7D4 :  ΔS and the TS Diagram for Ideal Gas Processes  8 pts 

An ideal
gas is contained in a pistonandcylinder device in which the system moves from state 1 to state 2. 

a.) If T_{2} is greater than T_{1}, show that the ΔS_{12} is greater if the process is isobaric than if it is isochoric. Sketch
the isobaric and isochoric process
paths on PV and TS diagrams. 

b.) Use your TS Diagram from part (a) to show that an isochoric path passing through a
state has a greater slope than an isobaric path passing through the same state. 

c.) If P_{2} is greater than P_{1}, show that the ratio of ΔS_{12} for an isothermal process to ΔS_{12} for an isochoric process is (1  g). Sketch the isothermal and isochoric proces paths on PV and TS diagrams. 

Read :  Sketch the process in parts (a), (b) and (c) first to get a better understanding of the processes.  
For part (a) use equations relating entropy to C_{p} and C_{V}.  
For part (b) recall that the slope on a TS Diagram is (dT/dS).  
For part (c) determine DS for each process and determine the ratio.  
Given:  A closed system consisting of an ideal gas with constant specific heat ratio g.  ^{}  
Find:  Part (a)  For the process where the T increases from T_{1} to T_{2}: show that DS is greater if the change in state occurs at constant P than if it occurs at constant V.  
Sketch PV and TS Diagrams for the process.  
Part (b)  Show on a TS Diagram that a line of constant specific volume passing through a state has a greater slope than a line of constant P.  
Part (c)  For the process where the P increases from P_{1} to P_{2}: show that the ratio of DS for an isothermal process to DS for a constant specific volume process is (1  g).  
Sketch PV and TS Diagrams for the process.  
Diagram:  
Part (a) and (b): 

Part (c): 


Part (a) and (b): 

Part (c): 


Assumptions:  1   The system consists of an ideal gas with constant specific heats.  
Equations / Data / Solve:  
Part a.)  There are two key equations for calculating the entropy change of an ideal gas.  

Eqn 1 

Eqn 2  
For Process 1A, specific volume is constant. For Process 1B, pressure is constant.  
We can apply Eqn 1 to Process 1A and Eqn 2 to Process 1B.  

Eqn 3 

Eqn 4  
Because the specific heats are constant, Eqns 3 & 4 can be integrated to obtain :  

Eqn 5 

Eqn 6  
Notice that both the intial and final temperatures are the same: T_{A} = T_{B} = T_{2}.  
Next, we can take the ratio of Eqn 2 to Eqn 1 : 

Eqn 7  
Cancelling terms leaves us with : 

Eqn 8  
For ideal gases : 

Eqn 9  
Use Eqn 9 to eliminate C_{P} from Eqn 8 : 

Eqn 10  
Because R and C_{V} are both positive numbers, we can conclude that :  (S_{B}  S_{1}) > (S_{A}  S_{1})  Eqn 11  
Part b.)  Here, we compare, at state 1 (dT/dS)_{V} to (dT/dS)_{P}.  
Since (dT/dS) at fixed V (or fixed P) is : 

Eqn 12  
In part (a), we showed that, for the same DT, DS at constant P is greater than DS at constant V.  
Consequently :  Eqn 13  
On a TS Diagram, a constant specific volume line passing through State 1 has a greater slope than a constant pressure line passing through the same state.  
Part c.)  For Process 1A, temperature is constant. For Process 1B, volume is constant.  
Apply Eqn 2 to Process 1A and Eqn 1 to Process 1B.  

Eqn 14  

Eqn 15  
We need to consider the ratio of Eqn 14 to Eqn 15 and compare its value to 1 to determine which is greater, DS_{1A} or DS_{1B}.  

Eqn 16  
But, for ideal gases undergoing a constant volume such as Process 1B :  

Eqn 17  and : 

Eqn 18  
Therefore : 

Eqn 19  
Now, we can use Eqns 9 & 19 to simplify Eqn 16 : 

Eqn 20  
Verify:  The assumptions made in this solution cannot be verified with the given information.  
Answers :  Part a.)  (S_{B}  S_{1}) > (S_{A}  S_{1})  
Part b.) 

Part c.) 
