Answer:
(a) 705 kg
Explanation:
it is given that volume =500[tex]m^3[/tex]
pressure = 200 kPa
it is given that 80% mass is vapor so dryness fraction =x=0.8
from the standard table at 200 kPa
[tex]v_f=0.001061 \frac{m^3}{kg}[/tex]
[tex]v_g=0.88578\frac{m^3}{kg}[/tex]
specific volume [tex]v=v_f+x\left ( v_g-v_f \right )[/tex]
[tex]v=0.00106+0.8\left ( 0.88578-0.00106 \right )=0.7088\frac{m^3}{kg}[/tex]
we know that [tex]specific\ volume =\frac{volume}{mass}[/tex]
[tex]0.7088=\frac{500}{mass}[/tex]
[tex]mass=\frac{500}{0.7088}=705.456 \ kg[/tex]