Thermodynamics, Isothermal Question

Click For Summary

Homework Help Overview

The discussion revolves around a thermodynamics problem involving an ideal gas undergoing a process represented on a P-V diagram. The original poster presents a scenario with specific initial conditions and seeks to calculate the work done during a cycle, questioning the correctness of their result.

Discussion Character

  • Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the nature of the process, questioning whether it is isothermal and discussing the implications of the straight-line path on the P-V diagram. There are inquiries about the graphical integration approach and the definition of isothermal processes.

Discussion Status

The discussion is active, with participants providing clarifications regarding the nature of the process and suggesting alternative methods of interpretation. Some guidance has been offered regarding the graphical approach, which has been acknowledged as helpful by the original poster.

Contextual Notes

There is a mention of the original poster's time away from academic studies, which may influence their understanding of the concepts discussed. The problem involves specific constraints related to the ideal gas law and the characteristics of the thermodynamic process.

kalbuskj31
Messages
16
Reaction score
0

Homework Statement



10 moles of an ideal gas, in the initial state P1 = 10 atm, T1 = 300K are taken around the following cycle.

a) A reversible change of state along a straight line path on the P-V diagram to the state P = 1 atm T = 300K. How much work is done during the cycle?

Homework Equations



Isothermal, ΔU = 0 so w = p dV

The Attempt at a Solution



w = p dV and p = nRT/V so w = nRT (dV/V)

w = nRT ln(Vf/Vi) = nRT ln(Pi/Pf)

w = [(10 moles)* (8.3144 J/ K *mole) * (300K)] ln(10atm/1atm)

w = 57.4 kJ but the correct answer is 123.4kJ. Suggestions??
 
Physics news on Phys.org
The process is not isothermal. You are not taking into account that the graph is a straight line.

You are finding the area under the isothermal curve between the beginning and ending points (PV = constant).

AM
 
Andrew Mason said:
The process is not isothermal. You are not taking into account that the graph is a straight line.

You are finding the area under the isothermal curve between the beginning and ending points (PV = constant).

AM

Thanks. Can you please explain why this process isn't considered isothermal?
 
kalbuskj31 said:
Thanks. Can you please explain why this process isn't considered isothermal?

An isotherm has the same temperature during the whole process, not just at the start and end points. It will look like an exponential curve on the PV diagram. Adiabats will be "steeper" because they cross the isotherms.

Since it's a straight line, did you try a graphical integration?
 
Thank you for the help. the graphical interpretation helped give me the correct answer. I've been out of school for 3 years, so I'm brushing up on everything.
 

Similar threads

Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
9K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
9K
  • · Replies 3 ·
Replies
3
Views
9K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K