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Chapter 8 Solutions Engineering and Chemical Thermodynamics 2e - PDF Drive.
Since this change in energy can be attributed to a change of the macroscopic position of the system and is not related to changes on the molecular scale, we determine the form of energy to be potential energy.
This argument can be enhanced by the form of the expression that the increased energy takes. If we consider the spring as the system, the energy it acquires in a reversible, compression from its initial uncompressed state may be obtained from an energy balance.
Assuming the process is adiabatic, we obtain:. We have left the energy in terms of the total energy, E. The work can be obtained by integrating the force over the distance of the compression:. It should be noted that there is a school of thought that assigns this increased energy to internal energy. This approach is all right as long as it is consistently done throughout the energy balances on systems containing springs. In the second situation, the gas is the system.
If heat transfer, potential and kinetic energy effects are assumed negligible, the energy balance becomes. Since work must be done on the rubber band to stretch it, the value of the work is positive. From the energy balance, the change in internal energy is positive, which means that the temperature of the system rises. When a gas expands in a piston-cylinder assembly, the system must do work to expand against the piston and atmosphere.
Therefore, the value of work is negative, so the change in internal energy is negative. Hence, the temperature decreases. In analogy to the spring in Problem 2. At sufficiently high temperatures, a portion of the water droplet is instantly vaporized. The water vapor forms an insulation layer between the skillet and the water droplet. At low temperatures, the insulating layer of water vapor does not form. The transfer of heat is slower through a gas than a liquid, so it takes longer for the water to evaporate at higher temperatures.
HOT System. If the entire apartment is treated as the system, then only the energy flowing across the apartment boundaries apartment walls is of concern. In other words, the energy flowing into or out of the refrigerator is not explicitly accounted for in the energy balance because it is within the system.
By neglecting kinetic and potential energy effects, the energy balance becomes. The W term represents the electrical energy that must be supplied to operate the refrigerator. To determine whether opening the refrigerator door is a good idea, the energy balance with the door open should be compared to the energy balance with the door closed. In both situations, Q is approximately the same. However, the values of W will be different. With the door open, more electrical energy must be supplied to the refrigerator to compensate for heat loss to the apartment interior.
Q Q Heater. Heater W. State 1, when you leave in the morning, and state, the state of your home after you have returned home and heated it to the same temperature as when you left. The case where more heat escapes will require more work and result in higher energy bills. When the heater is on during the day, the temperature in the system is greater than when it is left off. Since heat transfer is driven by difference in temperature, the heat transfer rate is greater, and W will be greater.
Hence, it is cheaper to leave the heater off when you are gone. The mass can be found from the known volume, as follows:. As in Example 2. The specific enthalpy is found from values in Appendix B. The values in parts a and b agree very well. The answer from part a will serve as the basis for calculating the percent difference since steam table data should be more accurate. Therefore, the work will be negative and. Pa th 2. Since internal energy is a function of temperature only for an ideal gas Equation 2.
Equation 2. During the constant volume part of the process, no work is done. The work must be solved for the constant pressure step. Since it is constant pressure, the above equation simplifies to. No work is done, and the kinetic and potential energies can be neglected. The energy balance reduces to. Potential and kinetic energy effects can be neglected. Therefore, the energy balance becomes. The value of the work will be used to obtain the final temperature. The definition of work Equation 2.
In this process the gas is initially at 2 bar, and it expands against a constant pressure of 1 bar. Therefore, a finite mechanical driving force exists, and the process is irreversible. To solve for the final temperature of the system, the energy balance will be written. The piston- cylinder assembly is well-insulated, so the process can be assumed adiabatic.
Furthermore, potential and kinetic energy effects can be neglected. The energy balance simplifies to. Open navigation menu. Close suggestions Search Search. User Settings. Skip carousel. Carousel Previous. Carousel Next. What is Scribd? Explore Ebooks. Bestsellers Editors' Picks All Ebooks. Explore Audiobooks. Bestsellers Editors' Picks All audiobooks. Explore Magazines.
Editors' Picks All magazines. Explore Podcasts All podcasts. Difficulty Beginner Intermediate Advanced. Explore Documents. Uploaded by Abolfazl. Did you find this document useful? Is this content inappropriate? Report this Document. Alternatively, if we use Equation E2.
Such a process is illustrated below: States 1 and 2 are at the higher temperature. The process from state 1 to state 2 is a reversible, isothermal expansion. The value of P1 is not known, but recognizing that the process from state 4 to 1 is an adiabatic compression provides an additional relation. The polytropic relationship can be employed to find P1. This equation will be used to calculate the work for the remaining processes.
The coefficient of performance is defined in Equation 2. Its sign is positive. The power obtained from the turbine is the area under the curve from state 1 to 2.
Its sign is negative. The area under the latter curve is much larger remember the log scale ; thus the net power is negative. Therefore, no work is done. Therefore, the randomness decreases. Problem 2. Moreover, P2 is known, so we only need to find T2 in order to calculate the entropy change. Since the process is reversible and adiabatic, we know from Table 3. First, we will determine s 2. The first law can be applied to constrain state 2.
We wish to use the steam tables to calculate the entropy change of liquid water as it goes from its freezing point to its boiling point. The steam tables in Appendices B. However, there is data for saturated water at 0.
If we believe that the entropy of water is weakly affected by pressure, then we can say that the entropy of water at 0. The molar volumes of most liquids do not change much with pressure at constant temperature.
Thus, the molecular configurations over space available to the water molecules do not change, and the entropy essentially remains constant. We do not need to consider the molecular configurations over energy since the temperature difference is so slight. There are many ways to reconcile this difference, but think about it from a molecular point of view. In process i , the available molecular configurations over energy are increased as the temperature increases.
As the temperature increases, the molar volume also increases slightly, so the 17 available molecular configurations over space also increase. Now consider process ii , where the molecules are being vaporized and entering the vapor phase.
The molecular configuration over space contribution to entropy is drastically increased in this process. In the liquid state, the molecules are linked to each other through intermolecular interactions and their motion is limited. In the vapor state, the molecules can move freely.
Refer to Section 3. Let the mixture of ice and water immediately after the ice has been added represent the system. Now, we can calculate the change in entropy. From Equation 3. We also know the entropy change for the universe is zero for reversible processes. Because the system is insulated, the work done by the mass being added to the piston is transformed into molecular kinetic energy.
The process in part a can be drawn as follows: If the process were irreversible, more work is required, and since the process is adiabatic, the change in internal energy is greater. Since the change in internal energy is greater, so too is the change in temperature. Therefore, the final temperature would be higher than the temperature calculated in Part b.
Since we know the pressure and internal energy at state 2, the entropy is constrained. Information given in the problem statement also constrains state 1. If you take the system to be the entire tank both sides of the partition , then no net work is performed as the gas leaks through the hole. Furthermore, the tank is well insulated, and kinetic and potential energy effects can be neglected.
The ideal gas law can be used to calculate the moles of gas present. After the partition ruptures, the pressure and temperature will remain constant at 1 bar and K, respectively. This can be shown by employing mass and energy balances. During the throttling process, no shaft work is performed and the rate of heat transfer is negligible.
Now test the first law of thermodynamics by writing an energy balance. For the second law to be valid, the rate of change in entropy of the universe must greater than or equal to 0, i. For constant heat capacity, we can calculate the entropy differences using Equation 3.
We must be careful, however, to select an expression which does not assume a constant heat capacity. Before this problem is solved, a few words must be said about the notation used. First, the mass of water present in each part of the system will be calculated. The following equation shows how the change in internal energy can be calculated. It has two components: component a and component b. To find the final temperature, first find the volumes using the ideal gas law.
The minimum pressure is required for a reversible process. The process is adiabatic, and potential and kinetic energy effects can be neglected. The initial state is the same as for Problem 3.
Since the water only expands against 1 bar, the work is lower than that for the differential process described in Problem 3. Thus, this adiabatic process looses less energy, leading to a higher final temperature. Another way to view this argument is to look at this process as a closed system. This depiction is the expansion analog of the compression process depicted for Example 2.
We can represent this process in terms of two latches, one that keeps the process in its initial state at 60 bar and one that stops the expansion after the pressure has reduced to 20 bar.
The process is initiated by removal of the first latch and ends when the piston comes to rest against the second latch. The corresponding reversible process of Problem 3. Clearly the process on the left does less work, resulting in a greater final temperature. Problem 3. First, find quality of the water. First, a summary of known variables is provided.
First, start with the turbine. For the entire Rankine cycle, Equation 3. This can be done in a variety of ways: 1. Increase the degree of superheating of steam in the boiler. This process is sketched in the upper left hand Ts diagram below. This change reduces the moisture content of steam leaving the turbine. This effect is desirable since it will prolong the life of the turbine; however, if the steam is heated too high, materials limitations of the turbine may need to be considered.
Lower the condenser pressure. Lowering the pressure in the condenser will lower the corresponding saturation temperature. This change will enlarge the area on the Ts diagram, as shown on the upper right below.
Thus, we may want to consider lowering the pressure below atmospheric pressure. We can achieve this since the fluid operates in a closed loop. However, we are limited by the low temperature off the heat sink that is available. Increase the boiler pressure. This will increase the boiler temperature which will increase the area as shown on the bottom left Ts diagram.
We can be more creative about how we use the energy available in the boiler. One way is to divide the turbine into two stages, Turbine 1 and Turbine 2, and reheat the water in the boiler between the two stages. This process is illustrated below. The pressure in Turbine 1 will be higher than the pressure in turbine 2. This process is schematically shown on the bottom right Ts diagram. This process leads to less moisture content at the turbine exit desirable and limits the temperature of the superheat desirable.
First start with the turbine state 2. The enthalpy of state 4 can be calculated from Equation 3. State Thermodynamic State Sat. Vapor Sat. And Vapor Sat. Mixture 0. Vapor 0. Liquid 0. Equation 3. Ra l at 0. Ra v at 0. Using Equation 3. An isentropic turbine adds significant level of complexity to the cycle.
Turbines are expensive and wear over time. The cost of the turbine is not justified by the increase in COP. R liquid at 0. R vapor at 0. Ra vapor at 0. One possible refrigeration cycle is presented below.
A number of fluids will work sufficiently for this system, but the design process will be illustrated using Ra. For states 3 and 4, the pressure is constant, and for states 1 and 2, the pressure is constant at a different value.
The listed saturation temperatures are the temperatures at which the fluid evaporates and condenses. Now that the pressures are known, we can compute the required flow rate required in order to provide 20 kW of cooling. To define each state, we need to thermodynamically constrain each state. Now, we need to find the liquid and vapor compositions of states 1, 2, and 3 to completely constrain the states. The four states of the magnetic material are shown of the sT diagram below.
Note the axis are shifted from the usual manner. Work is supplied to magnetize the material and to spin the wheel. When stretched the polymer chains tend to align. The alignment decreases the spatial configurations the polymer can have, and therefore, reduces that component of entropy.
If the process is adiabatic, the entropy of the system cannot decrease. Consequently, its thermal entropy must increase. The only way this can be accomplished is by increased temperature. The following data was taken from Table A. This problem shows that the entropy change of the system is negative, but nothing has been said about the entropy change of the universe.
We must look at the change in entropy of the surroundings to determine if the second law is violated. By looking at the enthalpies, we see that the reaction is exothermic, which means that heat is transferred from the system to the surroundings.
Therefore, the entropy of the surroundings will increase during this reaction. Therefore, the only remaining entropy contribution, the randomness of the atomic arrangements, must be considered. The randomness does not increase when CdTe forms.
In pure crystals of Cd and Te, the location of each atom is known because the crystal lattice constrains the atomic locations. In CdTe, the crystal lattice still defines the location of each atom, so the randomness has not increased. Therefore, the change in entropy is zero. Morris is arguing that since evolution results in more order, the second law of thermodynamics is violated, so evolution must be impossible.
However, the flaw in this argument is caused by ignoring the entropy change of the entire universe. The second law states that the entropy of the universe will remain constant or increase for any process. A system can decrease in entropy if the entropy of the surroundings increases by at least that much.
The greater the number of configurations, the more probable the state is and the greater the entropy. We can qualitatively relate this concept to the possible hands in a game of poker, but it is more interesting to quantify the results using some basic concepts of probability. We consider a hand of poker containing 5 cards randomly draw from a deck of 52 cards.. There are a finite number of permutations in which we can arrange a 52 card deck in 5 cards.
For the first card in the hand, we select from 52 cards, the second card can be any other card so we select from 51 cards, the third card has 50 cards, and so on.
We do this math in a similar way. For a given hand there are five cards we can pick first, four we can pick second, …. There are 13 different possible ranks of four of a kind, one for each number A, 2, 3, 4, 5, 6, 7,8, 9, 10, J, Q, K.
The fifth card in the hand could be any of the other 48 cards. The coulombic forces between the gas molecules affect the system pressure. This modification is similar to the van der Waals equation.
Since we are limited to 1 parameter, we need to choose the most important interaction. Since net electric point charges exert very strong forces, this effect will be more important than size. Therefore, the system pressure increases, i. Coulombic potential energy is proportional to r For both O 2 and propane, the dipole moments are zero. The value of similar. The intermolecular distance of molecules is greater at lower pressures.
Therefore, fewer intermolecular interactions exist, which cause less deviation from ideality. At higher temperatures, the kinetic energy of the molecules speed is greater. The molecules interact less; thus, the compressibility factor is closer to unity.
Dispersion is the controlling intermolecular force in this system, and its magnitude is directly related to the size of the molecules polarizability component of dispersion.
For the 5-monomer long polymer chain, a is 5 times the a value in Part b. The b parameter is also related to the size of the molecule since it accounts for the volume occupied by the molecules. Again, the b parameter for the reduced polymer chain is 5 times the b value from Part b. To illustrate the principles in Chapter 4, two of the simplest solution methods will be illustrated.
Method 1. Polarizability of each atom The polarizability of a molecule scales with the number of atoms; the polarizabilities of individual atoms are additive. The polarizability of C-Cl bonds is calculable with the polarizability of chloroform. This value predicts the polarizabilities of the other species in the table reasonably well. More accurate values for the polarizabilities can be calculated using more of the data given in the problem.
Because these are non-polar, diatomic molecules, only dispersion forces are present. Dispersion forces depend on the first ionization potential and polarizability.
Ionization energy is approximately equal for each molecule. Diethylether and n-butanol have the same atomic formula and similar spatial conformations. Therefore, they should be about equal in size. Methyl ethyl ketone has fewer atoms, but has two exposed electron pairs on the double-bonded oxygen.
Now we must determine if there is greater attraction in n-butanol or methyl ethyl ketone. Since induction and dispersion forces are similar in these molecules, we must consider the strength of dipole-dipole forces. Intermolecular attractions are greater, so the magnitude of molecular potential energy is greater. The potential energy has a negative value for attractive interactions.
The molecular kinetic energy is identical since the temperature is the same. Hence, the internal energy, the sum of kinetic and potential energies, is less at 30 bar. At K and 30 bar, isopropanol and npentane are both liquids.
The hydrogen bonding and dipole-dipole interactions are present in isopropanol, and dispersion is present in n-pentane. The intermolecular forces are greater in the isopropanol, so the compressibility factor is smaller for isopropanol.
At K and 30 bar, both species are gases. In the gas phase, hydrogen bonding does not play a significant role. The dispersion forces in n-pentane are stronger than the dipole-dipole forces of isopropanol. Therefore, the compressibility factor is smaller for n-pentane. For real NH 3 , the strong intermolecular forces dipole-dipole and dispersion cause the molar volume to decrease.
They outweigh the volume displaced by the physical size of NH 3 ; thus, z will be less than one. In the real gas, intermolecular attractions are present. Internal energy value is the sum of potential and kinetic energies of the molecules. The absolute values of the kinetic energies are identical at identical temperature: however, the potential energy decreases for real NH 3 due to attractive interactions - so the internal energy is less for the real NH 3.
Ammonia has an electric dipole in which positive and negative charge are separated. The intermolecular forces in the real gas cause the molecules to align so that the positive charge in one molecule is adjacent to a negative charge in a neighboring molecule to reduce potential energy. Therefore, fewer possible configurations exist, which creates less randomness and lower entropy.
With Ne, very weak intermolecular attractions are present, so volume displacement becomes important. The compressibility factor will be slightly greater than one. In NH 3 , the strong intermolecular attractions decrease the molar volume, so z is less than one.
The compressibility factor is greater for Ne. Both species are gases at these conditions. The intermolecular attractions present in NH 3 reduce the number of possible configurations. The weak forces present in Ne have a much smaller effect. However, NH 3 is asymmetrical, while Ne is symmetrical. The asymmetry of NH 3 results in more possible configurations that NH 3 can have.
Therefore, it is difficult to qualitatively show for which case the entropy is greater. Since both species are gases, intermolecular interactions are relatively weak, and we can guess that entropy is greater in NH 3. The molar volume can be found from the compressibility factor.
Clearly, London interactions are much more important, and gravitational effects can be neglected. As provided in the text, Equation 4. We need to choose reasonable criteria to specify. Other choices may be just as valid; you should realize that you have two parameters to fit and so must specify two features. The most stable configuration the bottom of the well occurs at a greater separation for the exp model. The Lennard-Jones potential increases more steeply at small radii, i.
The two models are in reasonable qualitative agreement 18 4. The interaction between the chlorine and sodium ions is Coulombic attraction. QNa QCl 4. The stronger the intermolecular attraction, the higher the boiling point. Dispersion and dipole-dipole interactions are present in all five species listed. The magnitude of the dipole-dipole interactions is similar so the pertinent intermolecular force in these molecules is dispersion.
The molecular size increases from left to right. Polarizabilities are greater in larger molecules, which manifests in larger dispersion forces. Therefore, the boiling point increases from left to right. The potential energy can be quantified with the Lennard-Jones potential function. The bond length is the r value where the potential is a minimum. From Table 4. The values of a for methane, ammonia, and water are of the same magnitude because the sums of the intermolecular attractions for each molecule are similar.
All three molecules have comparable dispersion forces; although slightly weaker in the ammonia and water. However, unlike methane, these two molecules also have dipole-dipole and induction forces. In fact, the strong dipole in water gives it the largest value The values of b are all of the same magnitude, as expected since b scales with size according to the number of atoms in the molecule. The values for the above equation were taken from Table A.
The basic potential result presented in the text assumes that the species are evenly distributed throughout the volume. It does not take into account the structure given to the fluid through intermolecular forces. In fact, a more careful development includes a radial distribution function, which describes how the molecular density of the fluid varies with r.
The radial distribution function depends on pressure and temperature of the fluid. They simply represent experimental data better. We can use our knowledge of intermolecular forces, however, to explain why these may work better. If we compare these equations to the van der Waals equation, we note that the first term on the right hand side of all three equation is identical; we accounted for this term as a correction for finite molecular size or alternatively repulsive interaction due to the Pauli Exclusion Principle.
This form represents a hard sphere model. The second term, that which deals with intermolecular attraction, is different in all three models.
Both of the later equations include a temperature dependence in this term. We have seen that if attractive forces depend on orientation dipole-dipole , they fall off with T as a result of the averaging process recall discussion of Equation 4.
Another explanation goes as follows: as T increases, the molecules move faster, reducing the effect of intermolecular forces. If we say that the potential energy between two molecules depends on the amount of time that they spend close to each other, then it would be inversely related to velocity The faster molecules are moving, the less time they spend in the vicinity of other species.
In this case, the correction term would go as V-1, where V is the molecular velocity. The inclusion of a "b" term in the second term may help relax van der Waal's "hard sphere" model with a more realistic potential function, i. It makes sense that this should be included in the force correction since this is taking into account repulsive forces. Thus it has more opportunity for attractive interactions than the larger species.
The Peng-Robinson equation exhibits the most complicated form in an attempt to better fit experimental data. Part a is not as accurate as Part b and Part c because water is not an ideal vapor.
The value from Part b is 1. From Equation 4.
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