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So, while reading about Quasi- Static process, I came across this question here:

Question about Quasi- static process

From the answers given there, what I understood is this: After irreversible adiabatic compression and expansion, even though the final pressure is same as initial pressure, final temperature of the gas is not the same as initial temperature. The final temperature will be more because more work has to be done by surroundings in compression than done by the gas in expansion.

My thoughts are: a) More work is done during compression because more weight is placed on the piston (two weights) during compression than expansion( only one weight is removed). So more work is transferred to the system during compression and then this work adds to system's internal energy, hence the temperature increase. Is this correct?

b) When one of the weights is reduced( for expansion), the pressure on the piston drops suddenly and the piston shoots up, so there will be temperature gradients inside the gas, which will give rise to entropy, hence making the process irreversible.

Is this correct? Thanks in advance!

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  • $\begingroup$ Thanks for replying!But why? Why is the compression work done in irreversible adiabatic more than expansion work done in irreversible adiabatic compression? Is my reasoning in point a) correct or not? $\endgroup$
    – Natasha J
    Feb 2, 2023 at 15:52
  • $\begingroup$ The proper question is Why is work done on the system in irreversible adiabatic compression > work done on the system in reversible adiabatic compression? (and vice versa for expansion) $\endgroup$
    – Poutnik
    Feb 2, 2023 at 16:01
  • $\begingroup$ I think because in irreversible compression more weights have to be placed on the piston and in reversible process the pressure difference is infinitesimal. $\endgroup$
    – Natasha J
    Feb 2, 2023 at 16:27

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