Efficiency of Carnot Engine w/ 0.75kg Ideal Gas

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Homework Help Overview

The problem involves calculating the efficiency of a Carnot engine using a 0.75 kg ideal gas undergoing a cyclic process consisting of two isobaric and two isometric processes. Participants are exploring the relationship between temperature, pressure, and volume in the context of thermodynamics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using the ideal gas law and various thermodynamic equations to find temperatures at different points in the cycle. There are attempts to calculate efficiency using different methods, with some questioning the validity of their approaches and the assumptions made about the processes involved.

Discussion Status

There is an ongoing exploration of different methods to calculate the Carnot efficiency, with some participants expressing uncertainty about their results. Multiple interpretations of the problem are being discussed, particularly regarding the nature of the processes (isobaric vs. isothermal) and the implications for temperature calculations.

Contextual Notes

Participants note the lack of specific information about the type of ideal gas and express frustration with the problem's wording. There is also confusion regarding the application of the ideal gas law during heat transfer processes.

  • #31
LOL! Well, I thought that too... But then it would have to have been curved... O_o I know you CAN calculate at any point the T because that is definitely a given on any Pressure versus Volume graph... So tell me, did I explain that part correctly now that we've cleared up the confusion?
 
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  • #32
I got it! BOOOO
 
  • #33
I thought your graph said 1-2 was a constant temperature line!

<Pulls my hair out>

None of it is constant temperature. You have to find the gas constant using:

PV=mRT

Then do as you guys tried by going around the entire process.

You will find that T_H = 835.714

and T_L = 390

use that in carnot and you will get:

\eta_{th,rev} = 1- \frac{T_L}{T_H} = 1 - .4666 = .5333

That was an AWFUL problem in my opinion.
 
  • #34
cyrusabdollahi said:
I got it! BOOOO

Oh darn it :-(

Now this is the part where you explain how you did it with your ever-knowing knowledge and mind.
 
  • #35
Oh wait.. Now I forget something... T(h) would be the hottest temperature and T(c) would be the coldest?

I kept thinking initial and final O_o Explain it to me again
 
  • #36
No, I just stole what you did. (with some refining) :smile:
 
Last edited:
  • #37
Wow, thanks a lot!
I kept taking the inital temperature as the hot too!
 
  • #38
Tell your teacher that problem is unclear.
 

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