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I have read in my textbook that maximum work is done by gas in a reversible expansion, but I do not know the reason behind it. The reason why more work is done by the system in a reversible process than an irreversible process is entropy is generated within the gas in an irreversible process. Whenever entropy is generated within the system there is a lost ...

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Thermodynamically the reversible heat is defined as TdS and dS>Q/dT according to the second law of thermodynamics. According to Transport phenomenon the heat of compression/expansion is given by $$\rho c_p \frac{\partial T}{\partial t} = - (\nabla \cdot q) -(\frac{\partial ln( \rho)}{\partial T})_p \frac{DP}{Dt}- \tau:\nabla v$$ so energy is lost as ...

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Since work done is integral of $P_{\rm external} \times ΔV$ , work will be maximum $P_{\rm internal}$ is almost equal to external pressure (using ideal gas rule) . And as the condition completely exists in reversible case, Reversible process has greater work done than irreversible process.

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The reason why more work is done in a reversible process than an irreversible process is in an irreversible process entropy is generated within the gas whereas in a reversible process entropy is not generated. That additional entropy has to be transferred to the surroundings in the form of heat, leaving less heat available to be converted to work. To ...

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Its almost true. Of course, Jamie would beg to differ. However, the general logic is simple. If you have something that's irreversible, energy is lost. That energy was work that could have been used to produce valuable work. If you consider this gas expansion as part of a cycle, then it is more clear that that irreversability calls for more work to be ...

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1st part-Since from the carnot cycle and 2nd law of thermodynamics, we know that efficiency for any cyclic process is maximum when it is a reversible carnot cycle .Since efficiency is maximum ,the work done is also maximum for reversible process 2nd part- No the magnitude of work done by gas in reversible compression is less than work done in irreversible ...

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In conventional thermodynamic language "adiabatic" process means being adiabatic (no heat exchnage) at every instant of the process not just in the sense that the total heat exchange is zero. The process itself can be reversible or irreversible, only the heat exchanged must be zero. The so-called adiabatic legs of the Carnot cycle (isothermal-...

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it seems we can have reversible isentropic processes that are not adiabatic.. No you cannot. An isentropic process is by definition a process that is both adiabatic and reversible. So you can't have an isentropic process that is not adiabatic. However, you can have an adiabatic process that is not isentropic, if it is not a reversible process. An example is ...

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Any reversible cyclic process will be isentropic (since entropy is a state function), whereas it's not necessary that it's adiabatic as well. A famous example is the Carnot cycle.

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The reverse reaction really is $${}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\ {}^1_0n + \gamma + 202.5\ \mathrm{MeV} \rightarrow {}^1_0n + {}^{235}_{92}\mathrm{U}$$ as you'd expect. Yes, your deduction from time symmetry is absolutely correct (as far as we know) - it has to be. The above will do the job. Yes, the four elements on the left will ...

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The reverse reaction, in which six particles join to make two, is allowed by time-reversal symmetry but forbidden by entropy considerations arising from classical thermodynamics. You just plain can't get those six particles to meet each other "just so"; there are too many ways for them to miss. The way that thermodynamic entropy breaks time ...

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