# Example Problem with Complete Solution

7D-4 : ΔS and the T-S Diagram for Ideal Gas Processes 8 pts
An ideal gas is contained in a piston-and-cylinder device in which the system moves from state 1 to state 2.
a.) If T2 is greater than T1, show that the ΔS12 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 P2 is greater than P1, show that the ratio of ΔS12 for an isothermal process to ΔS12 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 Cp and CV.
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 T1 to T2: 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 P1 to P2: 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 1-A, specific volume is constant. For Process 1-B, pressure is constant.
We can apply Eqn 1 to Process 1-A and Eqn 2 to Process 1-B.
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: TA = TB = T2.
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 CP from Eqn 8 : Eqn 10
Because R and CV are both positive numbers, we can conclude that : (SB - S1) > (SA - S1) 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 1-A, temperature is constant. For Process 1-B, volume is constant.
Apply Eqn 2 to Process 1-A and Eqn 1 to Process 1-B.
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, DS1-A or DS1-B.
Eqn 16
But, for ideal gases undergoing a constant volume such as Process 1-B :
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.) (SB - S1) > (SA - S1)
Part b.) Part c.)