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My teacher said that there is always wastage of work in irreversible process and that work in an irreversible process is always less than that in a reversible process even if one is adiabatic and the other is isothermal ....I am unable to understand as to why this happens...

Moreover I think I am unable to understand the very concept of irreversible works and it's calulation...so can someone guide with some examples or links so I can grasp it as I am ** unable to find any**..examples would be much appreciated

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Here is a link to an article I wrote that directly addresses your question: https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

It starts out by analyzing reversible vs irreversible work in a purely mechanical system involving a spring and a damper (dashpot) in parallel. This is like the spring and shock absorber system in your car. It then discusses the complete analogy between such a system and that of a gas experiencing reversible or irreversible compression or expansion. I hope it will give you a good fundamental understanding of the physical mechanisms that are involved.

ADDENDUM: Adiabatic irreversible expansion at constant pressure

In this case, you suddenly drop the external pressure from P1 to P2, and then hold the external pressure at P2 until the system equilibrates. This is what we call constant-pressure irreversible expansion.

$$nC_v(T_2-T_1)=-P_{ext}(V_2-V_1)=-P_2(V_2-V_1)=-P_2\left(\frac{nRT_2}{P_2}-\frac{nRT_1}{P_1}\right)$$

I replace $P_{ext}$ with $P_2$ because that is the final pressure that the gas equilibrates to. Solve the equation for $T_2/T_1$ as a function of $P_2/P_1$. Then determine $V_2/V_1$.

For reversible adiabatic expansion, we drop the external pressure gradually such that $$nC_vdT=-PdV=-\frac{nRT}{V}$$ In this case, the external pressure is always equal to that determined for the gas by the ideal gas law.

ADDENDUM #2

Solution for adiabatic reversible expansion or compression $$P_{ext}(V)=P(V)=P_1\left(\frac{V_1}{V}\right)^{\gamma}$$

Solution for isothermal reversible expansion or compression $$P_{ext}(V)=P(V)=P_1\left(\frac{V_1}{V}\right)$$

Solution for adiabatic reversible expansion or compression at constant external pressure from initial volume $V_1$ to final equilibrium volume $V_2$ $$P_{ext}(V)=\frac{P_1}{1+\gamma\left(\frac{V_2}{V_1}-1\right)}$$This equation applies for all $V_1<V<V_2$ in expansion or $V_2<V<V_1$ in compression

Solution for reversible expansion or compression of a gas held in contact with a constant temperature reservoir at the initial gas temperature $T_1$ and held at a constant external pressure from initial volume $V_1$ to final equilibrium volume $V_2$ $$P_{ext}(V)=P_1\frac{V_1}{V_2}$$ This equation applies for all $V_1<V<V_2$ in expansion or $V_2<V<V_1$ in compression

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  • $\begingroup$ What about the wastage of work thing my teacher said?..How can that be physically imagined? $\endgroup$ – Schwarz Kugelblitz Aug 24 at 13:10
  • $\begingroup$ The viscous dissipation of the mechanical energy (work) gets manifested as an increase in internal energy. If you feel the shock absorber, it feels warm. In the case of the gas with the punctured membrane, even though it has expanded adiabatically, it does not cool down. $\endgroup$ – Chet Miller Aug 24 at 14:02
  • $\begingroup$ How can we take pressure external in the formula as pressure final of gas to find work (while applying first law) for irreversible adiabatic? $\endgroup$ – Schwarz Kugelblitz Aug 24 at 15:08
  • $\begingroup$ My teacher has done this as a limiting case a couple of times..I am not sure how this is possible? $\endgroup$ – Schwarz Kugelblitz Aug 24 at 15:10
  • $\begingroup$ Your article was unable to clear all my quesries...what about computation and graphing of such work $\endgroup$ – Schwarz Kugelblitz Aug 24 at 16:43

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