7E3 :  Work and ΔS for IGs Undergoing Isothermal, Polytropic and Adiabatic Processes  8 pts 

Argon gas is compressed in a piston and cylinder device from 20 psia and 55^{o}F to 120 psia. The compression is internally reversible and the helium behaves as an ideal gas with a constant heat capacity of C_{P} = (5/2) R.  
Determine the work in Btu/lb_{m} and ΔS in Btu/lb_{m}^{o}R and sketch the process path on both PV and TS Diagrams assuming the compression is… a.) Polytropic with d = 1.5 b.) Adiabatic c.) Isothermal 

Read :  For part (a) start with the equation for PV work for internally reversible, polytropic processes for ideal gases. When determining DS assume the heat capacity is constant. Since argon is monatomic use C_{p }= (5/2) R.  
For part (b) determine work by applying an energy balance (where Q = 0). Assume constant heat capacity DU = C_{V} DT where C_{V} = C_{P}  R and C_{p} = (5/2) R. Determine T_{2} from an entropy balance. Recall that internally reversible, adiabatic processes are also isentropic.  
For part (c) start with the definition of PV work, substitute in the ideal gas EOS for pressure and integrate. Remember the process is isothermal (this simplifies the analysis of both the work and change in entropy).  
Given:  P_{1}  20  lb_{f }/ in^{2}  P_{2}  120  lb_{f }/ in^{2}  
T_{1}  55  ^{o}F  m  8.4  lb_{m}  
514.67  ^{o}R  Part (a)  d  1.5  
Find:  Part (a)  (c) :  W  ?  Btu  DS  ?  Btu/^{o}R  
Diagram: 





Assumptions:  1   As shown in the diagram, the system is the gas.  
2   The gas is modeled as an ideal gas.  
3   The compression is internally reversible.  
4   Boundary work is the only form of work that crosses the system boundary.  
5   Changes in kinetic and potential energies are negligible.  
6   Argon has a constant heat capacity of C_{P} = (5/2) R.  
R  1.986  Btu/lbmol^{o}R  
MW  39.948  lb_{m}/lbmol  
Equations / Data / Solve:  
Part a.)  In part (a) we must determine the work and the change in entropy for a polytropic process.  
For polytopic, internally reversible processes with ideal gases: 

Eqn 1  
The problem at this point is that we do not know T_{2}. But, we do know that the process is polytropic !  
In Lesson 7E we learned that : 

Eqn 2  
We can solve Eqn 2 for T_{2} : 

Eqn 3  
Substitute Eqn 3 for T_{2} into Eqn 1 and rearrange the result to get : 

Eqn 4  
Now, we can plug values into Eqns 12 & 10 or Eqn 13 :  n  0.2103  lb_{m}  
T_{2}  935.22  ^{o}R  
T_{2}  475.55  ^{o}F  
W_{12}  351.2  Btu  
DS can be determined by applying the
2nd Gibbs Equation
for ideal gases: 

Eqn 5  
If we assume the heat capacity is constant: 

Eqn 6  
Since argon is a montomic gas we can assume : 

Eqn 7  
C_{P}  4.965  Btu/lbmol^{o}R  
Subtituting C_{P} and other values into Eqn 6 yields :  DS  0.12471  Btu/^{o}R  
Part b.)  Here we must determine the work and the change in entropy for an adiabatic process.  
The 1st Law for an adiabatic process with negligible changes in kinetic and potential energies is : 

Eqn 8  
We can evaluate the change in the internal energy using C_{V} and : 

Eqn 9  
Assuming the heat capacity is constant, Eqn 9 simplifies to : 

Eqn 10  
The following relationship applies to ideal gases : 

Eqn 11  
We can combine this with Eqn 7 to get : 

Eqn 12  
C_{V}  2.979  Btu/lbmol^{o}R  
At this point, the only obstacle to using Eqns 18 & 16 to evaluate W_{12} is that we do not know T_{2}.  
We need to make use of the fact that the process is adiabatic and internally reversible to determine T_{2}.  
An adiabatic process that is also internally reversible is isentropic :  DS  0  Btu/^{o}R  
We can use this fact with the 2nd Gibbs Equation for ideal gases with constant heat capacities to determine T_{2} as follows :  

Eqn 13  
Solve Eqn 13 for T_{2} using ΔS = 0 : 

Eqn 14  

Eqn 15  

Eqn 16  
Hey, we already KNEW this ! 

Eqn 17  
Where : 

Eqn 18  
g  1.667  
T_{2}  1053.88  ^{o}R  
Now, we can plug T_{2} into Eqn 10 and DU_{12} into Eqn 8 :  DU_{12}  40.21  Btu/lb_{m}  
W_{12}  337.76  Btu  
Part c.)  Determine the work from the definition of boundary work : 

Eqn 19  
For an ideal gas substitute P = nRT/V into Eqn 19 : 

Eqn 20  
Integrate Eqn 2 (the process is isothermal, T_{1} = T_{2} = T) : 

Eqn 21  
We don't know V_{1} or V_{2} but we can determine the values from the Ideal Gas EOS (T_{1} = T_{2} = T):  

Eqn 22 

Eqn 23  
Dividing Eqn 22 by Eqn 23 we obtain : 

Eqn 24  
Now, substitute Eqn 24 back into Eqn 21 to get : 

Eqn 25  
Plug values into Eqn 25 :  W_{12}  385.10  Btu  
DS can be determined by applying the 2nd Gibbs Equation for ideal gases:  

Eqn 26  
For an isothermal process Eqn 26 reduces to: 

Eqn 27  
Now, we can plug values into Eqn 27 :  DS  0.74824  Btu/^{o}R  
Verify:  The ideal gas assumption needs to be verified.  
We need to determine the specific volume at each state and check if 


(Argon is a noble gas).  
Solving the Ideal Gas EOS for molar volume yields : 


Use :  R  10.7316  psiaft^{3} / lbmol^{o}R  V_{1} = V_{2C}  276.16  ft^{3}/lbmol  
V_{2A}  83.64  ft^{3}/lbmol  
V_{2B}  94.25  ft^{3}/lbmol  
The specific volume at each state is greater than 80 ft^{3}/lbmol, therefore the ideal gas assumption is reasonable.  
Answers :  a.)  W_{12}  351  Btu  
DS  0.125  Btu/^{o}R  
b.)  W_{12}  338  Btu  The isentropic process requires the least work !  
DS  0  Btu/^{o}R  
c.)  W_{12}  385  Btu  The isothermal process requires the most work !  
DS  0.748  Btu/^{o}R  
How can ΔS be negative in parts (a) and (c) ?  
Heat transfer from the system to the surroundings occurs. So, although ΔS_{system} < 0, ΔS_{surr} > 0 by an even larger amount so that ΔS_{universe} > 0 and the 2nd Law is not violated. 