Respuesta :
Answer:
Volume of Tank 2, V' = [tex]2.17 m^{3}[/tex]
Equilibrium Pressure, [tex]P_{eq} = 278.82 kPa[/tex]
Given:
Volume of Tank 1, V = 1 [tex]m^{3}[/tex]
Temperature of Tank 1, T = [tex]25^{\circ}C[/tex] = 298 K
Pressure of Tank 1, P = 500 kPa
Mass of air in Tank 2, m = 5 kg
Temperature of tank 2, T' = [tex]35^{\circ}C[/tex] = 303 K
Pressure of Tank 2, P' = 200 kPa
Equilibrium temperature, [tex]20^{\circ}C[/tex] = 293 K
Solution:
For Tank 1, mass of air in tank can be calculated by:
PV = m'RT
[tex]m' = \frac{PV}{RT}[/tex]
[tex]m' = \frac{500\times 1}{0.287\times 298} = 5.85 kg[/tex]
Also, from the eqn:
PV' = mRT
V' = volume of Tank 2
Thus
V' = [tex]\frac{mRT}{P}[/tex]
V' = [tex]\frac{5\times 0.287\times 303}{200} = 2.17 m^{3}[/tex]
Now,
Total Volume, V'' = V + V' = 1 + 2.17 = 3.17[tex]m^{3}[/tex]
Total air mass, m'' = m + m' = 5 + 5.85 = 10.85 kg
Final equilibrium pressure, P'' is given by:
[tex]P_{eq}V'' = m''RT_{eq}[/tex]
[tex]P_{eq} = \frac{m''RT_{eq}}{V''}[/tex]
[tex]P_{eq} = \frac{10.85\times 0.87\times 293}{3.17} = 287.82 kPa[/tex]
[tex]P_{eq} = 287.82 kPa[/tex]
