Answer:
Part a: The volume of vessel is 4.7680 and total internal energy is 3680 kJ.
 and total internal energy is 3680 kJ.
Part b: The quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.
Explanation:
Part a:
As per given data
m=2 kg
T=80 °C =80+273=353 K
Dryness=70% vapour =0.7
<em>From the steam tables at 80 °C</em>
Specific volume of saturated vapours=v_g=3.40527 
Specific volume of saturated liquid=v_f=0.00102 
Now the relation  of total specific volume for a specific dryness value is given as 
                                   
Substituting the values give

Now the volume of vessel is given as

So the volume of vessel is 4.7680 .
.
Similarly for T=80 and dryness ratio of 0.7 from the table of steam
Pressure=P=47.4 kPa
Specific internal energy is given as u=1840 kJ/kg
So the total internal energy is given as

The total internal energy is 3680 kJ.
So the volume of vessel is 4.7680 and total internal energy is 3680 kJ.
 and total internal energy is 3680 kJ.
Part b
Volume of vessel is given as 1.6 
mass is given as 2 kg
Pressure is given as 0.2 MPa or 200 kPa
Now the specific volume is given as 

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives
Temperature=T=120 °C
Quality=x=0.903 ≈ 90.3%
Specific internal energy =u=2330 kJ/kg
The total internal energy is given as 

So the quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.