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Maximum work can be obtained from reversible process ?

  1. Dec 2, 2014 #1
    My book says 'Maximum work can be obtained only from thermodynamically reversible processes,' but why is it so? What is the cause?

    Actually to me the definition of reversible process is confusing. It says that at each step during the process, equilibrium is maintained. But, let's say,we have to expand a gas in a cylinder headed by piston from pressure Pi to Pf .Suppose,the process is done with innumerable infinitesimal steps. Then, say at each step, dp is changed in the pressure.

    Before the step, gas and the piston were exerting same pressure Pi . Now, when the piston is raised, gas does expand to equalize the pressure,which is Pi - dp . So, at the end of the step, the gas & the piston will exert same pressure and hence will be in mechanical equilibrium.

    During the step,the gas and the piston was not in equilibrium;it was at the end of the step the equilibrium is achieved. So, during each step of the process the gas would be at disequilibrium . So,will it really be a reversible process? It is contradictory with the definition. But it is true that during each step, the system is at disequilibrium though infinitesimally small by . So, why does the definition tell that at each step,there is equilibrium?

    So, I have three questions:

    Why is maximum work only obtained from reversible processes?

    According to the definition, at each step of the reversible process,there must be equilibrium. But during each step, there is infinitesimal disequilibrium. So, why does the definition say so in-spite of the disequilibrium during each step?

    How do infinitesimal steps make a process reversible? I will be very grateful if anyone answer these three questions.

    Please help.
     
  2. jcsd
  3. Dec 3, 2014 #2

    DrDu

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    Science Advisor

    A reversible process is not a physical process but rather a mathematical construction: You integrate over some path in the space of equilibrium states. The question how well a reversible process can be approximated by a quasi-static process is strictly speaking outside the realm of equilibrium thermodynamics and requires some non-equilibrium thermodynamics.
    However the mathematical statement that the work is maximal for a reversible process and that all real processes yield less work can be proven using only arguments from equilibrium thermodynamics.

    That's often the case with mathematical proofs, they give you powerful bounds but no recipe on how to best approximate these bounds.

    So let's say you have designed a motor and you get 16% out of work. If you know that a reversible process would yield 40 % you have probably much potential for optimization. While, if the maximal yield is 18%, investing in further optimization would be rather fruitless.
     
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