How Do You Calculate Entropy Change in a Reversible Cyclic Process?

In summary: I solved it by using the fact that the change in entropy for the entire cycle is equal to zero, since the gas returns to its original state. This means that the change in entropy for the A-C phase must be equal and opposite to the change in entropy for the A-B phase. So I set up the equation:\Delta S_{AC} = \Delta S_{AB} + \Delta S_{BC}And then substituted the equations for \Delta S_{AB} and \Delta S_{BC} from the previous steps, and solved for \Delta S_{AC}. This gave me the correct equation:\Delta S_{AC} = nC_v\ln\frac{T_C}{T_A} - nR\
  • #1
Knight
6
0
An ideal gas undergoes a reversible, cycli process. First it expands isothermally from state A to state B. It is then compressed adiabatically to state C. Finally, it is cooled at constant volume to its original state, A.

I have to calculate the change in entropy of the gas in each one of the three processes and show that there is no net change in the cyclic process.

O.K. From A to B Delta S is nRln(V2/V1) since the process from A to B is isothermal. Delta S is 0 from B to C since that process occurs adiabatically.
But I am having trouble with the Delta S for the process from C to A. So far I have Delta S is Cv ln(T1/T2) But I'm having trouble converting this to something similar to the Delta S for A to B

Could someone please tell me if I'm on the right track with this problem, and possibly give me a hint
 
Physics news on Phys.org
  • #2
Knight said:
An ideal gas undergoes a reversible, cycli process. First it expands isothermally from state A to state B. It is then compressed adiabatically to state C. Finally, it is cooled at constant volume to its original state, A.

I have to calculate the change in entropy of the gas in each one of the three processes and show that there is no net change in the cyclic process.

O.K. From A to B Delta S is nRln(V2/V1) since the process from A to B is isothermal. Delta S is 0 from B to C since that process occurs adiabatically.
But I am having trouble with the Delta S for the process from C to A. So far I have Delta S is Cv ln(T1/T2) But I'm having trouble converting this to something similar to the Delta S for A to B

Could someone please tell me if I'm on the right track with this problem, and possibly give me a hint
You are right on the first two parts:

[tex]\Delta S_{CA} = \int_C^A ds = \int_C^A dQ/T[/tex]

Since: [itex]dQ = dU + PdV[/itex] and dV = 0, [itex]dQ = dU = nC_vdT[/itex]

[tex]\Delta S_{CA} = \int_C^A nC_v\frac{dT}{T} = nC_v\ln\frac{T_A}{T_C}[/tex]

So the total change in entropy is:

[tex]\Delta S_{AA} = nR\ln\frac{V_B}{V_A} + nC_v\ln\frac{T_A}{T_C}[/tex]

So the question is whether:

[tex]nR\ln\frac{V_B}{V_A} = nC_v\ln\frac{T_C}{T_A}[/tex]

AM
 
  • #3
I solved it.

Thanks :smile:
 
  • #4
Knight said:
I solved it.

Thanks :smile:
I trust that you used the fact that the adiabatic condition applies to the compression phase. So for the B-C phase:

[tex]\left(\frac{V_C}{V_B}\right)^{1-\gamma} = \frac{T_C}{T_B}[/tex]
AM
 
Last edited:

1. What is a reversible cyclic process?

A reversible cyclic process is a thermodynamic process in which the system undergoes a series of changes and then returns to its initial state, without leaving any permanent effects on the surroundings. This means that the process is both reversible and cyclic, as it can be reversed and repeated indefinitely without any net change.

2. How is a reversible cyclic process different from an irreversible process?

In an irreversible process, some energy is lost to the surroundings in the form of heat or work, resulting in a decrease in the system's ability to do work. In contrast, a reversible cyclic process is idealized and does not involve any energy loss, making it a more efficient and ideal process.

3. What is the significance of a reversible cyclic process?

Reversible cyclic processes are important in thermodynamics because they allow us to define an ideal, theoretical maximum for the efficiency of a system. This is known as the Carnot efficiency, and it serves as a benchmark for the performance of real-world systems.

4. Can real-world processes be reversible and cyclic?

No, it is impossible to have a completely reversible and cyclic process in the real world. All real-world processes involve some degree of irreversibility, such as friction, which leads to energy loss. However, reversible cyclic processes serve as a useful theoretical model for understanding the behavior of real-world systems.

5. How is a reversible cyclic process represented on a thermodynamic diagram?

A reversible cyclic process is represented as a closed loop on a thermodynamic diagram, such as a P-V (pressure-volume) or T-S (temperature-entropy) diagram. This loop represents the system going through a series of changes and returning to its initial state, with no net change in the system or its surroundings.

Similar threads

Internal energy won't add up to 0 in cyclic process
    • Introductory Physics Homework Help
Replies
1
Views
184
Thermodynamics Problem: Reversible and Irreversible Processes
    • Introductory Physics Homework Help
Replies
1
Views
926
Cyclic Process for Ideal Gases
    • Introductory Physics Homework Help
Replies
7
Views
1K
Help Calculate Work Done in Adiabatic Process & Fill Table
    • Introductory Physics Homework Help
Replies
19
Views
1K
To find total work done from multiple reversible processes
    • Introductory Physics Homework Help
Replies
3
Views
949
Entropy change for reversible and cyclic process
    • Introductory Physics Homework Help
Replies
3
Views
4K
B Change in entropy of reversible isothermal process
    • Thermodynamics
Replies
11
Views
324
What is the entropy for an irreversible adiabatic process?
    • Introductory Physics Homework Help
Replies
1
Views
1K
How to know whether a given process is reversible?
    • Introductory Physics Homework Help
Replies
10
Views
2K
Closed cycle of an ideal gas
    • Introductory Physics Homework Help
Replies
3
Views
118

Hot Threads

Recent Insights

Back
Top