Can an Adiabatic Process be Isothermal?

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Homework Help Overview

The discussion revolves around the concepts of adiabatic and isothermal processes within the context of thermodynamics. The original poster is grappling with a problem involving a thermodynamic cycle that includes various types of processes, specifically questioning the relationship between adiabatic conditions and temperature changes.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand whether an adiabatic process can also be isothermal, given their observations of temperature change in their problem. Some participants question the validity of this assumption, while others seek clarification on the specific calculations related to the cycle.

Discussion Status

The discussion is ongoing, with some participants providing direct responses and others indicating a need for further exploration of the problem. The original poster has indicated that the issue has become complex enough to warrant a separate thread for detailed examination.

Contextual Notes

The original poster mentions specific conditions and parameters related to their problem, including pressures and volumes, which may influence the analysis of the processes involved. There is an indication of confusion regarding the temperatures calculated for different points in the cycle.

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In an adiabatic process, I know that that dQ is zero. I am doing a problem with a cycle that has different types of processes, and when I get to the adiabatic leg, I am finding that the change in temperature is zero. Therefore, the work done and the change in internal energy also equal zero.

Is this possible? Can an adiabatic process also have no change in temperature, therefore being isothermal?
 
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OK, so in the problem A->B is isobaric, B->C is adiabatic, C->D is isothermal, D->A is isochoric.

Pa = 2atm, Va = 2 m^3, n = 1mole, Vb = 1/2Va, Pa = 2Pd

I always end up with the temperatures of B and C both equaling 2.44x10^4 K. What am I doing wrong?
 
I'm going to make a new post of this since it's really no longer a "small question."
 
How come you end up with equal temp.?Post your work.

Daniel.
 
I posted my work in a new thread since it is not a small problem anymore.
 

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