Irreversible isothermal Process Work External pressure not provided

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Homework Help Overview

The discussion revolves around an irreversible isothermal process, specifically focusing on the work done when the external pressure is not provided. Participants are exploring the implications of this lack of information on calculating work and heat absorbed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to calculate work using different expressions but are questioning the validity of their approaches due to the absence of external pressure. There is also confusion regarding how external pressures are determined in the context of an irreversible process.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the assumptions related to external pressure. Some guidance has been offered regarding the sketching of a PV diagram, but there is no explicit consensus on how to proceed with the calculations.

Contextual Notes

There is a noted concern about the changing nature of external pressure in irreversible processes, which may affect the calculations. Additionally, the original poster's attempts to derive work from given values are met with skepticism regarding their correctness.

Aurelius120
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Homework Statement
An ideal gas is irreversibly isothermally expanded from ##(8bar ,4L)##to##(2bar,16L)##to##(1bar,32L) ##Find heat.
Relevant Equations
NA
Screenshot_20231231_031131_Chrome.jpg

It is clear that the process is isothermal else it is not possible to find heat absorbed.
$$W=-P_{ext}(\Delta V)$$

However ##P_{ext}## is not given. How do I proceed?
I tried taking ##W=-(P_2V_2-P_1V_1+P_3V_3-P_2V_2)=\Delta(PV)## but it is wrong for obvious reasons.
 
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The one solution I found uses
##W_1=-(2)(16-4)## and ##W_2=-(1)(32-16)##
##W=(W_1+W_2) bar.Litre=-4000J##
And ##Q=4000J##

How external pressure becomes ##2\ bar## and ##1\ bar## is beyond me. It also seems wrong that ##P_{ext}## should change in an irreversible process? That is like the only thing that is good about irreversible calculations.
 
Sketch a PV diagram of the process. Do you know what an isotherm looks like for an ideal gas?
 
Mister T said:
Sketch a PV diagram of the process. Do you know what an isotherm looks like for an ideal gas?
images.jpeg

Something like this correct?
 
Looks like you've found the way!
Aurelius120 said:
How external pressure becomes ##2\ bar## and ##1\ bar## is beyond me.
A slow compression expansion.

Aurelius120 said:
It also seems wrong that ##P_{ext}## should change in an irreversible process?
A dramatic example would be an explosion.
 
Last edited:

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