Answer:
V2 = 1.77 m³
P3 = 222 kPa
Explanation:
Explanation is attached
The volume of the second tank and the final equilibrium pressure of air are respectively;
Vb1 = 1.77 m³
P2 = 222.02 KPa
We are given;
Volume of first tank; V_a1 = 1 m³
Initial volume of second tank; V_b1
Initial Temperature of first tank; T_a1 = 10°C = 283.15 K
Initial temperature of second tank; Tb1 = 35°C = 308.15 K
Surrounding temperature; T2 = 20°C = 293.15 K
Mass of air in second tank; m_b1 = 3 kg
Pressure in first tank; P_a1 = 350 KPa
Pressure in second tank; P_b1 = 150 KPa
Now, let us find the mass of air in the first tank from the formula;
m_a1 = (P_a1 × V_a1)/(R × T_a1)
Where R is a constant = 0.287 KJ/kg.k
Thus;
m_a1 = (350 × 1)/(0.287 × 283.15)
m_a1 = 4.31 kg
Now, let us find the initial volume in the second tank from the formula;
V_b1 = (m_b1 × R × T_b1)/P_b1
Thus;
V_b1 = (3 × 0.287 × 308.15)/150
V_b1 = 1.77 m³
Now, in state 2, we have;
For mass;
m2 = m_a1 + m_b1
m2 = 4.31 + 3
m2 = 7.31 kg
Also, for volume;
V2 = V_a1 + V_b1
V2 = 1 + 1.77
V2 = 2.77 m³
The final equilibrium pressure will be gotten from;
P2 = (m2 × R × T2)/V2
P2 = (7.31 × 0.287 × 293.15)/2.77
P2 = 222.02 KPa
Read more at; https://brainly.com/question/13844813
