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Thermodynamics of an Open System—C.E. Mungan, Summer Ref: JCE 89 , 968 (2012) 2012
focus on the open system consisting of the gas in the main cylinder only^ A^ classical monatomic ideal gas is indicated in yellow in the following diagram., excluding the amount in^ We will the small injector cylinder on the right-hand side. Other colors indicate regions of interaction with this system. The blue piston can be used to vary the volume V of the system by reversible adjusting the pressure. Normally this piston supports a fixed weight to maintain a constant pressure p , but if desired we can clamp the piston so that V is held constant, or we can add or subtract grains of sand to quasistatically change which normally maintain both the system and the gas in the injector at a constant temperature p. The red surfaces are thermal conductors, T , set by adjusting the electrical heater in the grey water bath. If desired, however vary the power supply to quasistatically transfer heat into or out of the ideal gas., we can slightly Alternatively, we can evacuate the entire bath, so that the gas will be adiabatic (thermally insulated). Finally the green vent can either be closed, sealing off the system from the injector, or it can be opened, to quasistatically vary the number piston. Normally, however, that piston will be adjusted so that the pressure in the injector N of gas atoms in the system by slightly adjusting the injector matches that in the system and there is no net flow of atoms one way or the other of the injector as allowing particle transfer into our open system, effectively controlled by the. We can think difference in chemical potential μ of the ideal gas on the two sides of the vent.
V , and number of atoms^ By choosing to express the internal ener N , we can write its difgy^ ferential in the usual way as U^ of our system in terms of its entropy^ S , volume
dU = TdS − pdV + μ dN. (1) Now in the figure above, we allowed for three kinds of interaction with the system. describe them in terms of mechanical work dW at the blue piston, heat dQ through the portions We will of the red sides that surround our system (i.e., excluding the heat transfer into the injector), and the energy dM associated with matter transfer through the green vent. The goal of the present exercise is to figure out how the three terms on the right-hand side of Eq. (1) relate to these three interaction energies.
grey water bath (to shut of^ We will sneak up on the answer by turning on one process at a time. Let’s first evacuate thef dQ ) and close the green vent (to shut off dM ). It is then clear that both closed system, as described by dS and dN are zero. We can however perform a reversible adiabatic compression of our pV γ (^) = constant where γ = 5 / 3. In this case we have
isothermal.^ dW^ =^ −^ pdV This time^. Next, let’s refill the water bath but leave it in its normal state, so our system is now dU and dN are zero. Now if we quasistatically increase p , we still do positive work transfer out of the gas in the form of a heat flow dW = − pdV on the system, but it is balanced by an equal and opposite ener dQ = TdS. Both dV and dS are negative. gy
T. This time we will also put the blue piston in its normal state, exerting a fixed pressure^ Finally we will open the green vent, while leaving the water bath in its normal state at fixed p. But we will quasistatically compress the injector piston. It is tempting to simply jump to the conclusion that dM = μ dN , dQ = TdS , and dW = − pdV but now we come to the big surprise that this conclusion is wrong! and see if you can spot the problem? Why don’t you think it over for a minute before reading further
heat that flows across the red walls surrounding our system is NOT^ Actually there’s nothing wrong with^ dW^ =^ −^ pdV^.^ The really big blooper is the only factor that changes^ dQ^ =^ TdS^.^ The the entropy of our system. from the injector. Once you realize that’s wrong, then you’ll also agree that There is ALSO an entropy flow from the particles entering the system dM = μ dN must be wrong, because we defined the quantities above such that dU = dQ + dW + dM. (2) [For what it’s worth, I follow Schroeder’s lead and call Eq. (2)—which is a statement of ener conservation—the “first law of thermodynamics.” But I call Eq. (1)—which is merely a gy mathematical statement of how to take a dif definitions of T , p , and μ in terms of derivatives offerential of 3 independent variables, combined with U —the “thermodynamic identity for internal ener think the present example nicely illustrates the difgy.” Many people confuse these two equations. But one is physics and one is merely math. Iference.]
piston all the way in and so transfer all of its contents to our system. Specifically^ A^ clear way to proceed is to actually compute everything of interest, if we push the injector, let’s spell out some initial conditions. Our system initially contains and volume V N i atoms in it, at pressure p , temperature T , with N atoms in it, at pressurei^ =^ N i kT^ /^ p^ where p , temperature^ k^ is Boltzmann’s constant; meanwhile, the injector starts out T , and volume V = NkT / p. The system ends up
