Final answer:
After half the initial mass of helium escapes the tank, the final pressure is halved to 1 MPa due to the reduction of mass while maintaining the same volume. The final temperature remains the same as the initial temperature, which is 80°C, based on the assumption of the isochoric process for the remaining gas.
Step-by-step explanation:
To determine the final temperature and pressure of helium in the tank after half of its initial mass has escaped, we can utilize the ideal gas law and the principle of conservation of energy. As no heat is transferred due to the tank being insulated, the process is adiabatic for the escaping helium.
However, since we are only interested in the final condition inside the tank, and we assume the remaining helium is not doing work on its surroundings, we can consider this part of the process to be isochoric (constant volume).
Initially, we have a certain mass of helium 'm' at a pressure 'P1', volume 'V1', and temperature 'T1'. After half of the mass escapes, the new mass is 'm/2'.
We can assume the process to be isochoric for the gas remaining in the tank. The ideal gas law is given by:
PV = nRT
For the initial state, we can write:
P1V1 = (m/M)RT1
Where 'M' is the molar mass of helium. For the final state (with 'P2' being the final pressure, 'V2 = V1' due to constant volume, and 'T2' being the final temperature):
P2V2 = ((m/2)/M)RT2
Since V1 = V2 and m/M is constant, we can deduce that:
P2/P1 = (m/2)/m * T2/T1 = 1/2 * T2/T1
So, T2 = 2 * P2/P1 * T1
Given the initial conditions (P1 = 2 MPa, T1 = 80°C + 273.15 = 353.15K), we can find T2 once we determine P2. P2 can be found from the ideal gas law, considering the final mass is m/2, we have:
P2 = P1 * (m/2) / m = P1 / 2 = 1 MPa
Now knowing P2, we can determine T2:
T2 = 2 * (1 MPa/2 MPa) * 353.15K
T2 = 353.15 K
So the final temperature remains the same as the initial temperature, which is 80°C. The final pressure is halved to 1 MPa.