When two chemicals are mixed together do they react? This is an existential question for a synthetic chemist. Not much chemistry is performed without making products. From the Zeroth and First Laws of Thermodynamics we have defined a state variable (temperature T) and two state functions (internal energy U and enthalpy H) that we know exist and are rather fundamental. Certainly, temperature changes upon mixing and subsequent reaction are indicative of chemical changes. The temperature increases when an exothermic reaction occurs in a thermos flask and decreases during the course of an endothermic reaction. We quantify the changes in U and H by measuring temperature changes at either constant volume or constant pressure.

We have an instinctive feeling that systems tend to lower their energy. Raise a coffee mug one meter off the ground. Let go. The result is inevitable. The system (ceramic plus finely brewed dark roasted aqueous extract of Arabica beans) spontaneously falls to the ground to lower its energy. Gravitational potential energy is converted into translational energy as the system accelerates toward the floor. Then a fraction of this energy is converted into vibrations and sound waves, while another portion is used to overcome the cohesive energy of the ceramic as the mug strikes the floor. There must be a chemical potential – a concept introduced by J. Willard Gibbs much like the gravitational potential – that indicates which direction is ‘downhill’ for chemical reactions.

Does U or H act as this chemical potential? This was one of the first arguments used by chemists to explain reactivity. Very often it is the case. Many reactions that proceed spontaneously are exothermic. Add H₂SO₄ to water. It dissolves with great evolution of heat – remember always add acid to water not water to acid when diluting from a concentrated solution. The ΔU and ΔH values for dissolving H₂SO₄ in water are negative. The system moves toward a state of lower energy. Similarly, when KCl is added to water, it dissolves spontaneously. And not just a little bit. Over 35 g of KCl can be dissolved in 100 g of water at room temperature. Nonetheless, the temperature drops upon the dissolution of KCl. The reaction is endothermic, ΔH is positive. Take a beaker full of water, it will remain in the liquid phase at atmospheric pressure and room temperature once it has established its equilibrium vapor pressure above the liquid. But if we add heat to the system so that the temperature reaches the boiling point, the water spontaneously evaporates. Again, the process is spontaneous even though it is endothermic.

‘Sometimes’ is not good enough when establishing a law of Nature. Sometimes, spontaneous reactivity corresponds to minimization of ΔH, but not always. Sometimes, an equilibrium that favors the reactants is established when a process exhibits a positive ΔU, but not always. Therefore, neither ΔU nor ΔH plays the role of the chemical potential. We need to discover one or more new state functions along the way to finding what truly is the chemical potential. The Zeroth and First Laws are insufficient for determining the direction of spontaneous change. They do not determine when a reaction will occur. They are also insufficient to determine whether an equilibrium exists and whether that equilibrium favors the reactants or products.

We need a new state function that describes the evolution of chemical and physical parameters. Some systems react completely, others establish equilibria. Some compounds dissolve completely, others establish reagent-favored equilibria. But even in systems with minimal or no chemical interactions, there are spontaneous changes that occur. We always observe mixing of gases upon opening a valve. Once mixed, ideal gases do not spontaneously unmix even though such a process conserves energy. Once a body or solution is brought into thermal equilibrium at a uniform temperature, spontaneous temperature polarization into hot and cold regions does not occur even, though it conserves energy. There must be another parameter, another state function that remains to be discovered that will unlock the key to describing spontaneity. That state function, as we shall see, is entropy and unlocking it will require the Second Law of Thermodynamics.

8.1 The second law of thermodynamics

The First Law is a statement of energy conservation. However, it does not predict the spontaneous direction of change. It does not explain why heat spontaneously flows from a hot to a cold object, or why a gas fills all available space in a container. In our introduction to statistical mechanics, we defined the state function called entropy. Entropy serves as a measure of the number of microstates accessible to a macroscopic system at a given energy. For an isolated system, equilibrium is achieved when entropy has a maximum value. However, if we do not have direct access to quantum states and their relative populations, how can we detect their influence?

Thermochemistry gives us direct access to the energy changes that occur during chemical and physical processes. We will now show that thermochemistry can be used to define and monitor entropy changes. We will establish that the thermodynamic and statistical mechanical definitions of entropy are equivalent. The distribution of population among energy levels is responsible for changes in entropy. Why is all this importance laid at the feet of entropy, energy levels and populations? Because quantifying entropy changes allows us to predict the spontaneous direction of change for a process.

The history of the development of the concept of entropy is long and fraught with the human complications that can accompany the proposal of imaginative ideas and their confirmation into the canon of accepted concepts. Carnot established the fundamental idea of entropy while studying heat engines. His work went largely unrecognized, in part because he described the idea literally not mathematically. Clapeyron developed his conception of entropy in obtaining mathematical relationships to describe the temperature dependence of vapor pressure. His reanimation of the concept allowed Kelvin and Clausius to refine the thermodynamic definition of entropy, and to restate its importance in the form that is now known as the Second Law of Thermodynamics.

Kelvin–Planck statement of the Second Law:

No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.

Clausius statement of the Second Law:

It is impossible for an engine to perform work by cooling a portion of matter to a temperature below that of the coldest part of the surroundings. Heat does not flow spontaneously out of a cold reservoir and into a hot reservoir.

In other words, no engine is 100% efficient. Great, but we are interested in the effect of entropy on chemical systems, not heat engines. The Second Law is particularly important since it tells us about the evolution of the universe. Thus, we restate it a third time in a form with more relevance to chemistry.

Second Law of Thermodynamics

There exists a property called entropy S. The entropy of the universe increases in any spontaneous process, ΔS_tot > 0. The entropy of a system is determined by the number of different ways to distribute the available energy over the various states of the system.

The deep importance of entropy for chemical and physical systems was developed, in particular, in the work of Boltzmann and Gibbs who were able to connect thermodynamics and statistical mechanics. By establishing statistical mechanics as the bedrock upon which the evolution of all chemical and physical systems is based, they set the stage for the understanding of reversible and irreversible chemical changes as well as ushering in the quantum mechanical revolution.

8.2 Thermodynamics of a hurricane

We will now demonstrate the discovery of a state function and proceed to define it thermodynamically. We will then show that this thermodynamic definition is equivalent to the statistical mechanical definition. To do so, we begin by looking up in the sky and pondering a cycle that describes how a tropical storm (hurricane or typhoon) is capable of generating such a large destructive force. In other words, how it is able to do work. The work is done by the expansion and compression of gases.

The processes involved are depicted in Fig. 8.2. In panel (a) the system starts at point A. The xy-plane lies on the surface of the Earth. The z-axis corresponds to altitude. A high-pressure system H is located over a body of water such as the Caribbean Sea. At the same altitude, but removed by some lateral distance, the low-pressure system L is located at point B, which is known as the ‘eye’ of the storm. A plug of gas moving from a region of high pressure to a region of low pressure expands. Since this gas is in contact with a massive water bath (the sea or ocean), its temperature will remain constant as it expands. Heat flows from the sea into the gas to maintain isothermal conditions in the plug of gas.

You will most likely have seen satellite images of the low-pressure system at B such as that in Fig. 8.1. Circulation of the atmosphere occurs around this point. The Coriolis force associated with the rotation of the Earth forces the atmosphere to rotate counterclockwise about the eye of the storm in the Northern Hemisphere (but clockwise in the Southern Hemisphere). When the plug of gas reaches the eye it is forced upward. The plug is surrounded by a more stagnant layer of air that exchanges little heat with the plug. Essentially, this step from the eye at sea level to some point C in the upper atmosphere happens adiabatically, q = 0. The pressure is also lower in the upper atmosphere; therefore this step is an adiabatic expansion.

If the plug of gas that is drawn into the eye is not replaced, a vacuum would appear at A. Therefore, gas must be returned to this point. This occurs in two steps. First, in the upper atmosphere the plug is isothermally compressed at a temperature of about 200 K. Finally, from this point D the gas is returned to the circulating high-pressure system at A. This final compression step occurs adiabatically.

In Fig. 8.2(b) the changes that occur to the plug of gas are depicted in terms of a plot of pressure versus volume. A succession of two adiabats and two isotherms defines a closed cycle. Adiabats are pathways connecting two states of the system for which q = 0. Isotherms are pathways connecting two states of the system for which ΔT = 0. In a closed cycle, mass and energy are conserved. Indeed, any state function does not change its value in the course of completing one complete cycle. The requirements for a closed system are more or less met by a hurricane. In the laboratory a system can be constructed to meet them more rigorously.

We now define rigorously what is occurring with p, V and energy in this system. What is the energy balance in this cycle? Where does the energy come from to create the winds? How much work is done? How much heat flows? The progression of our system in Fig. 8.2(b) is known as the Carnot cycle, which we describe quantitatively in the next section. You will have the opportunity to calculate these parameters for realistic hurricane conditions in the exercises at the end of the chapter.

8.2.1 Carnot cycle

Consider the process depicted in Fig. 8.2 with four reversible steps involving one mole of ideal gas, n = 1 mol. We want to calculate the internal energy change, heat and work of each step as well as the cycle in total. Each step can be calculated independently and then summed to obtain the values for the overall process. Since we are dealing with 1 mole of gas, all of the quantities calculated below are equal to the molar values but the subscript m will be dropped for simplification.

Step 1: Reversible isothermal expansion at T_h from A to B and V₁ to V₂. T_h is the constant temperature of our hot reservoir (i.e., the sea in our example). The internal energy change is

(8.1)

• 
  

  We calculate the two easiest parameters and then use Eq. (8.1) to obtain the third. The internal energy of an ideal gas depends on temperature alone. Therefore, for the isotherm expansion of an ideal gas

(8.2)

• 
  

  Per mole of ideal gas the work done in a reversible, isothermal expansion is

(8.3)

• 
  

  Thus, by the First Law, that is, applying Eq. (8.1)

(8.4)

• 
  

  The system does work on the surroundings. The energy exchanged as work is exactly balanced by the heat flow into the system from the surroundings; therefore, the internal energy of the system does not change.

Step 2: Reversible adiabatic expansion from B to C and V₂ to V₃. No heat leaves the system as the step is adiabatic,

(8.5)

• 
  

  The temperature falls from T_h to T_c, the temperature of the cold sink. The internal energy change in reversible adiabatic ideal gas expansion is given by

(8.6)

• 
  

  Then from the First Law, the internal energy change must equal the work

(8.7)

• 
  

  The cold temperature is lower than the hot, thus the work is negative. The system does work on the surroundings and the work is done at the expense of the internal energy of the system.

Step 3: Reversible isothermal compression from C to D and V₃ to V₄. Analogous to Step 1, we have

(8.8)

• 
  

(8.9)

• 
  

(8.10)

Step 4: Reversible adiabatic compression from D to A and V₄ to V₁. Analogous to Step 2, we have

(8.11)

• 
  

(8.12)

• 
  

(8.13)

• 
  

  Now we add up the internal energy changes, the heat flow and the work performed in going around the whole cycle. To evaluate these, we need to find relationships between the various volumes. Steps 2 and 4 are reversible adiabatic expansions, for which the volume change is related to the temperature change and the ratio of the heat capacities

(8.14)

• 
  

  by

(8.15)

8.2.1.1 Example

An ideal gas has a molar heat capacity at constant volume of C_{V, m} = 3/2 R, where R is the gas constant. Given one mole of ideal gas and the pressure and volume conditions in Table 8.1, calculate the work, heat, internal energy change and entropy change associated with each step as well as the whole process.

First, we collect the work and heat expressions for each step in the Carnot cycle, as shown in Table 8.2. Then, we use these expressions to calculate the work, heat, internal energy change and entropy change associated with each step and the whole process. The results are listed in Table 8.3.

Point	Pressure/bar	Volume/L
A	1.00	24.8
B	0.498	50.0
C	0.398	57.0
D	0.802	28.3

Step	Process	Work	Heat
1	isothermal expansion	wA → B = −RThln (V2/V1)	q = −w
2	adiabatic expansion	wB → C = CVΔT	q = 0
3	isothermal expansion	wC → D = −RTcln (V4/V3)	q = −w
4	adiabatic expansion	wD → A = CVΔT	q = 0

Step	w/J mol−1	q/J mol−1	T/K	ΔU/J mol−1	ΔS = q/T/J K−1 mol−1
1	−1740	1740	298	0	5.84
2	−312	0		−312	0
3	1590	−1590	273	0	−5.82
4	312	0		312	0
Sum	−150	+150		0	0.02

The cycle performs 150 J per mole of gas worth of work on the surroundings. The cycle extracted 150 J mol⁻¹ of heat from the surroundings, just as a hurricane would take heat from the water. The internal energy change is zero as expected. A state function must have a constant value for one complete circuit of the cycle. Neither w nor q exhibits this characteristic because they are path functions.

Have we discovered a new state function that corresponds to q/T? The value obtained from the calculation above is close to zero but not exactly zero. It is important to note that this is due to round-off error. We check this by performing the algebra explicitly,

This can be evaluated algebraically by applying Eq. (8.15) to determine the ratios of the volumes, from which we find

If two ratios raised to the same power are equal, the ratios are also equal. Rearranging, we obtain

Now we substitute into the expression for ΔS_cycle to find

Note the change of sign caused by inverting the ratio in the natural logarithm term. Thus, a new state function has been identified. We will define it more explicitly in the next section as not just q/T but q_rev/T, where q_rev is the heat evolved along a reversible path.

8.2.1.2 Directed practice

Show explicitly by performing the algebra that the net internal energy change, entropy change, heat and work for the Carnot cycle with reversible steps are given by

(8.16)

(8.17)

(8.18)

(8.19)

8.3 Heat engines, refrigeration, and heat pumps

The foundations of thermodynamics were largely set in studies of mechanical work and how it related to heat – either the generation of heat during the performance of work or conversion of thermal energy into work in heat engines. This is an important application that allows us to draw important conclusions regarding processes that affect the energy sector.

8.3.1 Heat engine efficiency

Return now to the Carnot cycle and its relevance to heat engines. The Carnot cycle with reversible steps acts as a limiting case for the perfect heat engine. The efficiency ϵ of a heat engine is given by the total work performed w divided by the heat transfer with the hot reservoir q_h,

(8.20)

But the work is just the difference between the heat supplied by the hot source and returned to the cold sink, thus

(8.21)

The expressions for q_c and q_h for a reversible Carnot engine are found in Table 8.2, in which the cold step is identified at Step 3 and the hot step is Step 1. Substituting we find

(8.22)

8.3.1.1 Directed practice

Derive the expression for the efficiency of a reversible Carnot engine, Eq. (8.22).

This is a statement of Carnot’s theorem, and it is true for all materials (not just ideal gases) as long as the engine operates reversibly. This is the highest possible efficiency that can be obtained in a heat engine. By equating Eqs (8.21) and (8.22) we find for a reversible engine

(8.23)

8.3.1.2 Example

At a power plant, superheated steam at 560 °C is used to drive a turbine for electricity generation. The steam is discharged to a cooling tower at 38 °C. Calculate the efficiency of this process.

Convert the temperatures to kelvin: T_h = 833 K, T_c = 311 K. Assuming that the generator is reversible,

Ideally, the generator is 63% efficient, but in practice friction, heat loss and other inefficiencies reduce the maximum efficiency to about 40% or a little better. These efficiency limitations do not apply for fuel cells, and this is one of the reasons that there is great interest in their development for more efficient and cleaner power production.

8.3.1.3 Example

What is the efficiency of the Carnot engine in Example 8.2.1.1 and to what does it apply?

It applies to the work done on the whole cycle. Note how the efficiency has dropped in comparison to Example 8.3.1.2. The highest efficiency is obtained by the greatest difference between the hot and cold temperatures. In practical systems, materials and environmental considerations limit the values of T_h and T_c. The T_h of an engine must lie below the melting point of the components that make up the engine. T_c can be no lower than the ambient temperature outside of the engine because refrigeration would simply add inefficiency.

8.3.2 Heat pumps and refrigerators

The physical processes of evaporation and condensation can be harnessed to provide both heating and cooling. Furthermore, the same apparatus is capable not only of doing both, but also performing more efficiently than simple combustion for heating.

The key to these processes is found in Fig. 8.3, and the realization that since condensation and evaporation are the reverse of one another, the enthalpy change accompanying them is equal and opposite (at the same temperature). Evaporation is always an endothermic process. Water has an enthalpy of vaporization of 43.98 kJ mol⁻¹ at 298 K, which allows for large heat transfers. We know this instinctively as the evaporation of sweat is used to cool our bodies. Conversely, condensation is always an exothermic process. We do not always make this natural connection but sit in a sauna, throw some water on the rocks, and you rapidly feel the heat released as the vaporized water condenses on your body. Similarly, steam at 100 °C is much more dangerous than water at 100 °C because the transfer of heat to your skin is much more efficient when the enthalpy of condensation is available.

The principle of operation of a heatpump/refrigerator is shown in Fig. 8.4. By performing evaporation on one side of a thermally conductive material, a fluid on the other side of the conductor is cooled. Performing condensation on the same conductor heats the fluid on the other side. Thus, a refrigerator is turned into a heat pump simply by reversal of the processes.

In order to cool the inside of a building where Reservoir 2 is located, a liquid must be made to evaporate from the reservoir. The walls of the reservoir are thermally conductive and in contact with a fluid that is cooled before it circulates to the rest of the building. The evaporated vapor is then transported to Reservoir 1, where it condenses and deposits heat exterior to the building. The direction of evaporation, vapor transport and condensation is controlled by a compressor. The compressor pumps gas out of Reservoir 2. This lowers the pressure above Reservoir 2 to below the vapor pressure of the liquid. Evaporation ensues. The vapor is then compressed above Reservoir 1. When the pressure exceeds the vapor pressure of the liquid, the vapor condenses.

To heat the building the process is reversed simply my making the compressor work in the opposite direction. The pressure above Reservoir 1 is lowered below the vapor pressure of the liquid, which turns Reservoir 1 into the evaporator. The vapor is compressed in Reservoir 2, turning it into the condenser. The heat liberated by condensation is used to heat the gas in contact with the outside of Reservoir 2, which then flows to the rest of the building to heat it.

Heat pumps and refrigerators require electrical power to run the compressor. Nonetheless, they offer the possibility to deliver heating and cooling extremely efficiently. Say our refrigerator needs to remove 1000 J from its interior at 4 °C and exhaust it into the room at 25 °C. From Eq. (8.4.4), the heat dumped by the condenser into the kitchen is

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