Water vapor is cooled in a closed, rigid tank from T1 = 480°C and p1 = 100 bar to a final temperature of T2 = 320°C. Determine the final specific volume, v2, in m3/kg, and the final pressure, p2, in bar.

v1 = v2 = 0.0347m³/kg the tank is rigid (constant volume)

P2 = 78.8bar

Explanation:

Assuming ideal gas behavior, the relationship holds for the water vapor

PV = RT/M

Where P = pressure in the tank

V = Specific volume in m³/kg

R = gas constant = 8.314 J/mol•K

T = absolute temperature in K

M = molar mass in kg/mol

The detailed solution steps can be found below in the attachment.

V2 = V1 because the tank is rigid so it is a constant volume process. Also at the final temperature the water molecules are still in the gaseous phase so V2 is still the same as v1.

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