liquid–vapour mixture.
Important Questions from Phase Change of a Pure Substance
- Which of the following exists in a P-V diagram for water?
What the region (2) in the p-V diagram given below is called?
The saturation temperature of steam with increase in pressure increases _________.
A rigid tank contains 10 kg of water at 90°C. If 8 kg of water is in liquid form and the rest is in vapour form, what will be the volume of the tank?
[Given: At 90°C, saturation pressure is 70 kPa , specific volume at saturated liquid state = 0.001 m3/kg , specific volume at saturated vapour state = 2.36 m3/kg.]
- The triple point on a T-s diagram is _________.
