THE SECOND LAW THERMODYNAMICS
There are several ways in which the second law of thermodynamics can be stated. Listed below are three that are often encountered. As described in class (and as derived in almost every thermodynamics textbook), although the three may not appear to have much connection with each other, they are equivalent.
- No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of this heat into work. [Kelvin-Planck statement of the second law]
Figure 5.1: This is not possible (Kelvin-Planck) |
- No process is possible whose sole result is the transfer of heat from a cooler to a hotter body. [Clausius statement of the second law]
- There exists for every system in equilibrium a property called entropy, , which is a thermodynamic property of a system. For a reversible process, changes in this property are given by
The entropy change of any system and its surroundings, considered together, is positive and approaches zero for any process which approaches reversibility.
For an isolated system, i.e., a system that has no interaction with the surroundings, changes in the system have no effect on the surroundings. In this case, we need to consider the system only, and the first and second laws become:
For an isolated system the total energy ( ) is constant. The entropy can only increase or, in the limit of a reversible process, remain constant.
The limit, or , represents the best that can be done. In thermodynamics, propulsion, and power generation systems we often compare performance to this limit to measure how close to ideal a given process is.
All of these statements are equivalent, but 3 gives a direct, quantitative measure of the departure from reversibility.
Entropy is not a familiar concept and it may be helpful to provide some additional rationale for its appearance. If we look at the first law,
the term on the left is a function of state, while the two terms on the right are not. For a simple compressible substance, however, we can write the work done in a reversible process as , so that
Two out of the three terms in this equation are expressed in terms of state variables. It seems plausible that we ought to be able to express the third term using state variables as well, but what are the appropriate variables? If so, the term should perhaps be viewed as analogous to where the parentheses denote an intensive state variable and the square brackets denote an extensive state variable. The second law tells us that the intensive variable is the temperature, , and the extensive state variable is the entropy, . The first law for a simple compressible substance in terms of state variables is thus
(5..1) |
Because Eq. 5.1 includes the second law, it is referred to as the combined first and second law. Because it is written in terms of state variables, it is true for all processes, not just reversible ones.
We summarize below some attributes of entropy:
- Entropy is a function of the state of the system and can be found if any two properties of the system are known, e.g. or or .
- is an extensive variable. The entropy per unit mass, or specific entropy, is .
- The units of entropy are Joules per degree Kelvin (J/K). The units for specific entropy are J/K-kg.
- For a system, , where the numerator is the heat given to the system and the denominator is the temperature of the system at the location where the heat is received.
- for pure work transfer.
Clausius inequality
Let system I be an arbitrary thermodynamic system, which is able to exchange heat with thermal reservoirs R1 and R2, having the temperatures T1 and T2 respectively. In this derivation we will not specify which of these two reservoirs is a heater and which is a cooler. We will consider the heat lost by a thermal reservoir to be positive and the heat gained by a reservoir to be negative. This will allow formulating the final result symmetrically with respect to both reservoirs.
Let the system I perform a cyclic process – reversible or irreversible – during which it received heat Q1 from reservoir R1 and Q2 from reservoir R2. Since the system has returned to its initial state the total amount of heat gained by the system Q1+Q2 is equal to the work it has performed.
Lets now take a Carnot engine and make it work using the same heat reservoirs R1 and R2. In order to the presence of the Carnot engine not to influence the amounts of heat Q1 and Q2 the system I has received from the heat reservoirs during its cycle, we can run the Carnot engine after the system I has completed its cycle. If at this point we thermally isolate the system I, the reservoirs R1 and R2 will exchange heat with the Carnot engine only. The presence of the Carnot engine does not affect the cyclic process in the system I since it has already finished by the time the Carnot engine was introduced.
Let the Carnot engine perform a cyclic process, during which it receives heat Q1’ from reservoir R1 and Q2’ from R2 respectively. It is essential for the further derivation to note that the Carnot engine is reversible. It can work either as an engine or a refrigerator. Apart from that the isotherm 12 in Carnot cycle can be taken as short as desired (see Fig. 1). Therefore, the work done by the Carnot engine can be as small as desired. One can also make the Carnot engine perform as much work as desired via performing many cycles. Thus the Carnot engine can perform any specified amount of positive or negative work. This allows choosing the value of either Q1’ or Q2’. It is always possible to make one of these heats have a chosen positive or negative value.
Figure 1. Carnot cycle
According to Carnot theorem
(Q1’/T1) + (Q2’/T2) = 0
Let us combine the system I and the Carnot engine into one complex system. The cycles performed consecutively by the system I and Carnot engine can be combined into one cyclic process. During this process the combined system
has received the heat Q1 + Q1’ from reservoir R1,
has received the heat Q2 + Q2’ from reservoir R2,
has performed the work W = Q1 + Q1’ + Q2 + Q2’.
Let us choose Q1’ so that Q1’ = -Q1 then
Q2’ = T2 (Q1/T1).
As a result of the cycle the state of the reservoir R1 remains unchanged. The reservoir R2 looses the amount of heat
Q2 + Q2’ = Q2 + t2 (Q1/T1) = T2 [(Q1/T1) + (Q2/T2)]
An equivalent work W = Q2 + Q2’ has been performed using this amount of heat. If this work were positive, we would have the Thomson-Plank process, which is forbidden by the II law of thermodynamics. Therefore, the work should be negative or equal to zero (W ≤ 0). Since the absolute temperature T2 is always positive this leads to
(Q1/T1) + (Q2/T2) ≤ 0.
This inequality is called Clausius inequality.
Clausius inequality and an efficiency of a heat engine
Let T1 be a temperature of a heater,
T2 – a temperature of a cooler,
we will consider the heat Q2 to be positive if the cooler receives it.
Then the Clausius inequality will read
(Q1/T1) – (Q2/T2) ≤ 0
or
(Q1-Q2)/Q1 ≤ (T1-T2)/T1.
On the left hand side of the last inequality we have the efficiency of a heat engine. Therefore, we have shown that a efficiency of any heat engine cannot be higher than of an ideal engine performing the Carnot cycle with the same temperatures of the heater and the cooler.