# Example Problem with Complete Solution

4A-3 : Quasi-Equilibrium Compression of R-134a 4 pts
R-134a vapor is held in a piston-and-cylinder device at 30oC and 65 kPa. The R-134a vapor is compressed as small weights are slowly added to the back of the piston until the pressure inside the cylinder reaches 400 kPa.
Determine :
a.) The boundary work done by the R-134a
b.) The final temperature of the R-134a
The following measurements have been made during the process:
P(kPa), V(L) = (65,1.94), (104,1.57), (167,1.04), (218,0.79), (255,0.67), (311,0.51), (350,0.35)

Read : The key concept here is that boundary or PV work is represented by the area under the process path curve on a PV Diagram.  So, once we plot the given data on a PV Diagram, all we need to do is numerically integrate to determine the area under the curve and we will know the work !
We can use the R-134a Tables in the NIST Webbook to determine the final temperature because we know both Pfinal and Vfinal.  But we don't know the number of moles in the system.  Fortunately, we can use the initial state P1, V1, and T1 to determine the number of moles in this closed system.
Diagram:
Given: P (kPa) V (L) Find: a.) Wtotal ??? J
65 1.94
104 1.57 b.) T2 ??? oC
167 1.04
218 0.79
255 0.67
311 0.51
400 0.35
T1 30 oC
Assumptions: 1 - Each state in the data table is an equilibrium state.
2 - The process is a quasi-equilibrium process.
3 - The system is a closed system.
4 - The trapezoidal rule gives an acceptable estimate of the area under the process path in the PV Diagram.
Equations / Data / Solve:
Part a.) The area of each trapezoid under the process path in the PV Diagram is the product of the average pressure for that trapezoid and the change in volume across the trapezoid.
Eqn 1
Where  subscript a refers to the left side of each trapezoid and subscript b refers to the right side of each trapezoid in the PV Diagram.
Trapezoid Pavg (kPa) ΔV (L) W (J)
A 85 -0.37 -31.3
B 136 -0.53 -71.8
C 193 -0.25 -48.1
D 237 -0.12 -28.4
E 283 -0.16 -45.3
F 356 -0.16 -56.9 Wtotal = -281.7 J
Part b.) If we knew the specific volume of the R-134a in the final state, we could use the R-134a tables to determine the temperature.  But, at this point, all we know is the total volume in the final state.
The trick here is that  the
mass is the same in the initial and final states.
Therefore, we need to determine the
mass of R-134a in  the system.
We can use the information we have for the
initial state to determine the mass of R-134a in the system.
First, look up the specific volume of R-134a in the initial state in the isothermal thermodynamic tables:
V1 0.37538 m3/kg
Then, to calculate mR134a, use: Eqn 1 mR134a 0.00517 kg
Finally, calculate V2 using: Eqn 2 V2 0.067723 m3/kg
Now, use the isobaric thermodynamic tables at a pressure of 400 kPa.  Specify a temperature range that you are sure brackets V2.  Here is a table that I cut-and-pasted from the NIST WebBook.
Temperature (C) Pressure (MPa) Volume (m3/kg) Phase
72 0.4 0.066907 vapor
73 0.4 0.067138 vapor
74 0.4 0.067369 vapor V2 falls between 75oC and 76oC
75 0.4 0.067599 vapor So, now, we must interpolate.
76 0.4 0.067830 vapor
77 0.4 0.068060 vapor T2 75.54 oC
Verify: None of the assumptions can be verified from the data given in the problem statement.
Answers : a.) Wtotal = -282 J b.) T2 75.5 oC