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It is for the same reason that you need to use a dashed line for irreversible process which is higher than reversible process. This is because we know it is higher but don't know how high. If the dashed line is lower, we will conclude this is not feasible. This is the first thing what we want to know from the diagram.


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This is a really deep question. My explanation will maybe be not so rigorous, but I hope it can help shed some light. Let's start by saying that reversible work is indeed path-dependent, so it is not a state function. Consider for example the two reversible transformations $A$ and $B$ in the picture: They both are composed by an isobaric and an ...


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If you allow a cylinder with a piston where there is friction present between the walls of the cylinder and the piston to expand, some of the work done is used in overcoming the frictional force (released as heat). Now if you supply the same amount of heat which was released as work and try to compress the piston back to its original state, some of the work ...


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Start from the first principle of thermodynamics : $d U = \delta W + \delta Q$ where $\delta Q_{\text{rev}} = T dS$ so $\delta W_\text{rev} = d U - T d S$ hence at least for an isotherm, reversible work only depends on internal energy and entropy, both of which are state functions. So yes, in this particular case, reversible work between two states is a ...


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Reversible work is done by conservative forces and so doesn't depend on the path. Non-conservative forces like friction, generate irreversibility and in presence of those forces, we cannot have a reversible process. Hence, if we want to determine reversible work, we should remove all irreversibilities i.e. all non-conservative forces. In thermodynamics (and ...


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Is this taken to be an additional (and apparently implicit) assumption? You are correct. Take two arbitrary points $A,B$ on the $PV$ (or any other) plane, and draw an arbitrary curve connecting them: you have just defined a reversible transformation connecting $A$ and $B$. This is because every point in the $PV$ (or any other) plane represents an ...


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Because, dissipative forces convert some of the work to heat. If we want to have a reversible process we must be able to return system and its environment back to their initial states without any change in universe. For this purpose, we must extract heat from environment and convert whole of that to work and it is impossible due to the second law of ...


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So, you want to prove that between any arbitrary two states of a system, it exists at least one reversible path. You can prove this if you accept continuity of properties of substances. I.e. for example, if we have an ideal gas in equilibrium at initial state $(P_i,T_i)$ and final state $(P_f,T_f)$; then certainly there are infinite equilibrium states ...


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I believe that the answer can be made very simple. If the engine is reversible, its internal processes are, by definition of "reversibility", all quasistatic, regardless of the internal details of the machine. On the other hand, because the machine acts on the gas through a quasistatic process, then by definition this process must be reversible.


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A reversible process is characterized by a continuous sequence of thermodynamic equilibrium states for whatever system you are considering. So, for your system to experience a reversible process, its pressure and temperature must differ only slightly from that of its surroundings throughout the entire process. And there can be no spatial temperature or ...


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It depends on what you consider to be the system. If the system is the entire container, then there are no thermodynamic operations, quasistatic or not, on the system by the external environment. And as you said, the system is not in thermal equilibrium. If you talk about a thermodynamic operations you need to define a system and an environment, in this ...


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Net entropy change means entropy change of the world (world means system plus environment). Carnot cycle is a reversible cycle. For a reversible cycle, world entropy change is zero. Because both of system and environment return to their initial states when cycle is completed. But, for an irreversible cycle, world entropy change (net entropy change) isn't ...


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In a sense, you can consider chromatography to be a kind of 'filter', and it can certainly separate out coffee components, including water. There's no lack of other ways, though, to extract water. Freeze drying would carry the water vapor away, and you can sell the residue as instant coffee...


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Coffee is a homogeneous solution hence it can not be separated by usual methods. I have few suggestions may not be very accurate but good for brainstorming. Distillation : Although you have mentioned not to mention it but I would like to add that coffee has several aromatic organic compounds that make the smell of coffee hence you can not get rid of ...


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Does the fact that the engine exchanging heat with the gas is reversible imply that the trasformation of the ideal gas (no matter what trasformation is) is a reversible trasformation? Yes. A reversible engine is an engine that only performs reversible transformations. A reversible transformation in which two systems interact (the engine and the gas-...


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If you want to filter out the grains then certainly you could using normal filter papers in a filter funnel and repeat until the solution is clear of bits. You could also use a sintered glass filter. However, there will still be compounds from the coffee dissolved in the water and so a molecular sieve could be used and/or a chromatography column to separate ...


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I'd think of effusion effect, centrifugal techniques or other methods used for uranium enrichment. You could possibly select particles with the right mass (mass of the water) out of this mixture in a lot of iterations. It would be long and very expensive but probably possible.


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Turbine is a device with steady flow process and it has one inlet and one exit. First law of thermodynamics for a steady flow process with one inlet and one exit is as below $$q-w=(h_e-h_i)+\frac 12\left(V_e^2-V_i^2\right)+g(z_e-z_i)\;\tag 1$$ Potential energy change is negligible for this device. Usually, kinetic energy change isn't negligible but it seems ...


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A heat engine is reversible if all processes related to working system in that are reversible. So, in this case, the heat exchanging between system and hot source (ideal gas) must be reversible. On the other hand, we know that In a reversible process, the entropy of the world maintains constant I.e. $$\mathrm dS_{\textrm{world}}=0$$ In a reversible ...


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You want to read the classic paper by Richard Beth, Mechanical detection and measurement of the angular momentum of Light, Physical Review 50 115 (1936). Beth used bright circularly polarized light to drive a torsion pendulum in a vacuum chamber, and was able to observe torques due to circular polarization of order $10^{-16}\rm\,N\,m$. This was with a ...


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All of your claims are essentially true. The angular momentum of light, in both its orbital and spin varieties, is indeed angular momentum that can be transferred to matter to make it spin and give it the garden variety of mechanical angular momentum. This is well explained in the relevant Wikipedia section, with good references for experiments that show it. ...


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The reason we talk about heat engines in thermodynamics is that they capture the most general thermodynamic process possible with two bodies. Every transformation can be summed up as "object A put in/took out this much heat, object B put in/took out this much heat, this much work came in/out", which is what happens if you stick a heat engine between them. ...


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Is this true for every reversible cycle? Is the efficiency of all reversible cycle equal to the efficiency of a Carnot Cycle? Yes. They are indeed. The equality in Clausius' Inequality $$\oint \frac{đq_\textrm{sys}}{T_\textrm{source}}=0$$ is strictly valid for all reversible cycles. Temperature of a reversible engine is at all times equal to the ...


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Thank you to everyone who contributed to this thread. Now, at least the first question, has an answer. As clearly described at this related question, the answer to the first question is: IF adiabatic and isentropic THEN reversible. In other words, for adiabatic processes, "isentropic" and "reversible" are equivalent notions. As an aside, I notice here ...


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We know that: irreversible+adiabatic = $\Delta S>0$, thus, if $\Delta S=0$ the process is either: irreversible+non-adiabatic, reversible+adiabatic, or reversible+non-adiabatic. Then you can conclude that: $\Delta S=0$+adiabatic=reversible. Regarding you example of an irreversible adiabatic cycle: It is impossible and the example is flawed. The ...


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A process does not have to be adiabatic or reversible to be isentropic. If the difference in entropy between State A and State B is zero, since entropy is a state function, any oddball path (either reversible or irreversible) that can take you from State A to State B will be isentropic.


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Entropy is a state function. This means that in every cyclical transformation (i.e. a transformation in which the initial and final states are the same) we have $$\Delta S=0$$ That is to say: every cyclical transformation is isentropic. But clearly there are cyclical transformation that are neither adiabatic nor reversible. Let's take the following ...



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