Thermodynamics and Propulsion

4.4 Muddiest Points on Chapter 4

MP 4..1   Why is the ability to do work decreased in B? How do we know?

In state A, the energy is in organized form and the molecules move along circular paths around the spinning flywheel. We could get work out this system by using all of the kinetic energy of the flywheel and for example lift a weight with it. The energy of the system in state B (flywheel not spinning) is associated with disorganized motion (on the molecular scale). The temperature in state B is higher than in state A. We could also extract work from state B by running for example an ideal Carnot cycle between $T_B$ and some heat reservoir at lower temperature. However the work we would get from this ideal Carnot cycle is less than the work we get from state A (all of the kinetic energy), because we must reject some heat when we convert heat into work (we cannot convert heat into 100% work). Although the energy of the system in state A is the same as in state B (we know this from 1^st law) the ``organization'' of the energy is different, and thus the ability to do work is different.

MP 4..2   With the isothermal reversible expansion, is $P_\textrm{external}$ constant? If so, how can we have $P_\textrm{system}\approx P_\textrm{external}$ ?

For a reversible process, if the external pressure were constant, there would need to be a force that pushed on the piston so the process could be considered quasi-equilibrium. This force could be us, it could be a system of weights, or it could be any other work receiver. Under these conditions the system pressure would not necessarily be near the external pressure but we would have $P_\textrm{system} \cong P_\textrm{external} + F_\textrm{work receiver}/A_\textrm{piston}$ . We can of course think of a situation in which the external pressure was varied so it was always close to the system pressure, but that is not necessary.

MP 4..3   Why is the work done equal to zero in the free expansion?

In this problem, the system is everything inside the rigid container. There is no change in volume, no ``$dV$ ,'' so no work done on the surroundings. Pieces of the gas might be expanding, pushing on other parts of the gas, and doing work locally inside the container (and other pieces might be compressed and thus receive work) during the free expansion process, but we are considering the system as a whole, and there is no net work done.

MP 4..4   Is irreversibility defined by whether or not a mark is left on the outside environment?

A process is irreversible when there is no way to undo the change without leaving a mark on the surroundings or ``the rest of the universe.'' In the example with the bricks, we could undo the change by putting a Carnot refrigerator between the bricks (both at $T_M$ after putting them together) and cooling one brick down to $T_L$ and heating the other brick to $T_H$ to restore the initial state. To do this we have to supply work to the refrigerator and we will also reject some heat to the surroundings. Thus we leave a mark on the environment and the process is irreversible.

MP 4..5   Is heat transfer across a finite temperature difference only irreversible if no device is present between the two to harvest the potential difference?

If we have two heat reservoirs at different temperatures, the irreversibility associated with the transfer of heat from one to the other is indeed dependent on what is between them. If there is a copper bar between them, all the heat that comes out of the high temperature reservoir goes into the low temperature reservoir, with the result given in Section 5.5. If there were a Carnot cycle between them, some (not all) heat from the high temperature reservoir would be passed on to the low temperature reservoir, the process would be reversible, and work would be done. The extent to which the process is irreversible for any device can be assessed by computing the total entropy change (device plus surroundings) associated with the heat transfer.