Answer:
Volume of the second tank is 3.21m³
Final eguilibrium presssure is 284.13kpa
Explanation:
Volume of the tank A, ![V_A=1m^3](https://tex.z-dn.net/?f=V_A%3D1m%5E3)
Temperature of air in tank A,![T_A = 25^0C = 298K](https://tex.z-dn.net/?f=T_A%20%3D%2025%5E0C%20%3D%20298K)
Pressure of air in tank A, P_A = 500kPa
Mass of air in tank B, m_b = 5kg
Temperature of air in tank B,![T_B = 35^0C = 308k](https://tex.z-dn.net/?f=T_B%20%3D%2035%5E0C%20%3D%20308k)
Pressure of air in tank B, ![P_B = 200kPa](https://tex.z-dn.net/?f=P_B%20%3D%20200kPa)
Surrounding temperature,![T_s_u_r_r = 20^0c = 293K](https://tex.z-dn.net/?f=T_s_u_r_r%20%3D%2020%5E0c%20%3D%20293K)
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
![m_A = \frac{P_AV_A}{RT_A} \\\\m_A = \frac{500\times1}{0.287\times298} \\\\m_A = 5.846kg](https://tex.z-dn.net/?f=m_A%20%3D%20%5Cfrac%7BP_AV_A%7D%7BRT_A%7D%20%5C%5C%5C%5Cm_A%20%3D%20%5Cfrac%7B500%5Ctimes1%7D%7B0.287%5Ctimes298%7D%20%5C%5C%5C%5Cm_A%20%3D%205.846kg)
Volume of the tank B can be determined from
![V_B=\frac{m_BRT_B}{P_B} \\\\V_B = \frac{5\times0,287\times308}{200} \\\\V_B = 2.21m^3](https://tex.z-dn.net/?f=V_B%3D%5Cfrac%7Bm_BRT_B%7D%7BP_B%7D%20%5C%5C%5C%5CV_B%20%3D%20%5Cfrac%7B5%5Ctimes0%2C287%5Ctimes308%7D%7B200%7D%20%5C%5C%5C%5CV_B%20%3D%202.21m%5E3)
so,when the valve is opened
Total volume,
![V = V_A+V_B\\V= 1+2.21\\V=3.21m^3](https://tex.z-dn.net/?f=V%20%3D%20V_A%2BV_B%5C%5CV%3D%201%2B2.21%5C%5CV%3D3.21m%5E3)
Volume of the second tank is 3.21m³
Total mass of air,
![m= m_A+m_B\\m = 5+5.846\\m= 10.846kg](https://tex.z-dn.net/?f=m%3D%20m_A%2Bm_B%5C%5Cm%20%3D%205%2B5.846%5C%5Cm%3D%2010.846kg)
The final equilibrium pressure (P) can be obtained from the ideal gas equation applied to total volume
![P = \frac{mRT_s_u_r_r}{V} \\\\P=\frac{10.846\times0.287\times293}{3.21} \\\\P= 284.13kpa](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BmRT_s_u_r_r%7D%7BV%7D%20%5C%5C%5C%5CP%3D%5Cfrac%7B10.846%5Ctimes0.287%5Ctimes293%7D%7B3.21%7D%20%5C%5C%5C%5CP%3D%20284.13kpa)
Final eguilibrium presssure is 284.13kpa