4A2 :  QuasiEquilibrium Expansion of a Gas  4 pts 

A pistonandcylinder device is oriented horizontally and a compressed spring exerts a force on the back of the piston, as shown below. The gas inside the cylinder is cooled slowly and, as the cooling proceeds, the spring pushes  


the piston farther into the cylinder. During the cooling process, the spring exerts a force that decreases linearly with position. Initially, the gas occupies a volume of 8.4 L while the spring exerts a force of 1.2 kN on the  
back of the piston. At the end of the process, the gas occupies 3.7 L and the spring exerts a force of 0 kN on the piston. The area of the piston face is 200 cm^{2} and atmospheric pressure is 100 kPa.  
Assume the gas behaves as an ideal gas and friction is negligible. Determine : a.) The initial and final pressures inside the cylinder, in kPa b.) The boundary work, in kJ. 

Read :  The key to solving this problem is to determine the slope and intercept for the linear relationship between the force exerted by the spring on the piston and the volume that the gas occupies. This relationship is linear because, for a cylinder of uniform diameter, gas volume varies linearly with respect to the position of the piston.  
Diagram:  Initial State 


Final State 


Given:  V_{1} =  8.4  L  P_{atm}  100  kPa  
V_{2} =  3.7  L  A_{piston}  0.020  m^{2}  
F_{1} =  1200  N  
F_{2} =  0  N  
Find:  P_{1} =  ???  kPa  W =  ???  kJ  
P_{2} =  ???  kPa  
Assumptions:  1  The gas in the cylinder is a closed system.  
2  The process occurs slowly enough that it is a quasiequilibrium process.  
3 There is no friction between the piston and the cylinder wall.  
4 The spring force varies linearly with position.  
Equations / Data / Solve:  
In the initial and final states, the piston is not accelerating. In fact, it is not moving. Therefore, there is no unbalanced force acting on it. This means that the vector sum of all the forces acting on the piston must be zero.  

Eqn 1  
Initial State:  P_{1}  160  kPa  

Eqn 2  
Final State:  P_{2}  100  kPa  
For a quasiequilibrium process, boundary or PV work is defined by:  

Eqn 3  
Because F_{spring} varies linearly with the position of the piston AND volume also varies linearly with the position of the piston, we can conclude that F_{spring} must vary linearly with respect to the volume !  

Eqn 4  
m  2.5532E+05  N/m^{3}  b  944.68  N  

Eqn 5  
W  611.0  J  
Verify:  None of the assumptions can be verified using only the information given in the problem statement.  
Answers :  P_{1}  160  kPa  W  611  J  
P_{2}  100  kPa 