Answer:
Volume of the second tank is 3.21m³
Final eguilibrium presssure is 284.13kpa
Explanation:
Volume of the tank A,
Temperature of air in tank A,
Pressure of air in tank A, P_A = 500kPa
Mass of air in tank B, m_b = 5kg
Temperature of air in tank B,
Pressure of air in tank B,
Surrounding temperature,
Assuming, at given conditions air behaves as an ideal gas.
For air, gas constant R = 0.287 kJ/kmol K
From ideal gas equation, mass of air in tank A is determined by
Volume of the tank B can be determined from
so,when the valve is opened
Total volume,
Volume of the second tank is 3.21m³
Total mass of air,
The final equilibrium pressure (P) can be obtained from the ideal gas equation applied to total volume
Final eguilibrium presssure is 284.13kpa