DU 5 2088 J This is an example where the change in enthalpy does not equal the change in internal
3.4 Entropy and the Second Law of Thermodynamics
We define entropy, S, as an additional thermodynamic state function. The infinitesi- mal change in entropy, dS, is defined as
dS5dqrev
T (3.12)
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where “rev” on the infinitesimal for heat, dq, specifies that it must be the heat for a reversible process. The temperature, T, must be in kelvins. Integrating equation 3.12, we get
DS53dqrev
T (3.13)
where DS is now the change in entropy for a process. As indicated in the previous section, for the Carnot cycle (or any other closed cycle) DS must be zero.
For an isothermal process, the temperature can be taken out of the integral and the integral can be evaluated easily:
DS5 1
T3dqrev5qrev
T (3.14)
Equation 3.14 demonstrates that entropy has units of J/K. These may seem like unu- sual units, but they are the correct ones. Also, keep in mind that the amount of heat for a process depends on the amount of material, in grams or moles, and so some- times the unit for entropy becomes J/mol#K. Example 3.2 shows how to include amount in the unit.
ExamplE 3.2
What is the change in entropy when 1.00 g of benzene, C6H6, boils reversibly at its boiling point of 80.1°C and a constant pressure of 1.00 atm? The heat of vaporiza- tion of benzene is 395 J/g.
Solution
Because the process occurs at constant pressure, DvapH for the process equals the heat, q, for the process. Because vaporization is an endothermic (that is, energy-in) process, the value for the heat is positive. Finally, 80.1°C equals 353.2 K. Using equation 3.14:
DS5 1395 J
353.2 K5 11.12 J K
for 1 g of benzene. Because this represents the entropy change for 1 g of benzene, we can also write this DS as 11.12 J/g?K. The entropy of the system—the benzene—
is increasing in this example.
Other cyclic processes having different steps or conditions can be defined. How- ever, it has been found that no known process is more efficient than a cycle that is defined in terms of reversible steps. This means that any irreversible change is a less efficient conversion of heat to work than a reversible change. So, for any arbitrary process:
earb#erev
where earb is the efficiency for that arbitrary cycle and erev is the efficiency of a re- versible cycle. If the arbitrary process is a reversible cycle, then the “equals” part of the sign applies. If the cycle is an irreversible cycle, the “less than” part of the sign applies. Substituting for efficiency:
11qout,arb
qin,arb #11 q3,rev
q1,rev qout,arb
qin,arb # q3,rev q1,rev
Remember that DH 5 qp. Temperatures must be in kelvins.
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where the 1s have canceled. The fraction on the right is equal to 2Tlow/Thigh, as demonstrated earlier. Substituting:
qout,arb
qin,arb # 2Tlow Thigh and rearranging:
qout,arb
qin,arb 1 Tlow
Thigh#0
This equation can be rearranged to get the heat and temperature variables that are as- sociated with the two reservoirs into the same fractions (that is, qin with Thigh and qout with Tlow). It is also convenient to relabel the temperatures and/or the heat values to emphasize which steps of the Carnot cycle are involved. Finally, we will drop the “arb”
designation. (Can you reproduce these steps?) The above expression thus simplifies to q3
T31 q1 T1 #0
For the complete cycle of many steps, we can write this as a summation:
all stepsa qstep Tstep#0
As each step gets smaller and smaller, the summation sign can be replaced by an integral sign, and the above expression becomes
3dq
T #0 (3.15)
for any complete cycle. Equation 3.15 is one way of stating what is called Clausius’s theorem, after Rudolf Julius Emmanuel Clausius, a Pomeranian (now part of Poland) and German physicist who first demonstrated this relationship in 1865.
Consider, then, the two-step process illustrated in Figure 3.3, where an irrevers- ible step takes a system from a set 1 of conditions to a set 2 of conditions, and then a reversible step takes it back to the original conditions. As a state function, the sum of the steps equals the overall change for the entire process. But from equation 3.15, the overall integral’s value must be less than zero. Separating the integral into two parts:
3
2
1
dqirrev T 13
1
2
dqrev T ,0
The expression inside the second integral is, by the definition in equation 3.12, dS.
If we reverse the limits on the second integral (so both terms refer to the same proc- ess going in the same, not opposite, directions), it becomes 2dS. We therefore have
3
2
1
dqirrev T 13
2
1
(2dS),0
Figure 3.3 A representation of a process that has an irreversible step. See text for discussion. Most real processes can be described like this, giving entropy a meaningful place in the understanding of real processes.
System with intermediate set of conditions
(p2, V2, T2) System with initial
set of conditions (p1, V1, T1)
Step 1: IRREVERSIBLE
Step 2: REVERSIBLE
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or
3
2
1
dqirrev T 23
2
1
dS,0 The integral of dS is DS, so for this step we have
3
2
1
dqirrev
T 2 DS ,0
3
2
1
dqirrev T , DS
Reversing and generalizing for any step, we simply remove the specific limits:
DS.3dqirrev
T (3.16)
If we want to keep this in terms of infinitesimals (that is, without integral signs) as well as include the original definition of dS from equation 3.12, this becomes
dS$ dq
T (3.17)
where again the equality is applicable to reversible processes, and the inequality is applicable to irreversible processes.
But consider that a spontaneous process is an irreversible process. Sponta- neous processes will occur if they can. With that in mind, we have the following generalizations:
dS.dq
T for irreversible, spontaneous processes dS5 dq
T for reversible processes Equation 3.17 also implies
dS, dq
T not allowed
The last statement is particularly important: The infinitesimal change in S will not be less than dq/T. It may be equal to or greater than dq/T, but it will not be less than that.
Consider, then, the following description. A process occurs in an isolated system.
Under what conditions will the process occur? If the system is truly isolated (there is no transfer of energy or matter between system and surroundings), then the pro- cess is adiabatic, because isolation implies that q50, and by extension dq50.
Therefore, dq/T is equal to zero. We can therefore revise the above statements:
dS.0 if the process is irreversible and spontaneous dS50 if the process is reversible
dS,0 is not allowed for a process in an isolated system
We conceptually collect the above three statements into one, which is the second law of thermodynamics:
For an isolated system, if a spontaneous change occurs, it occurs with a concurrent increase in the entropy of the system.
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If a spontaneous change does occur, entropy is the sole driving force for that change because both q and w are zero—and therefore DU is zero—under the stated conditions.
So far, we have focused on the entropy change of an isolated system and can properly label the DS as DSsys. What about the entropy change of nonisolated systems, and the concurrent change in the entropy of the surroundings, labeled DSsurr? We suggest that, parallel to equation 3.12, the entropy change of the sur- roundings is
dSsurr;dqsurr Tsurr
Typically, the surroundings are much larger than the system of interest, and the temperature of the relatively large surroundings can be taken as constant. Thus, this equation integrates to
DSsurr5 qsurr
Tsurr (3.18)
Finally, we submit that the total entropy change in the universe, DSuniv, equals the sum of the entropy change of the system and the surroundings:
DSuniv 5 DSsys 1 DSsurr (3.19) Now we consider two cases, a reversible change and an irreversible change.
For a reversible change, DSsurr equals 2DSsys, so the overall DSuniv equals zero.
However, for an irreversible change, DSsurr always has a larger magnitude than 2DSsys, and the overall DSuniv will always be greater than zero. The following examples illustrate.
ExamplE 3.3
A 1.00-g sample of H2O(g) condenses reversibly at 100.0°C, releasing 2260 J of heat into the surroundings. Determine DSsys,DSsurr,and DSuniv.
Solution
The entropy change for the system can be calculated using equation 3.14:
DSsys5 qrev
T 5 22260 J 373.1 K DSsys526.06 J
K
To determine the entropy change of the surroundings, we need to know what tem- perature the surroundings have. However, if this change is truly reversible, the tem- perature of the surroundings must also be 100.0°C! Thus, we have
DSsurr5qsurr
T 5 12260 J 373.1 K DSsurr516.06 J
K The sum of these two entropy changes is DSuniv:
DSuniv 5 26.06 J
K 1 6.06 J K 5 0 The heat q is negative because heat is
being lost by the H2O(g) in the system.
Here q is positive because heat is going into the surroundings.
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Contrast this result with the following irreversible process.
ExamplE 3.4
A 1.00-g sample of H2O(g) at 100.0°C with a pressure of 1.00 atm has a volume of 1.70 L. It is isothermally compressed by an external pressure of 2.00 atm until it reaches a volume of 0.850 L. Determine DSsys,DSsurr,and DSuniv. Assume ideal gas behavior.
Solution
Because the process is isothermal, DU 5 0, so q 52w. The work can be calcu- lated by
w 52pextDV
5 2(2.00 atm)(0.850 L21.70 L)a101.32 J L#atm b w 5 172 J
Therefore, q 52172 J. Because this heat went into the surroundings, we have qsurr 5 172 J, so we can calculate DSsurr:
DSsurr5 172 J
373.1 K 5 10.461 J K
However, the DSsys must be calculated using the reversible q, and this process is not reversible. But, because DS is a state function, we can use whatever process(es) we need to go from initial conditions to final conditions. If the process were reversible, we could calculate the work performed by
w5 2nRT ln Vf Vi The reversible work, under the given conditions, is
w5 2° 1.00 g 18.02 g
mol
¢ a8.314 J
mol#Kb1373.1 K2 ln 0.850 L 1.70 L w 5 119 J 5 2qrev
Thus, qrev 5 2119 J. Now calculating the entropy change of the system:
DSsys5 2119 J
373.1 K5 20.319 J K Now calculating DSuniv:
DSuniv520.319 J
K1 0.461 J K DSuniv510.142 J
K
Thus, for this irreversible change, we see that the overall entropy of the universe has gone up.
Because the change is isothermal, the temperature of the surroundings is also 100.0°C or 373.1 K.
This is the expression for reversible work with an ideal gas from Chapter 2.
The process is isothermal, so DU 5 0 and w 52q.
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86 Chapter 3 | The Second and Third Laws of Thermodynamics
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Like the first law of thermodynamics, the second law of thermodynamics can be expressed in terms of the system and surroundings interacting, a much more useful form. It is:
for any spontaneous (that is, irreversible) change, the entropy of the universe increases.
Because spontaneous processes are occurring all the time, the entropy of the uni- verse is constantly increasing.