A rigid vessel of 1m^3 contains steam at 20 bar and 400 degree C. the vessel is cooled until the steam is just dry saturated. Calculate the mass of steam in the vessel.

To calculate the mass of steam in the vessel, you need to use the properties of steam at the initial and final states and apply the ideal gas law.

The ideal gas law is expressed as:

PV = mRT

Where:
- P: Pressure
- V: Volume
- m: Mass
- R: Ideal gas constant
- T: Temperature

First, we need to find the properties of steam at both the initial and final states. At the initial state, the pressure is 20 bar and the temperature is 400°C.

To convert the temperature from Celsius to Kelvin, we add 273.15.
Therefore, the initial temperature (Ti) in Kelvin = 400 + 273.15 = 673.15 K.

To find the specific volume (v) at the initial state, you can use steam tables or steam property calculators. For example, at 20 bar and 673.15 K, the specific volume of steam could be 0.098 m³/kg (hypothetical value for explanation purposes).

Next, we will determine the final state when the steam is just dry saturated. When steam is dry saturated, it is in a state where it exists as a mixture of liquid and vapor.

At this state, the pressure (Pf) remains the same as the initial state, which is 20 bar.

To calculate the specific volume (vf) at the final state, we can use steam tables or property calculators. For dry saturated steam at 20 bar, the specific volume could be 0.239 m³/kg (hypothetical value).

Now, let's calculate the mass of steam (m) in the vessel:

Using the ideal gas law, we can equate P, V, and m, and rearrange the equation to solve for mass:

(Pi * Vi) / Ti = (Pf * vf) / Tf

Rearranging, we have:

m = (Pi * Vi * Tf) / (Pf * vf)

Plugging in the values:

m = (20 * 0.098 * 673.15) / (20 * 0.239)

m ≈ 138.96 kg

Therefore, the mass of steam in the vessel is approximately 138.96 kg.