Irreversible isothermal Process Work External pressure not provided

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The discussion centers on the challenges of calculating work in an irreversible isothermal process without provided external pressure. The formula for work, W = -P_ext(ΔV), is highlighted, but the absence of P_ext complicates the calculations. Participants express confusion over how external pressures of 2 bar and 1 bar are derived and question the validity of changing P_ext in irreversible processes. The conversation also touches on the importance of sketching a PV diagram to visualize the isothermal behavior of an ideal gas. Overall, the participants seek clarity on the relationship between external pressure and work in this thermodynamic context.
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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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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