# Answer in Mechanics | Relativity for Martin Navarro #138438

Two rigid tanks connected by a valve to each other contain air at specified conditions. The volume of the second tank and the final equilibrium pressure when the valve is opened are to be determined.

At specified conditions, air behaves as an ideal gas.

The gas constant of air is R = 0.287 kPa.m3/kg.K

Let’s call the first and the second tanks A and B.

“V_A=1;m^3”

“T_A = 25 +273 = 298;K”

“P_A = 500;kPa”

“m_B = 5;kg”

“T_B = 35+273 = 308;K”

“P_B = 200;kPa”

“T_S = 20+ 273 = 293;K”

Treating air as an ideal gas, the volume of the second tank and the mass of air in the first tank are determined to be

“V_B = frac{m_BRT_B}{P_B}”

“V_B = frac{(5;kg)(0.287 kPa; m^3/kg;K)(308;K)}{200;kPa}=2.21;m^3”

“m_A = frac{P_AV_A}{RT_A}”

“m_A = frac{(500;kPa)(1;m^3)}{(0.287 kPa; m^3/kg;K)(298;K)} = 5.846;kg”

“V = V_A+V_B = 1.0 + 2.21 = 3.21;m^3”

“m = m_A+m_B = 5.846 + 5.0 = 10.846;kg”

The final equilibrium pressure

“P = frac{mRT_s}{V}”

“P = frac{(10.846;kg)(0.287 kPa; m^3/kg;K)(293;K)}{3.21;m^3}=284.1;kPa”

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