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2.15 Reversibility

In the experiment presented in Section 2.14, it is impossible to reverse the direction of heat flow without a gross change in the relative temperatures, Th and TC. It is an irreversible process.

Let us examine another case (suggested by Professor D. Baganoff of Stanford University) designed to illustrate reversibility:

Consider again an adiabatic cylinder with a frictionless piston. It contains 10 m3 of an ideal monatomic gas (y = 1.67) under a pressure of 100 kPa at 300 K. We will follow the behavior of the gas during compression and subsequent expansion by means of the p versus V and the T versus V diagrams in Figure 2.5. The initial state is indicated by "1" in both graphs.

When the piston is moved so as to reduce the volume to 2 m3, the pressure will rise along the adiabatic line in such a way that the pVY product remains constant. When State 2 is reached, the pressure will have risen to 1470 kPa and the temperature to 882 K. The work done during the compression is J pdV between the two states and is proportional to the area under the curve between 1 and 2.

If the piston is allowed to return to its initial position, the gas will expand and cool, returning to the initial state, 1. The process is completely reversible. Now assume that inside the cylinder there is a solid object with a heat capacity equal to that of the gas. If we compress the gas rapidly and immediately expand it, there is no time for heat to be exchanged between the gas and the solid, which will remain at its original temperature of 300 K. The process is still reversible. However, if one compresses the gas and then waits until thermal equilibrium is established, then half of the heat generated will be transferred to the solid and the gas will cool to (882 + 300)/2 = 591 K, while its pressure falls to 985 kPa according to the perfect gas law. This is State 3.

When the gas expands again to 10 m3, it cools down to 201 K and the pressure falls to 67 kPa (State 4). Immediately after this expansion, the solid will be at 591 K. Later, some of its heat will have been transferred Figure 2.5 Pressure-volume and temperature-volume diagrams for the experiment described in the text.

back to the gas whose temperature rises to 396 K and the pressure will reach 132 kPa (5). The gas was carried through a full cycle but did not return to its initial state: its temperature (and internal energy) is now higher than initially. The increase in internal energy must be equal to the work done on the gas (i.e., it must be proportional to the shaded area in the upper portion of Figure 2.5, which is equal to the area under curve 1 to 2 minus that under curve 3 to 4. The process is irreversible.

What happens if the compression and the expansion are carried out infinitely slowly? Does the process become reversible? You will find that it does when you do Problem 2.2. An electric analogy may clarify the situation. When a real battery (represented by a voltage source, V, with an internal resistance, R) is used to charge an ideal energy accumulator, part of its energy will dissipate as heat through the internal 12R losses, leaving only part to reach the accumulator. Clearly, the relative loss decreases as the current decreases—that is, as the charge time increases. If the energy is transferred from the battery to the accumulator infinitely slowly (I ^ 0), there are no losses. The system is reversible in the sense that all the energy transferred to the accumulator can later be returned to the battery.

2.15.1 Causes of Irreversibility

Among the different phenomena that cause thermodynamic processes to become irreversible one can list the following.

### 2.15.1.1 Friction

Of all the causes of irreversibilities, friction is perhaps the most obvious. For example, in the cylinder-piston case, if some energy is lost by friction during compression, it is not returned during expansion; on the contrary, additional losses occur during this latter phase.

2.15.1.2 Heat Transfer across Temperature Differences

Consider a metallic wall separating a source of heat—say a flame—from the input of a heat engine. All the heat, Q, absorbed from source is transmitted without loss through the wall, yet for this heat to flow there must be a temperature difference across the wall. The source side is at T1, the engine side is at T2, and T1 must be larger than T2. The entropy on the source side is Q/T1, and on the engine side, it is Q/T2, which is, of course larger than Q/T1. So, in passing through the wall, the entropy was increased—the heat became less "noble" on the engine side.

In Chapter 3, we will show that the maximum efficiency of a heat engine is n = ThtHT° . If the engine could have operated without the wall, its efficiency could have reached Ti[TC and would be larger than when operated on the other side of the wall when it would be limited to T2[TC.

2.15.1.3 Unrestrained Compression or Expansion of a Gas

In the subsection on adiabatic processes, we dealt with an example of abrupt expansion of a gas and found that it led to irreversibilities. 