4A3 :  QuasiEquilibrium Compression of R134a  4 pts 

R134a vapor is held in a pistonandcylinder device at 30^{o}C and 65 kPa. The R134a 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 R134a b.) The final temperature of the R134a 

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 R134a Tables in the NIST Webbook to determine the final temperature because we know both P_{final} and V_{final}. But we don't know the number of moles in the system. Fortunately, we can use the initial state P_{1}, V_{1}, and T_{1} to determine the number of moles in this closed system.  
Diagram: 


Given:  P (kPa)  V (L)  Find:  a.)  W_{total}  ???  J  
65  1.94  
104  1.57  b.)  T_{2}  ???  ^{o}C  
167  1.04  
218  0.79  
255  0.67  
311  0.51  
400  0.35  
T_{1}  30  ^{o}C  
Assumptions:  1  Each state in the data table is an equilibrium state.  
2  The process is a quasiequilibrium 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  P_{avg} (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  W_{total} =  281.7  J  
Part b.)  If we knew the specific volume of the R134a in the final state, we could use the R134a 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 R134a in the system. We can use the information we have for the initial state to determine the mass of R134a in the system. 

First, look up the specific volume of R134a in the initial state in the isothermal thermodynamic tables:  
V_{1}  0.37538  m^{3}/kg  
Then, to calculate m_{R134a}, use: 

Eqn 1  m_{R134a}  0.00517  kg  
Finally, calculate V_{2} using: 

Eqn 2  V_{2}  0.067723  m^{3}/kg  
Now, use the isobaric thermodynamic tables at a pressure of 400 kPa. Specify a temperature range that you are sure brackets V_{2}. Here is a table that I cutandpasted from the NIST WebBook.  
Temperature (C)  Pressure (MPa)  Volume (m^{3}/kg)  Phase  
72  0.4  0.066907  vapor  
73  0.4  0.067138  vapor  
74  0.4  0.067369  vapor  V_{2} falls between 75^{o}C and 76^{o}C  
75  0.4  0.067599  vapor  So, now, we must interpolate.  
76  0.4  0.067830  vapor  
77  0.4  0.068060  vapor  T_{2}  75.54  ^{o}C  
Verify:  None of the assumptions can be verified from the data given in the problem statement.  
Answers :  a.)  W_{total} =  282  J  b.)  T_{2}  75.5  ^{o}C 