Answer to Question #138438 in Mechanics | Relativity for Martin Navarro

Question

A 1mcubed container of air at 25 Celsius and 500kpa is connected through a valve to another container holding 5kg of air at 35 Celsius and 200kpa. The valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20celcius. Determine the volume of the second tank and the final equilibrium pressure of air.

Answer 1

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

$P = \frac{mRT_s}{V}$

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