Carnot cycle, heat and monatomic ideal gas

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Homework Help Overview

The discussion revolves around a problem involving a monatomic ideal gas used in a Carnot cycle, specifically focusing on the heat expelled during one of the processes in the cycle. The original poster is seeking assistance with understanding the relationship between work done and heat transfer in this thermodynamic context.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses confusion about the heat expelled during the process C => D, noting that it must be less than 400 J. Some participants discuss the efficiency of the Carnot cycle and its implications for heat transfer, while others reference the adiabatic equation as a means to approach the problem.

Discussion Status

The discussion includes various attempts to clarify the problem, with some participants offering insights into the efficiency of the Carnot cycle and its equations. There is an indication that productive dialogue is occurring, although no consensus has been reached regarding the specific heat value.

Contextual Notes

The original poster mentions a specific value of work done (400 J) and provides a reference to a PV graph, suggesting that visual aids are part of the discussion context. The problem appears to be constrained by the need to apply thermodynamic principles without providing complete solutions.

frznfire219
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Hi, I would appreciate any help with this:

A monatomic ideal gas is used as the working substance for
the Carnot cycle. Processes A => B and C => D
are isothermal, while processes B => C and D => A are adiabatic.
During process A => B, there are 400 J of work done by the gas on
the surroundings. How much heat is expelled by the gas during process C => D?

So I'm completely stuck, all I know is that it's less than 400 J, obviously.
There's a picture of the corresponding PV graph actually at http://www.compadre.org/psrc/evals/Physics_Bowl_2003.pdf (page 12).

Thanks for any help!
 
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I'm not sure about this explanation, but whatever. You know that the carnot cycle runs at perfect efficiency ie. 1-(Ql/Qh) and you know that efficiency of the carnot cycle is also 1-(Tl/Th). I get 100 J.
 
Thanks! I actually figured it out later with the adiabatic equation (PV^gamma is constant) but your way is much more elegant.
 

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