Insulated vessel has two compartments separated by a membrane. One side is 1 kg of steam at 400 C and 200 bar. The other side is evacuated. The membrane ruptures filling the entire volume. The final pressure is 100 bar. Determine the final temperature of the steam and the volume of the vessel.

To determine the final temperature of the steam and the volume of the vessel, we can use the principle of energy conservation and ideal gas laws.

First, let's consider the initial state of the system. We have 1 kg of steam at 400°C and 200 bar on one side of the membrane, while the other side is evacuated. Since the vessel is insulated, we can assume that there is no heat exchange with the surroundings.

Now, when the membrane ruptures, the steam expands to fill the entire volume. The mass remains constant at 1 kg, but the final pressure is given as 100 bar. We need to find the final temperature and volume of the vessel.

To solve this problem, we can use the ideal gas law for steam:

PV = mRT

where P is the pressure, V is the volume, m is the mass, R is the specific gas constant for steam, and T is the temperature.

Given:
Initial pressure (P1) = 200 bar
Final pressure (P2) = 100 bar
Initial temperature (T1) = 400°C = 400 + 273.15 = 673.15 K
Mass (m) = 1 kg

Using the ideal gas law, we can rearrange the equation to solve for the final temperature (T2) as follows:

T2 = (P2/P1) * (V1/V2) * T1

Where V1 and V2 are the initial and final volumes, respectively.

Since the vessel is evacuated initially, we can assume that the initial volume V1 is zero. Therefore, the equation simplifies to:

T2 = (P2/P1) * T1

Substituting the given values:

T2 = (100 bar / 200 bar) * 673.15 K
T2 = 0.5 * 673.15 K
T2 = 336.58 K

Hence, the final temperature of the steam is approximately 336.58 K.

To find the volume of the vessel, we need to use the ideal gas law again. Rearranging the equation, we get:

V2 = (m * R * T2) / P2

Substituting the known values:

V2 = (1 kg * R * 336.58 K) / 100 bar

Here, we need to use the specific gas constant (R) for steam, which is approximately 0.4615 kJ/(kg·K).

V2 = (1 kg * 0.4615 kJ/(kg·K) * 336.58 K) / 100 bar

Converting the pressure from bar to kPa:

V2 = (1 kg * 0.4615 kJ/(kg·K) * 336.58 K) / (100 * 100 kPa)

Simplifying the calculation:

V2 ≈ 1.556 m³

Therefore, the volume of the vessel after the membrane ruptures is approximately 1.556 m³.