Dryness fraction of water vapor=0.7,Occupies volume =0.197 m3m^3m3 ,P1=1.4P_1=1.4P1=1.4 MPa,
Vfg=xVgV_{fg}=xV_{g}Vfg=xVg
Vg=0.1970.7=0.2814m3V_g=\frac {0.197}{0.7}=0.2814 m^3Vg=0.70.197=0.2814m3
W=pdV=1.4×106×(0.2814−0.197)=0.118×106J=1.18×106JW=pdV=1.4 \times 10^6\times (0.2814-0.197)=0.118 \times 10^6 J=1.18 \times 10^6 JW=pdV=1.4×106×(0.2814−0.197)=0.118×106J=1.18×106J
As per thermodynamic law
dQ=dU+dW,dU=0dQ=1.18×106dQ=dU+dW,dU=0 dQ=1.18 \times 10^6dQ=dU+dW,dU=0dQ=1.18×106
Now on removing we get
1.25=x1.4
x=0.893
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