A 1-m3 tank containing air at 25 °C and 500 kPa is connected through a valve to another tank containing 5 kg of air at 35 °C and 200 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20 °C. Determine the volume of the second tank and the final equilibrium pressure of air.

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

v 154.9= 0.9 length length s little over 154.9cm

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s