3.2 Thermo-mechanical Equilibrium: Partitions in a Box#
Consider a box of \(5\:m^3\) volume composed of three isolated compartments A, B and C containing air in three diffrent states. Part A contains \(0.5\:kg\) air at \(200\:kPa\). Part B measure \(25 ^{\circ} C\) in temperature, \(100\:kPa\) is pressure and occupies \(1\:m^3\) of space. Finally, part C contains \(2\:kg\) of the same materials at a temperature of \(15 ^{\circ} C\) and a pressure of \(50\:kPa\).
Assuming ideal gas law for air in these states, calculate:
a) volume of air in partition C
b) volume of partition A
c) mass of air in partition B
d) temperature of air in partition A in \(^{\circ} C\)
e) now imagine all the seperators break and air in three compartments mix together. What would be the final pressure if the box is let to get in thermal stability with room temperature?
Solution Approach for a)#
Ideal gas equation can be used as EOS since the assumption is valid as per the statement, so
\(PV=mRT\)
\(V=mRT/P\)
#defining variables and looking up tables
R = 0.287 #air gas constant in kJ/kg.K
m_C = 2 #mass of air in partition c in kg
T_C = 20 + 273.15 #temperature of air in partc in K
P_C = 50 #pressure of air in partition c in kPa
#using ideal gas law equation of state
V_C = m_C * R * T_C / P_C #volume of partition c in m3
print('The air volume in partition C is:', f"{V_C:.1f}", 'm3')
The air volume in partition C is: 3.4 m3
Solution Approach for b)#
The box is composed of three partitions, therefore:
\(V_A+V_B+V_C=5\:m^3\)
\(V_A=5-V_B-V_C\)
V_B = 1 #volume of partition B in m3
V_A = 5 - V_B - V_C
print('The air volume in partition A is:', f"{V_A:.3f}", 'm3')
The air volume in partition A is: 0.635 m3
Solution Approach for c)#
from ideal gas law
\(m=PV/{RT}\)
#defining state variables
P_B = 100 #pressure of air in partition b in kPa
V_B = 1 #volume of partition b in m3
T_B = 25 + 273.15 #temperature of air in partition b in K
#using ideal gas law equation of state
m_B = P_B * V_B / (R * T_B) #mass of air in partiton b in kg
print('The mass of air in partition B is:', f"{m_B:.1f}", 'kg')
The mass of air in partition B is: 1.2 kg
Solution Approach for d)#
from ideal gas law
\(T=PV/{mR}\)
#defining state variables
P_A = 200 #pressure of air in partition a in kPa
m_A = 0.5 #mass of air in partiton a in kg
#using ideal gas law equation of state
T_A = P_A * V_A / (m_A * R) #temperature of air in partition a in K
#converting from Kelvins to Celsius
T_AC = T_A - 273.15 #temperature of air in partition a in C
print('The temperature of air in partition A is:', f"{T_AC:.1f}", 'C')
The temperature of air in partition A is: 611.4 C
Solution Approach for e)#
referring to ideal gas law equation of state,
\(P=mRT/V\)
The temperature would be in \(25 ^{\circ} C\) since the box is in thermal stability with room temperature.
#defining/calculating state variables
T = 25 + 273.15 #final temperature of air in K
m = m_A + m_B + m_C #final mass of the mixture
V = 5 #total volume of the box in m3
#using ideal gas law equation of state
P = m * R * T / V #final pressure of the mixture in kPa
print('The final pressure of mixed air is:', f"{P:.1f}", 'kPa')
The final pressure of mixed air is: 62.8 kPa