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Other related issues treatable in thermodynamics include how liquid water droplets grow in humid environments. Does the presence of air affect the vapor pressure above a liquid surface? Does the air dissolve appreciably in the liquid? Does the presence of salt dissolved in the liquid affect the vapor pressure? Does the size of a droplet affect its vapor pressure?

What is the role of moisture and cloud formation in this process? As parcels rise, they expand and their temperature drops why? Does this mean they are denser and they might return? What are the conditions for continued rising? Atmospheric chemistry Most chemical reactions in the atmosphere are between trace gases such as ozone and so-called air pollutants, but many occur between natural constituents. What are the criteria for a reaction to proceed one way or another?

How are chemical equilibria between reactants and products established and how do these equilibrium concentrations vary as the temperature varies? Notes A complete bibliography is given at the end of the book. All university level physics books contain a few chapters on thermodynamics; the numerous editions of Sears and Zemanski as well as those of Halliday and Resnick and the one by Giancoli are good examples. Many general chemistry books also contain a good description 18 Introductory concepts of the subject, for example, Whitten, Davis and Peck The little paperback titled simply Thermodynamics by Enrico Fermi has some marvelous descriptions of thermodynamic states, equilibria, etc.

Two newer books on applications of thermodynamics to atmospheric science and oceanography are those by Bohren and Albrecht and Curry and Webster Both are pitched at a higher level than the present text and both delve more deeply into many aspects of the subject. H is usually between 8 and 9 km. What is this in thousands of feet? What is the ratio of the atmospheric scale height Problem 1. If p0 is hPa, what is a typical pressure in Denver mile high city? Use the expression in Problem 1. Far away its potential energy is zero. It is called the escape velocity. How long does the round trip take?

Gases differ from liquids and solids in that the force between neighbors on the average over time is very weak, since the intermolecular force is of short range compared to the typical intermolecular distances for the individual gas molecules. If an imaginary plate is held vertically in a gas as shown in Figure 2. The forces on opposite sides of the inserted plane are equal; otherwise, if forces on the opposing sides were unbalanced, the plate would experience an acceleration.

These impulsive forces are so frequent that the resulting macroscale force is effectively steady. The force is perpendicular to the face and has the same value no matter how the face is oriented. This can be seen by considering the collisions with the wall and the tendency for no momentum to be transferred parallel to the plane surface. The perpendicular component of the force per unit area on the plane is called the pressure. Tangential components of the force cancel out when averaged over many collisions with the wall and therefore vanish when averages are taken over a large number of collisions with the surface.

The more appropriate unit would be the kPa, but this is not used much in practice. In general, there is a mathematical relationship between the three variables, or thermodynamic coordinates, called the equation of state. It is important to remember that the density is uniform throughout the volume of a gas in thermodynamic equilibrium. The condition for this is that the molecules spend a large fraction of their time apart from one another so that the intermolecular forces are acting only a small fraction of the time for a given molecule.

This will become clearer in the next few sections where some estimates of intermolecular spacings and distances between collisions are compared to the sizes of the molecules. Typical intermolecular forces for neutral molecules are appreciable only over distances of the order of the radius 22 Gases of a molecule. Example 2. What is its number density? This value is called the Loschmidt number. Place a cube around each molecule in the gas. Then each molecule sits at the center of a cube of side length d. The number of these cubes per unit volume is n0. In a liquid or a solid intermolecular distances are on the order of the molecular sizes see Table 2.

Most high school or college chemistry books describe the Bohr model of the hydrogen atom. To obtain an estimate of the mean free path imagine the background gas particles to be stationary. A collision between our prototype molecule and a background molecule will occur when their centers are within 2r0 of each other Figure 2.

We can think of the test molecule having radius 2r0 and the lattice composed of stationary points see Figure 2. Hence, as the test molecule moves through the lattice it sweeps out a cylinder of radius 2r0. The number of background 2.

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Actually, it is possible to solve the problem when all the particles are in motion, and the derivation can be found in books on the kinetic theory of gases. Data from Atkins and de Paula Ar—Ar 0. Our exponential model was just cooked up, but it turns out to be a very good approximation. When the dimensions of the body of gas say, a storm or a cold air mass are large compared 2.

This is usually the case in atmospheric science up to altitudes of about km. The mean free path also gives us an idea of the length scales over which transport of properties occurs. If a property is transported via collisions between molecules it has to happen over length scales comparable to the mean free path. Imagine a molecule going left and right across the box and bouncing back at the walls. The side dimension of the cubical box is L.

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This relationship says that the temperature expressed in kelvins is proportional to the average kinetic energy of individual molecules. Moreover, quantum mechanics tells us that there is motion even at this lowest of low temperatures. Fortunately, we never encounter these low temperatures in meteorology and the Ideal Gas Law virtually always applies to the gases of interest. Hence, the formula and its interpretation are correct. However, over 2. Now if we add the collisions of all the molecules, as in the derivation of pressure, we can be assured of a steady force perpendicular to the wall.

There are a few loose ends that must be addressed. First, not all molecules are traveling strictly in the x, y and z directions. This can be disposed of by noting that for the wall perpendicular to the x direction only the x component of the motion matters. The y and z components do not affect this wall. Do the collisions one by one cancel their y and z components before and after the collision with the wall? After all, the wall is not a smooth surface at the molecular level.

The answer is that over the long run and averaging over many particles this cancellation is complete. Lastly, the molecules do not travel uninterrupted from one end of the box to the other. They go only one mean free path a few tens of nanometers at STP before they suffer a collision with another molecule. The solution to this problem lies in the conservation of momentum. After a collision the x component of momentum is conserved for the colliding pair and it is the momentum change at the wall that matters, whether it is the same molecule or not.

Dry air is a mixture of different ideal gases. The value of Rd takes this mixture into account as will be explained shortly. Let the conditions be STP.

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Hence, there are about collisions per second with the 1 m2 wall at STP. So where did the divisor 6 come from? It comes from a careful integration over all the angles, etc. With or without the 6, this is a very large collision rate. In treating random variables we consider independent realizations of the variable like drawing values from a hat. Instead molecules will have instantaneous velocity components vx , vy , and vz. Consider the vx component for an individual molecule at a given time. The value of vx will take on a range of values from one time to the next because of collisions with other molecules it can be thought of as a random variable.

Computer simulations suggest that after only a hundred or so collisions per molecule the probability density of values of vx settles to a steady functional form. The normal sometimes called the Gaussian distribution occurs often in nature. It generally comes about when the variable is subjected to a long history of random jolts that add up to its current value.

After a long time many additive increments to the value of the variable its probability distribution approaches the normal distribution. This can be proved under rather general conditions in mathematical statistics under the heading of the Central Limit Theorem. The normal distribution has the familiar bell shape shown in Figure 2. These next two examples occur often. More cases can be found in elementary statistics books. The distribution of velocities has no dependence on direction, only on speeds i. A graph of this function is shown in Figure 2.

Recall that the escape velocity is The number of these molecules to escape even over the history of the planet 4. The median of the distribution moves to higher speeds if the temperature is raised or if the mass of the molecules is lowered. For example, Figure 2.


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This is a value large enough that if H reaches the upper atmosphere it will be depleted over planetary lifetimes. However, H is continually produced in the upper atmosphere by photodissociation see Chapter 8 of water vapor. This is probably large enough for H to escape, but small enough that water molecules steadily being disassociated by hard very short wave solar radiation can maintain a presence at very high altitudes. The maximum surface temperature on the Moon is about K.

Figure 2. We present one more here since the result comes up often. We want to know the number of molecules striking a wall perpendicular to the x-axis per unit time and area. This is simply the number density times the x component of velocity averaged over the velocity distribution. The molecular weight, M , of a pure gas is the sum of the atomic weights of the atoms making up the molecules. The chemical properties of the element are determined by the number of protons in the nucleus, which is designated the atomic number.

The atomic weight is determined by the sum of the number of protons and the number of neutrons. An element can have different isotopes, i. But the most abundant isotope found in nature is usually dominant, with only a small percentage of the other isotopes present. If we take a random sample from nature this leads to a weighted average of the atomic weight, and this is the value used in most computations. For our purposes, we can simply use the numbers given in Table 2. The mass in grams! The gram-molecular weight is the one used in most existing tables; hence, we use it here.

But in keeping with our SI units, we need to express m0 in kilograms in most formulas. MG , with M NA 2. It is useful to know that 1 mol of an ideal gas occupies Answer: Use 2. How many moles of O2 , N2 , and Ar are there? This is just larger than the scale height of the atmosphere. Note that the result is independent of the initial dropping height h. Is it any wonder that atmospheric pressure measures the weight of air in the column above a square meter? It states that the partial pressures of the individual constituent gaseous components are additive. Based upon the kinetic theory derivation above it is not surprising that the pressures would be additive for mixtures of ideal gases with different molecular masses m1 , m2 ,.

Now return to 2. Meff 2. What is RAr? What is the mass of air above 1 m2 at sea level? One which is very readable and includes a mix of elementary kinetic theory as well is Sears Another more recent but very readable account is by Houston A purely thermodynamic treatment is given in the physics text by Zemansky Modern physical chemistry books are perhaps best for discussions of gas thermodynamics, for example Atkins For a readable but rigorous discussion of constraints, etc.

In newtons, in pounds? Note: 1 kg weighs 2. Compare to the escape velocity. It is known to have 5 g of water vapor and the rest is dry air. What is the partial pressure of water vapor? What is the density of this moist air? Compare to the density of dry air at the same overall pressure. The surface pressure is hPa. What is the kinetic energy of the column? What is its rotational kinetic energy?

What is the total number of molecules in a column with unit crosssectional area? What are reasonable values for n0 0 and H? Problems 43 2. What is the total kinetic energy of the molecules in the column? The particular mass of gas may be thought of as a small parcel of matter that is transported through the environment by natural forces acting upon it. We could also imagine moving it virtually via an abstract thought experiment, for example to determine its stability under small perturbations.

As an air parcel rises for whatever reason in the real atmosphere, it will almost instantaneously adapt its internal pressure to the external pressure exerted by the local surroundings, but the temperature and composition adjust more slowly. In convection, entrainment of neighboring air also speeds up the process of equalizing the temperature between inside and outside air. This time scale separation has made the parcel concept a useful and even powerful tool in the atmospheric sciences. We will return to it often.

Thermodynamics is concerned with the state of a system an example of which is the parcel alluded to above now treated as an approximate thermodynamic system and the changes that occur in its state when certain processes occur such as its being lifted. The force exerted by the system on the movable wall is pA, where A is the area of the wall. One might imagine a cyclic process in which the system expands from VA to VB along one path and returns along another path.

The area between the two curves represents the net work 1 In this text we use the sign convention that positive W means work done by the system on its environment. Some textbooks use the opposite sign for work done by the system. The area under the curve is the work performed by the system on the environment. Example 3. Let the system expand isobarically from its initial volume of 1 m3 to 3 m3.

How much work is done by the system? See Figure 3. Obviously an isobaric process leading to a tripling of volume is very unlikely in the atmosphere. We are implicitly assuming that we are in a state of thermodynamic equilibrium at each step — in other words the system has time to come to equilibrium i. In real processes such as the compression of a piston in an internal combustion engine, the gas in the chamber might be highly nonuniform and locally disturbed by such things as shock waves during the next change in volume perhaps the equation of state does not even hold during this interval of time.

For an irreversible change such as in the internal combustion engine, an amount of work will be done, but it may not be calculable using p dV. In more advanced books on thermodynamics it is shown that when the system does work for example by expansion p dV is the maximum work that can be done. But when the system is compressed, the reversible calculable work p dV is the minimal work done by the system during the compression.

The unfortunate mechanical engineer simply cannot win in the face of irreversible processes. Luckily, most natural processes of interest to the atmospheric scientist are better behaved. The idealization of reversible work allows us to do calculations using p dV even though in reality it never quite works that way. In most applications that follow in this book the assumption of reversible work is adequate. During this process some energy might be transported into the system because of a temperature difference between the interior of the system and the surrounding environment.

This energy transported into the system is thermal energy 2 as described in Chapter 2. Thermal energy consists of all the modes of energy associated with individual molecules: translational kinetic energy, rotational energy for polyatomic molecules and vibrational energy potential and kinetic for polyatomic molecules that experience internal stretching oscillations. These individual energy terms each contribute to the thermal energy of a molecule but only the translational kinetic energy contributes to pressure.

In fact, the energy on the average is shared equally between the different modes translational kinetic, etc. If there is a gradient of temperature, molecules from the warmer region will penetrate a distance of the order of a mean free path before suffering a collision into the cooler region and vice versa , causing the cooler region to warm; molecules moving the other way cause the warmer region to cool through individual collisions. The news and conversion are brought about slowly but surely.

This is in stark contrast to the propagation of pressure differences which move via a sound-like wave distance of advance of the pressure front being proportional to time. It is possible to solve this problem analytically, but the details need not be given here. We will use the term heat to mean the integral over the heating rate with respect to time. Typically, rotational levels are closely enough spaced for them to be excited, but vibrational thresholds are much higher, requiring very high temperatures for excitation.

The pulse spreads out in the shape of a normal distribution as shown in Figure 3. The standard deviation of the spread of the elevated temperatures is only about 3 m after 12 hours. This means that the concept of parcel integrity for objects of the order of several hundred meters is safe for days if the only stirring mechanism is molecular diffusion. There are other mechanisms that can shorten the time of mixing depending on the conditions, but these are still usually slow compared to the adjustment of the interior to the exterior pressure.

Note that sound waves travel at several hundred meters per second. The adjustment of pressures should be accomplished in several hundred passes of sound waves back and forth across the parcel — still very fast compared to molecular and even eddy turbulent transport processes.

The sound waves are eventually dissipated into thermal energy. Thermal energy or heat as we have been discussing it can now be contrasted with the work being done by a system during a process. The thermal energy is at the molecular level and it migrates from place to place via gradients in the temperature say from the system to a reservoir , while work is at the truly macroscopic level. Returning to our system, heat can be transported into it because of small differences in temperature between the system and its environment. This is the sign convention followed by virtually all textbooks.

In this process no work is done by the parcel on its environment, since the volume of the parcel does not change. All the heat given to the system goes into its internal energy. Hence, a change in internal energy is equivalent to a change in the kinetic energy for an ideal monatomic gas. What is the mass of air in the room? How many joules are required to raise the temperature by 1 K?

So the result is 0. Note that the cost of 1 kWh is a few cents US. Let us take the walls to be wood and 1 cm thick. The total heat capacity of the solid matter is kJ, nearly twice that of the air contained. The last equation is a statement of the conservation of energy. The First Law of Thermodynamics actually goes much further and states that the internal energy U is a function only of the state of the system.

For the ideal monatomic gas this is obvious from our simple kinetic theory model since the internal energy is the total kinetic energy summed over all the molecules in the system and this is proportional to the Kelvin temperature. Note that neither Q nor W are functions of state.

We must specify its internal energy as a function of the thermodynamic coordinates. The constant f depends upon the internal structure of the molecules; it is known as the number of degrees of freedom in the molecule. For a diatomic molecule that is very stiff does not stretch and contract under the temperature conditions being considered such as O2 and N2 near room temperature, the value of f is 5 because there are two more degrees of freedom due to the ability of the molecule to rotate in a two-dimensional plane, but not about the axis joining the two atomic constituents its moment of inertia is too small about the axis joining the two atoms.

At high temperatures not naturally occurring in the lower atmosphere the number of degrees of freedom goes up by two because of molecular vibration due to the spring-like binding. Consider the constant volume heating case. If heating occurs in a box of monotonic gas, all the energy must go into increasing the linear translational kinetic energy of the molecules.

If the molecule is spatially extended such as a diatomic molecule, it can rotate as well as translate. The added energy can go into rotational energy as well as translational energy. Basically the heat energy that at the molecular level must be shared among all the degrees of freedom, but only the linear kinetic energy goes into causing pressure since it carries momentum to the walls or across boundaries.

Aremarkable theorem proved in the classical study of statistical mechanics shows that in equilibrium the energy will be shared equally between each of the rotational and translational modes and vibrational modes when applicable : the principle of equipartition of energy. In the case of a diatomic molecule only two rotational modes are available, the rotation about the axis joining the two atoms does not count. In the case of a triatomic molecule in which the atoms are not in a straight line e. On the other hand, such modes of vibration and rotation play an important role in the absorption and emission of infrared radiation as it passes through air.

In summary, each molecule on the average possesses 12 kB T for each of its mechanical degrees of freedom. Hence for argon it is 7. Hence the gravitational potential change is comparable to the internal energy for a lift of one scale height. The heat capacity at constant volume, cv , is proportional to the number of degrees of freedom. Another important process is the heating of the gas at constant pressure.

Answer: J. How much heat is required? This is because U is a function of state. Remember, however, that for many substances other than the ideal gas U might depend on more than just temperature; nevertheless, it is only a function of state. Assume the air above and the ground below are insulated from the well-mixed air in the boundary layer. This of course is a large rate of increase for normal conditions. The assumption of W is the problem. The next example shows a more realistic case in which the heating is accomplished by black carbon particles in the air.

Much of the heating in the atmospheric boundary layer comes from the absorption of infrared radiation as well as solar radiation by water vapor. This means the cross-sectional area of an individual particle is 3. Note that this is only a tiny fraction of the 1 m2 cross-sectional area of the cube of air. The heating rate of this cubic meter of aerosol-loaded air is then 0. If the air is at sea level its density is about 1. The parcel effectively is surrounded by a thermally insulating blanket. A key point here is that it takes a long time for temperature differences to diffuse into a parcel from outside compared to the relatively short time for the pressures inside and outside to equalize.

This means that in many situations we can regard the process as being adiabatic, that is, isolated from diathermal contact. It gives the temperature that a parcel of dry air would have if it were brought adiabatically to a pressure of hPa. If the parcel moves adiabatically its potential temperature will not change.

In practice parcels do move adiabatically in convective motions to a good approximation. Is p0 on top or not? This can help us to obtain the correct formula. Figure 3. Here Ha is called the scale height of the atmosphere. The temperature outside is K. What must be done to the air to bring it to an inside temperature of K?

The air must now be cooled at constant pressure to K by the air conditioning system. Its temperature is K. What is its potential temperature? It is lifted adiabatically to the hPa level. What is its temperature at the hPa level? How much is the temperature changed? By what ratio is the volume of such a parcel changed? These include heating of a parcel by solar radiation at a particular altitude, condensation heating, and contact heating at the surface.

Calculus refresher: partial derivatives Thermodynamic functions nearly always involve more than one variable as we have seen already, e.

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Sometimes especially in mathematics and physics, the partial derivative is denoted by a subscript. You simply take the ordinary derivative but hold all variables constant except the one being varied. What is the rate of change of enthalpy? What is the rate of change of T along the path of motion? Enthalpy is especially useful since these processes often take place at constant pressure.

An example is the evaporation of 1 mol of water. In this case the system the volume containing the water is heated by maintaining a small temperature differential with its surroundings at constant pressure and constant temperature. The heat now we can call it enthalpy absorbed to effect this transition is often called the latent heat 64 The First Law of Thermodynamics Table 3. In the table Tf is the freezing point and Tb is the boiling point at 1 atm of pressure. Many tables give values in terms of moles rather than kilograms of the substance. See Table 3. By convention the standard quantity is evaluated at the freezing point at 1 atm of pressure.

We wish to evaporate the water and raise its temperature to K at constant pressure. What is the change in enthalpy for the two steps? Answer: 36 g is 2. Step 2 is a heating at constant pressure. The temperature is K. What is the change in enthalpy of the system? Notation and abbreviations 65 Answer: The dry air is irrelevant. We must do several steps to accomplish our goal. To use the value in Table 3. This represents a 1. This freedom allows us to use standard tables.

Notes Most of the good thermodynamics books referred to in earlier chapters work well for this one. Evaluate the following. Calculate both for an ideal gas.


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  • Problems 67 3. Find the internal energy of 1 kg of dry air at STP. Now suppose a 1 kg parcel is lifted adiabatically one scale height H. How much work does the parcel do on the environment in the process? Isobaric process A 1 kg parcel of dry air has temperature K and pressure hPa. It is heated by contact with the dry ground to a temperature of K. Isothermal process 1 kg of dry air at K and hPa is expanded isothermally pretty unusual in the atmosphere from a volume of 2 m3 to twice that value.

    Isochoric process 1 m3 of dry air at hPa and K is enclosed in a rigid box. What are the changes in internal energy and enthalpy? Sketch a diagram of the change in the V —p plane. Adiabatic process A parcel of mass 1 kg is lifted adiabatically from hPa, where its temperature is K, to hPa. What is the new temperature? It is heated by infrared radiation being absorbed by some water vapor in the parcel. How does the potential temperature change per unit time? Do you recall from physics why the adiabatic compressibility gives the correct answer instead of the isothermal compressibility?

    See Problem 3. Sketch a graph. We will introduce a new function of state by considering a simple quasi-static series of changes for an ideal gas. First recall that some quantities are functions of the state only, while others depend on the path taken between two states. Recall that the enthalpy H is also a function of state. Changes in other quantities such as the amount of work done by the system on the surroundings in the quasi-static transition depend upon the path taken between the two states.

    For example, in the case of the ideal gas the internal energy and enthalpy depend only on the temperature, independent of the history of the system. For more complex systems U and H might depend on other state variables as well, but not on the history of the system.

    Note that the heat induced by a temperature difference between the system and its environment taken into the system during the transition must also depend on the 69 70 The Second Law of Thermodynamics p A B V Figure 4. Next we explore quasi-static transitions of an ideal gas to see whether there might be another function of the state of the system. Now multiply through by the integrating factor ebt. Integrating from one point in the x—y plane to another always yields the same answer.

    Of course, in nature the transition might be and often is an irreversible or spontaneous one from state A to state B, but there always exists a reversible path joining the two states so that the change in entropy can always be calculated. For a parcel of dry air, knowing the potential temperature is equivalent to knowing its entropy. Example 4. What is the change in entropy? Answer: First note that 1 kg of water is The subscript is again a reminder that the calculation must be conducted along a reversible path.

    Consider a system which is embedded in its surroundings. Together we say these comprise the universe. So far in this book we have only considered closed systems, but we will encounter open systems as well. In equilibrium, both the system and the surroundings have thermodynamic coordinates. A reversible change is one which is quasi-static taken in small slow steps in such a way that equilibrium is maintained; that is, there is enough time between steps for the pressure and temperature to homogenize throughout the volume of the system and which can be reversed at any point returning both system and surroundings to their former coordinate values without the expenditure of any additional work more on this below.

    Such a transition is irreversible. Under many of these spontaneous transitions the internal energy does not change, but the entropy does. Inside the chamber is a wall separating half the volume on each side. There is an ideal gas on one side of the partition, vacuum on the other. What are the changes in internal energy, enthalpy and entropy? Answer: The internal energy may be calculated from the First Law. The system does no work since the vacuum exerts no back pressure during the expansion. Also no heat is taken into the system because the walls are impermeable to such a transfer.

    Note that we were able to apply the First Law even though the path was irreversible. The fact that the internal energy is invariant means that in the free expansion, the temperature does not change true for an ideal gas — in a real gas there is some temperature change, even though no change in U occurs.

    Since the temperature does not change we can see that the enthalpy does not change for the ideal gas. The change in entropy must be computed by an alternative reversible path. It is zero. Hence, the change in entropy for the universe is MR ln 2 which is a positive number. This is an example of the second part of the Second Law. For an irreversible process the entropy of the system and its surroundings taken together, the universe will always increase.

    This means that some quasistatic processes are irreversible. So how could an adiabatic expansion ever be reversible? In this way a reversible approximation can be found to the adiabatic expansion curve. Hence, the entropies of the system and its surroundings are additive. The same additivity principle applies to the internal energy and the enthalpy and any other of the extensive parameters describing the composite system.

    The extensive parameters volume and mass also satisfy this principle. When we add up several systems to form a larger system in which the entropies, internal energies and enthalpies can be added up, we call this a composite system. Its pressure and temperature might be varying with altitude z but we can 4.

    Bringing two systems into contact amounts to removing or relaxing a constraint. For example, if two gases A and B at the same pressure are in a chamber, but separated by a partition, the removal of the partition will allow the gases to mix. No change in internal energy will occur if the chamber containing both subsystems is insulated from its environment. On the other hand, the change in the total entropy must be zero or positive. Processes either reversible or not in which the sum of the entropies of a system and its surroundings yield a negative change in total entropy do not occur in nature.

    This principle can be of great utility in determining which way a process will proceed when a constraint is relaxed or removed. Recall that removal of a constraint means that in order to restore the system to its original state additional work must be performed by some external agent. This last is really the essence of the Second Law of Thermodynamics. It turns out that the entropy change provides this additional information.

    Answer: First we would have to bring the ideal gas into contact with a reservoir of temperature T , then we would perform an isothermal compression of the gas from volume 2V0 to V0. Note that in the case of mixing two gases A and B mentioned above we would need to recompress each to its original volume by an isothermal route in order to restore them to their original states. We might in this case use a membrane 80 The Second Law of Thermodynamics which allows molecules A to pass into a new adjacent chamber but not B.

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    But suppose A and B are the same gas. Does the entropy increase when the partition is removed? To proceed consider the expression for an ideal gas dry air parcel. T0 p0 4. We immediately see that such a function S does exist for the simple ideal gas system. We also see that S is extensive proportional to M , and the extensive arguments U and V are expressed as ratios of standard states with the same mass.

    In addition we can see that S U , V is an increasing function of the internal energy U and the volume V. While we worked out the analytical expressions for the case of an ideal gas, the differential relations for dU and dS are general and hold for any substance. Therefore the last two equations for the partial derivatives also hold generally. We then learn two important facts. Mcp p0 4. For example, the equilibrium concentration of gaseous reactants and gaseous products in a chemical reaction will reach its equilibrium value when the entropy of the system is maximized see Chapter 5.

    The two subsystems come into thermal contact. The last equation is the answer to our problem, but let us look further to see what happens to the entropy. Figure 4. We illustrate it here for an ideal gas. This is a four step loop process as illustrated in Figure 4. The branch ab is along a hot isotherm at temperature Th during which heat Qh is transferred to the system from the hot reservoir.

    The next step is an adiabatic expansion bc to the cooler temperature Tc. Finally, there is the adiabatic compression da, which completes the cycle 84 The Second Law of Thermodynamics p hPa a Isotherm Th b d Isotherm c Tc V m3 0. The cycle proceeds as follows. Step ab is an isothermal expansion from a to b, at temperature Th , drawing in heat Qh. Step bc is an adiabatic expansion from b to c. Step cd is an isothermal compression at temperature Tc expelling heat Qc from the system.

    Finally, step da is an adiabatic compression from d to a. Then all other intersections are determined, e. Va Vd 4. This way both Qh and Qc are positive numbers. The difference is the work extracted from the system during the cycle. It would be nice to use all of the heat from the hot reservoir to create work, but this is impossible, since some heat is always rejected to the cold reservoir.

    We can never turn all of our heat extracted from the hot reservoir into work. This is a major consequence of the Second Law. The First Law says energy must be conserved, but the Second Law goes further to tell us that we not only cannot obtain something for nothing, we cannot even obtain all of the something to be used for work. The area enclosed is the work done by the system. In most cases in atmospheric science it is easier to deal with areas in the S—T plane than in the V —p plane.

    This is because the volume of a parcel is not easily observed, whereas its pressure same as the environmental pressure and approximately equivalent to altitude is easily observed. Isotherms and adiabats are straight lines for any system in the S—T diagram in Figure 4. This means that the Carnot cycle is useful in describing cycles even of a composite system composed of any substances. A particular example is the case of a composite system consisting of a liquid in equilibrium with its vapor, both enclosed in a chamber. Thermodynamic loop diagrams for atmospheric processes will be exploited in later chapters as we analyze the energetics involved in parcels undergoing transitions in the real atmosphere.

    We can show a few properties that hold for the general case that were derived above only for the ideal gas case. For example, consider the work done in the cycle. Finally, we can verify that the area enclosed in the rectangle is Wloop. There are also the intensive parameters, pressure p, and temperature T. It is best to think of the internal energy as a function of the entropy, volume and mass.

    Note that this last expression shows how the internal energy changes in terms of other state variables S and V. But we can take dQsurr sys since the heat gained by the surroundings has to be supplied by the system. The dQsys need not be reversible in this problem.

    As an example, consider the irreversible! Consider what happens to a system undergoing a spontaneous transition because some constraint has been relaxed or removed. The other three inequalities can be expressed similarly. It proves to be useful for processes which occur at constant pressure and constant temperature. We can use the Gibbs energy to help us in deciding the direction of a chemical reaction and in determining the equilibrium phases or concentrations of chemical species in equilibrium. The Gibbs energy is particularly useful for open systems those in which mass can enter or leave the system and for systems in which the internal composition might change due to chemical reactions.

    We will take up some of these cases later in this chapter. This actually happens in a phase transition. Let the chamber contain a liquid with its own vapor in the volume above it. If the piston is withdrawn isothermally and quasi-statically some of the liquid will evaporate into the volume above the liquid surface. We should note that the pressure in the liquid is the same as the pressure in the vapor we ignore gravity here. Different positions of the piston leading to different volumes of the vapor correspond to the same temperature and the same pressure in both liquid and vapor.

    Along this locus of points in the state space say the V —p diagram for this composite system the Gibbs energy is constant. We will return to the two-phase problem in Chapter 5. Returning to the general problem we see from 4. In this case the atmosphere which contains orders of magnitude more neutral background molecules nitrogen, oxygen and argon than the usually trace reactants acts as a massive thermal and pressure buffer holding the temperature and pressure constant.

    In these cases where T and p are held constant only the concentration of the species is allowed to change. This is the perfect setup for use of the Gibbs energy. Equilibrium will establish itself at the minimum of G T , p , much as it did in the last chapter for a maximum of S U , V only this time we have a function whose dependent variables are more under our control or more to the point those found in naturally occurring circumstances.

    These molar indicators are thermodynamic coordinates. We can obtain some insight into this equalizing of the G i by considering a system at constant pressure and temperature in which there are two chemical species, A and B. Further, suppose the two gases have the same pressure p and temperature T. Now suppose the two subsystems are brought into material contact with the volume being the sum of the original volumes, the pressures and temperatures also being the same.

    Once the gases are mixed into the larger volume, the total pressure will be the same, but the partial pressures will be only half as much since they occupy twice the volume but at the same temperature. For the internal energy and enthalpy, the job is easy. As expected, during the mixing of ideal gases the Gibbs energy decreases, while the enthalpy and internal energies do not change. To calculate the entropy change we choose a reversible isothermal path. Then, from 4. In the beginning each subvolume contains the same number of moles of identical gases. What changes after the mixing of the gases?

    Then the change in entropy should be zero.

    So we get two different answers for the same problem. This has become known as the Gibbs Paradox. The reason this paradox arises is that in classical physics we cannot consider the mixing of two identical gases as a limiting case of the mixing of two different gases. If we start our consideration for different gases, they have always to be different. It is impossible to get the answer for the entropy change of the mixing of two identical gases simply by equating the masses and the gas constants in equation 4. In classical physics the exchange of coordinates between two identical particles gas molecules in our case corresponds to a new microscopic state of the system two gases in the cylinder , although nothing changes with such an exchange at the macroscopic level.

    This paradox does not exist in quantum theory, where the exchange of two identical particles does not correspond to a new microscopic state of the system. Therefore, when two identical gases are mixed, the entropy does not change. Notes Aside from the books already mentioned in earlier chapters, a beautiful treatment of thermodynamics from an axiomatic point of view is given by Callen Thermodynamics and its applications in engineering has a long history. Problems 97 4. Find its change in potential temperature. What is the change in its enthalpy? What is the change in its entropy?

    Its potential temperature? All steps are performed at constant pressure. Compute the change in entropy for steps a , b , and c. Step 1: the volume is increased adiabatically until it is doubled. Step 2: the pressure is held constant and the volume is decreased to its original value. Step 3: the volume is held constant and the temperature is increased until the original state is recovered. It undergoes a process that is depicted in the V —p plane in Figure 4. Calculate the change of entropy, enthalpy and internal energy for this air parcel. Starting from point A describe how the temperature changes during this cyclic process.

    Va Vd Hint: Divide one of the four equations for work done along the different legs of the cycle by another one. The air is lifted adiabatically in the eye wall to a height above the tropopause where it begins to cool due to loss of infrared radiation to space. Finally the air descends to the surface adiabatically. The result holds for any system, not just an ideal gas. Energy transfers during transformations among these phases have important consequences in weather and climate. The system of redistribution of water on the planet constitutes the hydrological cycle which is central to weather and climate research and operations.

    Water is also an important solvent in the oceans, soils and in cloud droplets. These and other effects lead us into the fascinating role of water in the environment. Of course, thermodynamics is an indispensable tool in unraveling this very challenging puzzle. The chamber is to have a volume that is adjustable, as shown in Figure 5.

    In the following let the chamber have no air present — only the gas from evaporation of the liquid. We are to choose a volume V such that there is some liquid present at the bottom of the chamber we say here the bottom of the chamber, but otherwise we ignore gravity. There are gas molecules constantly striking the liquid from above and sticking. Molecules in the liquid phase must have at least a certain minimum vertical component of velocity inside the condensed phase to escape the liquid surface they have to overcome the potential energy necessary to leave the surface.

    At point B, liquid begins to form on the base of the chamber. If the rate of departures should exceed the rate of sticking arrivals, the number density of gas molecules n0 would steadily increase until the rates equalize. If the volume of the chamber is decreased slightly, the equilibrium will have to be re-established. In decreasing the volume we must condense a net amount of vapor molecules into the liquid phase under these conditions.

    This excess kinetic energy of the molecules entering the liquid is quickly shared with the other water molecules in the liquid, slightly raising its temperature. This tiny excess temperature over that of the reservoir in contact with the system is quickly wiped out with a heat enthalpy transfer to the reservoir , maintaining the isothermal constraint. As the volume is decreased, there is another form of energy being added to the system, because during the compression, work is being performed on the system by the piston maintaining constant pressure.

    Consider the situation in a V —p diagram Figure 5. We wish to trace an isotherm for this system. We start at point A where all the matter in the chamber is in the gas phase. We start at point A where the system is all vapor, and compress isothermally to point B, where liquid begins to appear. In going from B to C a mixture of liquid and vapor is present. Along this second stage the isotherm is also an isobar.

    The portion of the curve to the left of C represents the liquid phase. The critical isotherm is shown by the bold line. T1 of the curved droplet surfaces would complicate the energetics here; but also no gravity in this experiment — the water could congregate on the ceiling for all we care here. If you raise the temperature, you are effectively adding a product to the reaction, which will cause it to shift back to the reactant side. Be careful about this though because temperature can change equilibrium constants.

    Kinetics, on the other hand, does not depend in the slightest on what the situation looks like at equilibrium. The rate of the reaction has no dependence on the overall reaction equation but instead depends on the reaction mechanism, the elementary steps. This was the part of the reaction sequence that we ignored for thermodynamics. The molecules on the left of each elementary step must collide in order to react so that the products on the right are formed.

    Notice that in the first step of the reaction sequence above, the reactant A doesn't have to collide with anything. This is what is called a "unimolecular," "first order" elementary step because only one atom is involved. In the second step of the reaction sequence, C and D do have to collide in order to produce E. This is what is called a "bimolecular" step because two atoms have to come together for the reaction to occur. The rate of an elementary step depends on the concentration of species available to react. For example, in the second step, if there are many molecules of C and D around, then the likelihood of a molecule of C colliding with a molecule of D with sufficient energy and the right orientation to make the elementary step go is high.

    Therefore, the rate of the elementary step is proportional to the concentrations of the reactant molecules. Here are expressions for the rates of the two elementary steps for the reaction sequence above:. Notice that if for instance one of the elementary steps were to involve two molecules of C colliding, then the rate of that step would be proportional to [C] 2. Elementary steps of higher molecularity termolecular and on up are very rare because in any real scenario, it is unlikely that three molecules would hit each other in exactly the right way and with exactly enough energy for the step to happen.

    The rate constant k is a quantity that students love to confuse with the equilibrium constant K. K tells you the ratio of products to reactants at equilibrium while k tells you the rate of an elementary step in the reaction mechanism! Although it is most usual to find little k experimentally, it can also be found from the Arrhenius equation,.

    The Arrhenius equation does not tell you the rate of the reaction; it tells you the rate constant for an elementary step of the reaction. The variable Ea is the activation energy for the step, or the height of the hump on the reaction diagram at the beginning of the section. The constant R is our old friend the gas constant, and T is the temperature at which the elementary step is performed. The large sensitivity of k to T is the reason that it is extremely difficult experimentally to find rate constants.

    Most elementary steps either give off or take up heat, and the resulting temperature change changes the rate of the elementary step itself. Thus, the practical utility of the Arrhenius equation is limited. The constant A is the "Arrhenius factor. When you take Chem 33, you will learn that for some reactions classified as "SN 2 " the collision must involve one molecule putting electron density into an antibonding orbital on another molecule.

    This is a very precise place to have to put electron density! Therefore, only a small fraction of collisions result in reaction. The Arrhenius factor is also called an "entropic factor" to stress that it accounts for how random collisions can be if they are to result in a reaction. From experimental data, it is often possible to find the rate law for a reaction. For example, given the following numerical data, you can deduce that the overall reaction is first order in A, first order in B, second order in C, and that the rate constant is 6 x 10 2 M -3 s This is because the rate of reaction doubles when you double the concentration of either A or B leaving all other initial concentrations constant and quadruples when you double C leaving all other initial concentrations constant.

    Once you know the exponents, you can plug in to the equation to obtain k. Keep in mind that this procedure finds the initial rate of the reaction. Often, as the reaction progresses, the rate changes. Experiment [A] [B] [C] rate 1 0. We've discussed the rate of individual elementary steps of a reaction, but how do we find the rate of an overall reaction?

    One way to do this is to realize that the rate of the reaction will be determined by the rate of its slowest step. If there is a long line at the ATM but no line at the coke machine next to it, then the rate of your getting a coke is pretty much the same as the rate of your getting money out of the ATM. Consider again what is fast becoming our favorite reaction sequence:.

    If the first elementary step is standing in line at the ATM and the second step is getting a coke, and if it takes a long time to get money but relatively little time to get a coke, then we can write that the rate of the entire reaction here modeled by the coke-obtaining process is equal to the rate of the first elementary step:. Note that there is no way we could have predicted which step would be the slowest. The bottleneck step must be determined experimentally. If there were a long line at the coke machine but no line at the ATM, then the rate of the overall reaction would be the rate of the second elementary step:.

    However, it is considered bad form to write the rate of an overall reaction in terms of a reaction intermediate. Reaction intermediates are constantly being created and being consumed, so [C] varies greatly from time to time during the reaction. We can get around this by using what is called the "steady state approximation" to solve for [C] in terms of the concentrations of the other reactants.

    In order to understand the steady-state approximation, we have to realize that thus far we have only considered the rate of an elementary step going forward. Each elementary step has a corresponding back reaction that also has an associated rate. The real situation can be represented as follows:. This is why it is called the "steady state approximation. So the math for this scenario is as follows:.

    Pretty complicated rate expression, eh?! Depending on what tricks you use the steady state approximation is just one of them you can get some very crazy expressions. If you understood the preceding example, you already understand all of the important ideas behind Michaelis-Menton kinetics. Michaelis-Menton does the same steady-state approximation math for a biological enzyme-substrate system.

    Enzymes are special proteins that catalyze biological reactions. Many enzymes break down food molecules into material from which your body can get energy. The enzyme called E has a little niche into which the substrate food molecule called S fits just perfectly. We can represent the system as follows:. To solve this system, use the fact that the second step is the slow step to invoke the steady-state approximation.

    Note that there is no k -2 for this reaction: the enzyme will not catalyze the conversion of the product back to the substrate. Given all of these constraints, the math looks like what follows. To make sure that you understand it, try to reproduce the answer from the given reaction sequence and the constraints detailed in this paragraph. If [S] is really big i. The last topic to consider before we leave kinetics and go back for a last look at thermo is integrated rate laws. Given the elementary steps, it is possible to integrate the corresponding rate law, using calculus to solve for the concentration of some reaction species as a function of time.

    Test yourself heavily on both first order and second order rate law integration. From the equation for the elementary step, you should be able to figure out the concentration of the species as a function of time. Remember, this is just math. The chemical parts are only the first line and the last line of each derivation. Here are the answers you should be able to derive:. Actively test yourself on this!! You may be able to follow all the math, but could you reproduce it? Common examples of first order reactions are radioactive decay processes.

    As must be the case for first order reactions, the rate of decay depends on the amount of radioactive stuff that's around at any one time. This results in an exponential decrease in the decay rate with time. When less stuff is around, it does not decay anywhere nearly as fast. When chemists talk about radioactive decay, they typically like to talk about half-lives.

    The half life of a substance is how long it takes for half of whatever's there to decay. K Little k Thermodynamic, not kinetic Kinetic, not thermodynamic K doesn't really have units, though we often treat it as if it does. Rate constant changes with T and with catalyst. A catalyst would change the activation energy for the rate-determining step.