Irreversible Processes
Real processes are not reversible. They have friction and are carried out with fi nite driving forces. Such processes are irreversible processes. In an irreversible process, if the system is returned to its original state, the surroundings must be altered. The work obtained in an irreversible process is always less than that obtained in the idealization of a reversible process. Why?
To help solidify these abstract ideas, a concrete example is illustrative. We will compare the value of work for six processes. We will label these cases process A through process F. Three processes (A, C, and E) entail isothermal expansion of a piston– cylinder assembly between the same states: state 1 and state 2. The other three (B, D, and F) consist of the opposite process, isothermal compression between state 2 and state 1. An isothermal process results in the limit of fast heat transfer with the surroundings. We could perform a similar analysis on adiabatic processes where there is no energy transfer via heat between the system and the surroundings.
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2.3 Reversible and Irreversible Processes ◄ 49 Process A is illustrated in Figure 2.4. The system contains 1 mole of pure ideal gas.
A 1020-kg mass sits on the piston. The surroundings are at atmospheric pressure. The molar volume in state 1 can be found from the area 10.1 m22 and the height (0,4 m):
v15 ¢Az n ≤
1
50.04 Bm3 molR
Since the piston is originally at rest, the pressure inside the piston can then be found by a force balance:
P15mg
A 1Patm52 3bar4
where Patm is taken to be 1 bar. These two properties completely constrain the initial state. State 1 is labeled on the Pv diagram in Figure 2.4.
Process A is initiated by removing the 1020-kg mass. The pressure of the piston is now greater than that exerted by the surroundings, and the gas within the piston expands.
The expansion process continues until once again the pressures equilibrate. The piston then comes to rest in state 2 where the pressure is given by:
P25Patm51 3bar4
The ideal gas law can be applied to this isothermal process to give:
Pv5RT5const The volume of state 2 is then given by:
v25P1v1
P2 50.08 Bm3 molR
State 2 is now constrained by two independent, intensive properties and is also labeled in Figure 2.4.
To calculate the work, we must consider the “external” pressure upon which the gas must expand [see Equation (2.7) and discussion]. The piston only starts to move once the 1020-kg block is removed. Hence, at any volume larger than 0.04 m3, the external pres- sure is only that from the atmosphere. The path of external pressure vs. volume is illus- trated in Figure 2.4. Note, we are not saying anything about the pressure in the system but rather we are graphically illustrating what external pressure the gas expands against.
To fi nd the work, we apply Equation (2.7):
w5 23
v2
v1
PEdv5 2Patm3
v2
v1
dv5 24000 3J/mol4 (2.8) The same result can be found by graphically counting the area under the curve in the Pv diagram in Figure 2.4. The negative sign indicates we get 4 kJ of work out of the system from this expansion process.
In our analysis, we have idealized the process to stop precisely where the forces balance. In reality, the piston may undergo damped oscillations in its path toward its
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50 ► Chapter 2. The First Law of Thermodynamics
fi nal resting position in state 2. In that case, the kinetic energy of the piston causes it to overshoot its fi nal equilibrium position, leading to a molar volume greater than 0.08 3m3/mol4. Once past this volume, the pressure in the system will be less than the pressure of the surroundings. At some point, the motion stops and the piston reverses direction, back toward the equilibrium position. It may again overshoot the equilibrium position, leading to a molar volume less than 0.08 3m3/mol4, where it is then again turned around and expands, and so on and so on. This process will inevitably contain a fric- tional dissipative mechanism that causes the piston to come to rest at state 2. Since the external pressure is the same during these oscillations, the contribution of the oscillating expansions and contractions to the work will exactly cancel, leading to the same value as calculated by Equation (2.8).5,6 In this text, we will ignore the oscillatory behavior of these types of processes and approximate them in the simpler context where the system does not overshoot its fi nal equilibrium state. While it is only an approximation of the real behavior, this simplifi cation proves useful in allowing us to compare irreversible processes with reversible processes.
Figure 2.4 Schematic of an isothermal expansion process (process A). The corresponding plot of the process on a PEv diagram is shown at the bottom.
State 1
Weightless, frictionless piston
A = 0.1 m2
A = 0.1 m2
1 mol of pure, ideal gas
1 mol of pure, ideal gas m = 1020
kg
State 2 Process A:
Isothermal expansion
0.4 m
2
1
.04 1
2
Molar volume (m3/mol) .08
Work is given by the area under the curve
"external" pressure (bar) Process A
Patm
Patm
Fexternal/area or
5 We assume all the energy dissipation occurs within the system.
6 On the other hand, if there were no dissipative mechanism, the piston would oscillate forever. Its kinetic energy could be accounted for by the difference in force between the system pressure and the external pressure.
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2.3 Reversible and Irreversible Processes ◄ 51
We next want to calculate the work needed to compress the gas from state 2 back to state 1. This process is illustrated in Figure 2.5 and is labeled process B. In process B, we drop the 1020-kg mass back on the piston, originally in state 2. The external pressure now exceeds the pressure of the gas initiating the compression process. The piston goes down until the pressures equilibrate, at state 1. The external pressure vs. molar volume is plotted in Figure 2.5. In this case, the external pressure consists of contributions from both the block and the atmosphere. Again, we are not representing the system pressure in this graph, but rather the force per area that acts on the piston. The work is found by Equation (2.7):
w5 23
v1
v2
PEdv5 2aPatm1 mg Ab3
v1
v2
dv58000 c J mold
This value can also be found from the area under the curve. Comparing process A and process B, we see it costs us more work to compress the piston back to state 1 than we got from expanding it to state 2. The net difference in work 18000−400054000 3J4 2 in going from state 1 to state 2 and back to state 1 results in a “net effect on the surrounding.”
Figure 2.5 Schematic of an isothermal compression process (process B). The corresponding plot of the process on a PEv diagram is shown at the bottom.
State 2 State 1
0.8 m
0.4 m
2
1
.04
Molar volume (m3/mol) .08 Process B 1
2 A = 0.1 m2
A = 0.1 m2 1 mol of pure,
ideal gas
1 mol of pure, ideal gas m = 1020
kg
m = 1020 kg Process B:
Isothermal compression
Work is given by the area under the curve
"external" pressure (bar) Patm
Patm
Fexternal/area or
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52 ► Chapter 2. The First Law of Thermodynamics
Examining our defi nition of a reversible process, we see that processes A and B are irreversible.
Next we again consider expansion from state 1 to 2 (process C) and compression from state 2 to 1 (process D), but now we use two 510-kg masses instead of one larger 1020-kg mass. The expansion is carried out as follows: The system is originally in state 1 when the fi rst 510-kg mass is removed. This causes the gas to expand to an intermediate state at a pressure of 1.5 bar and a molar volume of 0.053 m3/mol. The second 510-kg mass is then removed, completing the expansion to state 2. These three states are shown in the Pv diagram in Figure 2.6. The expansion process is labeled process C and follows arrows from state 1 to the intermediate state to state 2. The work the system delivers to the surroundings is given by:
w5 23
v2
v1
PEdv5 2BaPatm1
m 2g
Ab3
vint
v1
dv1Patm3
v2
vint
dvR 5 24667 B J molR Again, the work can be found graphically from the area under the curve. Process C is
“better” than process A in that it allows us to extract more work from the system. Of course, we want to get the most work out of a system as possible.
The compression process is the opposite of the expansion. With the system in state 2, a 510-kg block is placed on the piston until it compresses to the intermediate state, followed by placement of the second block. The work is found by:
w5 23
v1
v2
PEdv5 2BaPatm1
m 2g
Ab3
vint
v2
dv1 aPatm1 mg A b3
v1
vint
dvR56667 B J molR In analogy to the expansion process, process D is “better” than process B in that it costs us less work to compress the system back to state 1. When we have to put work into a system, we want it to be as small as possible. However, it still costs us more work to compress from state 2 to state 1 than we get out of the expansion, so these processes are still irreversible.
We did “better” in both expansion and compression processes when we divided the 1020-kg mass into two parts. Presumably we would do better by dividing it into four parts, and even better by dividing it into eight parts, and so on. If we want to do the best possible, we can divide the 1020-kg mass into many “differential” units and take them off one at a time for expansion or place them on one at a time for compression. These pro- cesses are labeled process E and process F, respectively, and are illustrated in Figure 2.7.
At each differential step, the system pressure is no more than 'mg/A different than the external pressure. Thus, to a close approximation:
P5PE
The process paths are illustrated in the Pv diagram in Figure 2.7. To fi nd the work, we integrate over the external pressure. However, since the external pressure is equal to the system pressure, we get:
w5 23
v2
v1
PEdv5 23
v2
v1
Pdv5 23
v2
v1
P1v1
v dv5 2P1v1 ln ¢v2
v1≤ 5 25545 c J
mold (2.9)
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2.3 Reversible and Irreversible Processes ◄ 53
Similarly the work of compression is:
w5 23
v1
v2
PEdv5 23
v1
v2
Pdv5 23
v1
v2
P1v1
v dv5 2P1v1 ln ¢v1
v2≤ 55545 c J mold In processes E and F, we can return to state 1 by supplying the same amount of work that we got from the system in the expansion process. Hence, we can go from state 1 to 2 and back to state 1 without a net effect on the surroundings. From our defi nition, we see that these processes are reversible. In a reversible process, we are never more than slightly out of equilibrium. At any point during the expansion, we could turn the process around the other way and compress the piston by adding differential masses instead of removing them. Moreover, we get more work out of the reversible expansion than the irreversible expansions. Similarly, the reversible compression costs us less work than the irreversible processes. The reversible case represents the limit of what is possible in the real world—it Figure 2.6 Schematic of two-step isothermal expansion (process C) and compression (process D) processes. The corresponding plots of the process on a PEv diagram is shown at the bottom.
2
1
.04 .08
1
2
0.8 m
0.4 m
Molar volume (m3/mol) Process C
Process D
A = 0.1 m2
A = 0.1 m2
1 mol of pure, ideal gas 1 mol of pure,
ideal gas m = 510
kg
m = 510 kg
m = 510 kg
m = 510 kg
Process D:
Isothermal compression Process C:
Isothermal expansion
Work is given by the area under the curve
"external" pressure (bar) Patm
Patm
Fexternal/area or
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54 ► Chapter 2. The First Law of Thermodynamics
gives us the most work we can get out or the least work we have to put in! Moreover, only in a reversible process can we substitute the system pressure for the external pressure.
Why do we get less work out of the irreversible expansion (process A) than the reversible expansion (process E)? Work is the transfer of energy between the system and the surroundings. In this case, as the gas molecules bounce off the piston, their change in net z momentum between before and after a collision is determined by the movement of the piston. This change of momentum with time represents the net energy transferred between the system and surroundings; that is, it is the work. The irreversible expansion, process A, never has a mass on the piston; thus the molecules of the gas are running into something “smaller” and will not transfer as much energy. On the other hand, in the irreversible compression process, the mass on the piston is larger than the corresponding reversible process. It therefore imparts more energy to those molecules and costs more work. Those of you who are baseball fans may consider an analogy to the size of a hitter’s bat. The heavier the bat, the more energy can be transferred to the baseball. The greater the force the piston exerts on a given molecule as it rebounds off the piston, the more the molecule’s speed will increase and the higher its kinetic energy. If we sum up all the molecules, we see that the net energy transfer (work) is greater.
Figure 2.7 Schematic of infinitesimal-step, reversible isothermal expansion (process E) and com- pression (process F) processes. The corresponding plots of the processes on a PEv diagram is shown at the bottom.
2
1
1
2
.04 .08
∂m
∂m
0.8 m
0.4 m
Molar volume (m3/mol) Process E
A = 0.1 m2
A = 0.1 m2
1 mol of pure, ideal gas 1 mol of pure,
ideal gas
Process F:
Isothermal compression Process E:
Isothermal expansion
"external" pressure (bar)Fexternal/area or
PE = P Process F
mT= 1020 kg mT= 1020 kg
Patm
Patm
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Efficiency
We can compare the amount of work required in an irreversible process to that of a reversible process by defi ning the effi ciency factor, h. For an expansion process, we compare how much work we actually get out to the idealized, reversible process. Thus, the effi ciency of expansion, hexp, is:
hexp5 wirrev
wrev (2.10)
For example, the effi ciency of Process A would be:
hexp5 wA
wE5 24000 2554550.72
where wi represents the work of process i. Thus, we say process A is 72% effi cient; in comparison, the two-block process C is 84% effi cient. We have done better.
On the other hand, to determine the effi ciency of a compression process, hcomp we compare the reversible work to the work we must actually put in:
hcomp5 wrev
wirrev (2.11)
For example, the effi ciency of process B would be:
hcomp5 wF
wB 5 5545 800050.69
or 69% effi cient. In both cases, effi ciencies are defi ned so that if we can operate a process reversibly, we would have 100% effi ciency, while the real processes are less effi cient.
One strategy for actual, irreversible processes is to solve the problem for the idealized, reversible process and then correct for the irreversibilities using an assigned effi ciency factor.