The final mass of water vapor is "7.445 Kg" and liquid vapor is "24.665 Kg".
From the saturated water pressure,
At P = 1 MPa,
→ [tex]V_1 = V_f +x(V_g-V_f)[/tex]
[tex]= 0.001127+0.8(0.19436-0.00127)[/tex]
[tex]= 0.15571 \ m^3/Kg[/tex]
When the cylinder is rigid, then
→ [tex]V_1=V_2= 0.15571 \ m^3/Kg[/tex]
From the saturated water pressure,
At P₂ = 270.3 KPg,
→ [tex]V_2 = V_{f2}+x_2(V_{g2}-V_{f2})[/tex]
[tex]0.15571=0.00107+x_2 (0.668-0.00107)[/tex]
[tex]x_2 = 0.23186[/tex]
As we know,
→ [tex]m = \frac{V}{v}[/tex]
[tex]= \frac{5}{0.15571}[/tex]
[tex]= 32.1109 \ Kg[/tex]
hence,
→ The final mass of water vapor will be:
= [tex]x_2m[/tex]
= [tex]0.23186\times 32.11[/tex]
= [tex]7.445 \ Kg[/tex]
→ The final mass of liquid vapor will be:
= [tex]m-m_v[/tex]
= [tex]32.1109-7.445[/tex]
= [tex]24.665 \ Kg[/tex]
Thus the above answer is correct.
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