The properties of water and steam used to do work can be determined from the steam table, given input of the temperature, pressure, specific volume, as well as other thermodynamic variables
The responses to the required values of pressure and specific volume are as follows;
Part I
(a) Given T = 140°C, v = 0.5 m³/kg, to find p
From the steam tables, we have, at 140°, 
= 0.00107976, 
= 0.508519, 
= 3.61501
Given that the < v < , the fluid has two phases and the experienced pressure = 3.61501 bar
(b) Given that at 30 MPa, and 80°C the steam is in the superheated heated region
From the single phase region of the steam tables, we have;
v = 1.01553 × 10⁻³ m³/kg
(c) Given that at p = 10 MPa = 100 bar, and T = 600°C, we have from the single phase region of the steam table
v = 3.83775 × 10⁻² m³/kg
(d) At T = 80°C, and x = 0.4, we have;
v = 0.00102904 + 0.4 × (3.40527 - 0.00102904) ≈ 1.363
v = 1.363 m³/kg
Part II
(a) At T = 440 °C, p = 20 MPa. from the single phase region of the steam table, we have;
v = 1·22459 × 10⁻² m³/kg
(b) At T = 160°C, p = 20 MPa = 200 bar, the steam is in the superheated region, and we have;
v = 1.08865 × 10⁻³ m³/kg
(c) At T = 40°C and p = 2 MPa = 20 bar, we have
v = 0.00100700 m³/kg = 1.007 × 10⁻³ m³/kg
Learn more about the steam tables here;
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