# Answer to Question #66212 in Solid State Physics for Adeel

Question #66212

Water is contained in a rigid vessel of 5 m3 at a quality of 0.8 and a pressure of 2 MPa. If the a pressure is reduced to 400 kPa by cooling the vessel, find the final mass of vapor mg and mass of liquid mf.

Expert's answer

v= vf +x (vg – vf)

v= 0.001084 + 0.8 (0.46242 – 0.001084)

v= 0.3732104 m3 / kg

v =V/m → m = V/v

m= 5 m3 / 0.3732104 m3 / kg = 13.4 kg

mg = mx = 13.4 kg x 0.8 = 10.72 kg

mf = m - mg = 13.4 kg - 10.72 kg = 2.68 kg

Answer: 10.72 kg and 2.68 kg

v= 0.001084 + 0.8 (0.46242 – 0.001084)

v= 0.3732104 m3 / kg

v =V/m → m = V/v

m= 5 m3 / 0.3732104 m3 / kg = 13.4 kg

mg = mx = 13.4 kg x 0.8 = 10.72 kg

mf = m - mg = 13.4 kg - 10.72 kg = 2.68 kg

Answer: 10.72 kg and 2.68 kg

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## Comments

Juan Carlos29.05.18, 03:46hi, Vf is 0.001084 not 0.46242 because in this operation the result is 0

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