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Solution Manual for Introduction to Hogg R. Mathematical Statistics 6th Edition []. Mechanical Engineering Series K. Gupta auth. West Theory, Second Edition [2nd ed. Wade and Solution Manual [8th ed. D C Stage 3. Entropy, S Figure 3. The efficiency is defined in [3A. Equating them an expression for the net work is obtained.
For the process to be permissible by the Second Law the Clausius inequality [3A. Answer to discussion question Solutions to exercises E3B. As discussed in Section 3B. For vaporisation this becomes. Thus, the increase in the molar entropy of nitrogen gas is. Therefore the final temperature is. Although the above result may seem self-evident, the more detailed explaina- tion is as follows. The heat capacity at constant volume is defined in [2A.
As shown in Section 2A. The temperature variation of the entropy at constant volume is given by [3B. Thus the overall process can be broken down into steps that are easier to evaluate. First consider heating the initial system at constant pressure to the final temperature. The variation of entropy with temperature at constant pressure is given by [3B. Next consider an isothermal change in pressure. First consider heating the ice at constant pressure from the initial temperature to the melting point, Tm.
Next consider the phase transition from solid to liquid at the melting temper- ature. The entropy change of a phase transition is given by [3B. Then the liquid is heated to the boiling temperature, Tb. In analogy to the first step.
Solutions to problems P3B. First consider cooling the water at constant pressure to from the initial temper- ature T to that of freezing, Tf. Entropy variation with temperature at constant pressure is given by [3B. Consider the enthalphy change for the same path. The variation of the enthalpy with temperature at constant pressure is given by [2B. Thus for the first and third steps, respectively.
Because the change in the total entropy is negative, the Second Law implies that the process is not spontaneous. The variation of the entropy with temperature is given by [3B. This is because the temperature variation of the heat capacity over this range is relatively small. First consider condensation of vapour.
Given that all the vapour turns to liquid water, the heat released is the opposite of that of vaporization. The newly formed liquid water is initially at the boiling point. The next step is the water cooling from the boiling to final temperatures. The en- thalpy change associated with a temperature change is given by [2B. The total entropy change is therefore. DU as a function of temperature is given by [2A. This is rearranged to give a ratio of temperatures. As explained in Section 3B.
Thus, the entropy changes for the steps 2 and 4 are, respectively. TD TD. The empirical expression for the constant-pressure molar heat capacity is given by [2B.
Using this, the expression for the molar entropy variation becomes. Answer to discussion question Solutions to exercises E3C. The temperature dependence of the entropy T is given by [3C. For a given temper- ature T the change in molar entropy from zero temperature is therefore.
Standard reaction entropy is given by [3C. Therefore, using data from the Resource section. Solutions to problems P3C.
The temperature dependence of the entropy is given T by [3C. The entropy change of a phase transition is given by [3C. Finally, the contribution of the second phase transition is.
The Third-Law standard molar entropy at K is the sum of the above con- tributions. The temperature dependence of the T entropy is given by [3C. Thus for a given low temperature T the change in the molar entropy from zero is. The plot fits well to a polynomial of order 4. The standard Third-Law molar entropy at K is the sum of the both contri- butions. For entropy at K, the above integral in needs to be calculated to the required limit. The standard reaction enthalpy is given by [2C.
The temperature dependence of the reaction entropy is given by [3C. For the reaction at K. The plot is found to be well fitted by a polynomial of order 2. The integral of the fitted functions over the range x i to x f is y x f 5.
Assuming the heat capacity to be constant over the temperature range of inter- est, the change in entropy as a function of temperature is given by [3B. The difference is slight as expected because C p,m does not vary significantly in the given temperature range.
Thus, the Debye contribution. Answers to discussion questions D3D. Both the Helmholtz and Gibbs energies refer to properties of the system alone. The second term is the entropy change of the system.
Thus the sum of the two is the total en- tropy change, which the Second Law shows must be positive in a spontaneous process. Therefore, the change in the Helmholtz energy is an indicator of the total entropy change, even though the former refers only to the system. Similar considerations can be applied to the Gibbs energy.
It is also possible to express the criterion for spontaneity in terms of the change in H, U or S for the system. However, the variables which are being held constant here U and V do not correspond to such easily realizable conditions such as constant temperature and volume or pressure so such criteria are less applicable to chemical sys- tems. Solutions to exercises E3D. The standard reaction enthalpy is given in terms of the enthalpies of formation by [2C. The standard reaction Gibbs energy is given by [3D.
The standard reaction entropy is given by [3C. Therefore, for the formation of the compound. The standard reaction Gibbs energy is defined in [3D.
Solutions to problems P3D. The power density is the work that is done in a time interval for a unit volume. The standard reaction entropy is [2C.
The Gibbs en- ergy of solvation in water is given by Born equation [3D. Be- cause the Born equation [3D. The standard re- action Gibbs energy is given by [3D.
The model is derived considering only the net work needed to transfer a charged ion from vacuum to a given medium. This implicitly assumes no heat exchange during the process hence resulting in zero entropy contribution.
Answer to discussion questions D3E. Solutions to exercises E3E. This allows rewriting the previous equation for the change in Gibbs energy due to isothermal gas expansion. The change in entropy is thus.
Expressing for the change in enthalpy gives. Assuming that the volume of liquid water changes little over the range of pressures considered. Solutions to problems P3E. Similarly, the standard reaction enthalpy is [2C. The result is expected as the relative volume expansion for the van der Waals gases is always slightly greater than that of the perfect gas due to the fixed finite excluded volume.
Therefore, the entropy is expected to be more sensitive to the changes in volume, especially at the very small volume. Answers to integrated activities I3. In a system at high temperature the molecules are spread out over a large number of energy states.
Increasing the energy of the system by the transfer of heat makes more states accessible, but given that very many states are already occupied the pro- portionate change in W is small. In contrast, for a system at a low temperature fewer states are occupied, and so the transfer of the same energy results in a proportionately larger increase in the number of accessible states, and hence a larger increase in W.
This argument suggests that the change in entropy for a given transfer of energy as heat should be greater at low temperatures than at high, as in the thermodynamic definition. As discussed in Section 3C.
At a molecular level the absence of thermal motion in a perfectly localized crystalline solid is interpreted as there is only one way to arrange the molecules like that. Answers to discussion questions D4A. Physically, an incompressible system does not store energy like a spring, which is the analogy for a compressible gas; however, it can transmit energy in the same way that a hydraulic fluid does.
Furthermore, an incompressible system under pressure is under stress at a molecular level. Its bonds or intermolecular repulsive forces resist external forces without contraction. Then proceed on an clockwise path centred on the critical point. Eventually the path will reach the liquid—vapour phase boundary, and if the traverse is stopped at this point the sample will be found to consist of liquid at the bottom of the tube with vapour above, and a visible meniscus.
As the path is continued the system moves into the liquid phase in principle the meniscus would rise up the tube and then disappear and once more only one phase is present.
Eventually the path takes the system above the critical point and back to the starting point. The somewhat curious thing about this path is that it takes us from vapour to liquid via the usual process of condensation, but then returns the liquid to the vapour phase without crossing a phase boundary, and so with no visible boiling.
Solutions to exercises E4A. Point a lies within an area and therefore only one phase is present. Point d lies on the boundary between two areas, and therefore two phases are present. Points b and c each lie at the intersection of three phase boundaries, so in each case three phases are present. So three phases Ice I, liquid water, and gas will be present.
Solutions to problems P4A. On increasing the pressure, a point will be reached at which gas and solid are in equilibrium. Above this pressure only the solid form will exist. A single phase a supercritical fluid therefore exists at all pressures except perhaps at extremely high pressures when a solid might be formed. Below 5. Above this pressure, the phase diagram shows that only the solid phase will be present.
Note that in reality the phase boundaries may be curved rather than straight. There is one triple point which is marked with a dot. Answers to discussion questions D4B. Because the molar volume is always positive, the slope of the change in chemical potential with respect to change in pressure is positive: that is, the chemical potential increases with increasing pressure.
Solutions to exercises E4B. The chemical potential of the liquid rises by more than that of the solid. So if they were initially in equilibrium, the solid will be the more stable phase at the lower temperature. M At the melting temperature the entropy of fusion is 2. It does not matter that the pressure is given in units of Torr because only the slope of ln p is required.
K or The standard boiling point is therefore lower than the normal boiling point by The rate of vaporization is then rate of energy absorption 8. This is less than the partial pressure of water in the atmosphere, so yes , the frost will sublime. A partial pressure of 0. Solutions to problems P4B.
This temperature is then substituted back into one of the expressions for the pressure The difference in slope on either side of the normal freezing point of water is therefore. From the data, this is at The plot is shown in Fig. The data fall on a good straight line, the equation of which is. The Clayperon equation [4B. The fractional increase in vapour pressure is therefore 0. The second derivative is therefore. Heat capacity is invariably positive, so this expression implies a negative cur- vature since T cannot be negative.
For water, the curvatures of the liquid and gas lines at the normal boiling point Solutions to integrated activities I4. The The standard melting point is therefore estimated to be The expression for ln p is inserted and differentiated, and then evaluated at the standard boiling point found above. This value is substituted into the fitted function to give. Answers to discussion questions D5A. Changing the composition of a mixture gives rise to a change in Gibbs energy, given by [5A.
It therefore follows that. This equation relates the chemical potential of a component in a mixture to its mole fraction.
Solutions to exercises E5A. The total volume is calculated from the partial molar volumes of the two com- ponents, [5A. If it is assumed that the differential can be replaced by the small change. Therefore [5A. Because the separate volumes are equal, and at the same pressure and temperature, each compartment contains the same amount of gas, so the mole fractions of each gas in the mixture are equal at 0.
Under these conditions the Gibbs energy of mixing is given by [5A. As expected, the entropy of mixing is positive and the Gibbs energy of mixing is negative. If A is 1,2-dimethylbenzene and B is 1,3-dimethylbenzene the total pressure is. The composition of the vapour — that is the mole fractions of A and B in the vapour — is calculated from their partial pressures according to [1A. These two expressions for p A is equated to give.
The task is therefore to find the amount in moles, n A and n B , of A and B in a given mass m of solution. The total volume of a solution of A and B is calculated from the partial molar volumes of the two components using [5A. The above expression for VA evaluates as. A test of this law is to make a plot of p B against x B which is expected to be a straight line with slope K B ; such a plot is shown in Fig. Molality is the amount of solute per kg of solvent.
Solutions to problems P5A. The partial molar volumes are functions of the composition of the mixture which is specified by x A and x B. If the variation of VA with x A is known, then the integral on the right can be evaluated and hence a value found for VB.
Let component A be trichloromethane and component B be propanone. The data fall on a good straight line which implies that the partial molar vol- ume the slope is nearly constant with composition. Equation 5. Using these rela- tionships the following table is drawn up, and the data are plotted in Fig.
The constant is estimated from the partial pressure at the smallest mole frac- tion; these data are selected from the table. Answers to discussion question D5B. The volume of mixing of an ideal solution is zero, which can be understood at a molecular level as a result of the A and B molecules fitting together in just the same way that A or B molecules fit with one another.
A thermodynamic explanation is that for an ideal system the partial molar volume is not a function of composition. Melons are large relative to oranges. When melons pack together it is conceiv- able that oranges could fit into the interstices between the melons. As the number of oranges is increased, the interstices would eventually all be filled, presumably when the ratio of oranges to melons is around After this point, adding oranges will result in an increase in the volume and it is conceivable that poor packing of objects of different sizes could result in the volume of the mixture being greater than that of the separated species.
In this case the excess volume will be positive. When a melons are added to oranges the volume will always increase because there are no spaces for the melons to occupy. If just a few melons are added then these will not disrupt the packing of most of the oranges, so the excess volume will be zero.
However, as more are added the inefficiency of packing will result in a positive excess volume. Differences in boiling-point constants are therefore identified as being due to differences in the boiling points of the pure liquids.
Water and benzene have different boiling points and so have different boiling-point constants. The chemical potential of the solvent in the solution is lower than that of the pure solvent, therefore there is a tendency for the solvent to pass through the membrane from the side on which it is pure into the solution because this results in a reduction in Gibbs energy.
Solutions to exercises E5B. The task is to work out the mole fraction that corresponds to the given molality. The molar mass of A is The task is to relate the mole fraction of A to the masses of A the solvent and B the solute , and to do this the molar masses M A and M B are introduced. The molar mass of the solvent 2-propanol C3 H8 O, A, is From the data and the known value of K f it is possible to calculate b B.
The task is then to relate this to the given masses and the desired molar mass of the solute, M B. A volume cm3 of water has mass g to a good approximation.
Using these values with the data given and the value of the freezing-point constant from the Resource section gives the freezing point depression as. Kf mB 1. The new freezing point is therefore The task is to relate [B] to b B so that these two relationships can be used together. In this expression all of the quantities are in SI units therefore the temperature is expected to be in K, which is verified as follows. The total amount in moles is 1. As equal amounts in moles of the two components are mixed, the mole fractions of each are This is evident from Fig.
The task is to relate the mole fraction of A ethylbenzene to the masses of A and B benzene , and to do this the molar masses M J are introduced. The molar mass of solvent bismuth is The molar mass of solute Pb is With so much more solute Pb than solvent Bi, the solution cannot really be described as Pb dissolved in Bi. The mixture will boil when the sum of the partial vapour pressures of A and B equal the external pressure, here 19 kPa.
These relationships give rise to four equations. In these equations x A and p tot are the unknowns to be found. The expressions for p A are set equal, as are those for p B , to give. Solutions to problems P5B.
The ratio of b B,app to b B gives an indication of the number of species present per molecule dissolved b B,app 0. Therefore, each molecule of Th NO3 4 appears to dissociate into about 4 species in solution.
The first task is to relate the quoted solubility to the mole fraction of the solute. The data gives the solubility in g of solute per g of solvent: let this quantity be S and, for convenience, let the mass of solvent be m A.
With this the relationship for the ideal solubility is developed as. The data do not, even at the simplest level, conform to the predictions of the ideal solubility equation. To use this definition an expression for G as a function of the n J is required.
The excess Gibbs energy G E is defined in [5B. These quantities are all molar, so the Gibbs energy of a mixture of n A moles of A and n B moles of B is. The ideal Gibbs energy of mixing per mole is given by [5B. The final expression for G is.
The algebra used in going to the last line is used to make it easier to compute the derivative. In finding the derivative recall that terms such as n A ln n A require the application of the product rule. This function is plotted in Fig. As g increases the deviation from ideal behaviour the solid line increases, with the effect being larger at small x A , corresponding to larger x B.
As described in Section 5B. The data are plotted in Fig. It is evident from the graph that the data do not conform to this expecta- tion. One approach is to select only the data at the lowest concentrations because in this limit the expectation is that the two-term virial equation will be sufficient to describe the data. Such a plot is shown in Fig. The data fit reasonably well to a straight line with intercept 2. This may be attributed to both solvent and polymer being non-polar.
The agreement with the results from part d is modest. The relationship is plotted for several temperatures in Fig. For the plots shown in Fig. The plot shows a maximum at much higher and rather unrealistic temperatures. Answers to discussion questions D5C. These intermolecular interactions are determined by factors such as dipole moment polarity and hydrogen bonding.
Conversely, a high-boiling azeotrope has a boiling temperature higher than that of either component, so it is more difficult for the molecules to move into the vapour phase. This reflects the relatively unusual situation of components that have more favorable intermolecular interactions with each other in the liquid phase than with molecules of their own kind. Solutions to exercises E5C. The horizontal axis is labelled z A , which is interpreted as x A or y A according to which set of data are being plotted.
In addition to the data in the table, the boiling points of the pure liquids are added. The exact points of intersection can be found either from the graph or by using the fitted functions. The level rule shows that there is about 2. As the temperature is raised the B-rich phase becomes slightly less rich in B, and the other phase becomes richer in B.
The lever rule implies that the proportion of the B-poor phase increases as the temperature rises. At temperature T2 the vertical line intersects the phase boundary. The mole fraction of aniline A is. The shape conforms to the expected phase diagram for such a system. When the mole fraction of A is high a single phase forms, but as the mole fraction goes below about 0. Initially, according to the lever rule, the proportion of the B-rich phase is very small, but as more and more B is added the proportion of this phase increases.
When the x A is just over 0. Solutions to problems P5C. The horizontal axis is labelled z B , which is interpreted as x B or y B according to which set of data are being plotted. In addition to the data in the table, the boiling point of pure chlorobenzene is added.
The resulting phase diagram is shown in Fig. By the lever rule, [5C. An alternative way to explore the solutions to eqn 5. The values of x A at which the two curves intersect gives the position of the minima. Mathematical software can also be used to solve eqn 5. Answers to discussion questions D5D.
Incongruent melting means that the compound AB2 does not occur in the liquid phase. Solutions to exercises E5D. The solid points are the data given in the Exercise, and lines are simply plausible connections between these points. The dash-dotted lines are referred in to Exercise E5D.
The mixture cools until about K at which point solid first forms. The solid which forms is the compound and as this happens the liquid becomes richer in B2 H6. Cooling continues with more and more of the compound being formed until K, the second eutectic, at which point the system completely solidifies to a two-phase material consisting of the compound and solid B2 H6.
The cooling curves are shown in Fig 5. The break points, where solid phases start to form are shown by the short horizontal lines, and the dotted lines in- dicate the temperatures of the two eutectics K and K. The horizontal segments correspond to solidification of a eutectic. Cooling curve c corre- sponds to direct solidification of the complex. The incongruent melting point is marked T1.
For iso- pleth a the first break point is where the isopleth crosses the liquid curve be- tween temperatures T1 and T2 ; this is followed by a eutectic halt at T2. For isopleth b the first break point is somewhat above T1 where the isopleth crosses the liquid curve, there is a second break point where the isopleth crosses the boundary at T1 , and then a eutectic halt at T2. Although the liquid does not contain any AB2 units, the liquid can be thought of as a mixture of dissociated AB2 in A.
Let the amount in moles of the compound be n c and that of free A be n a. The mole ratio of compound to free A is given by nc x2 0. Solutions to problems P5D. Using this expression, mole fractions corresponding to the three compounds are: 1 P4 S3 , 0.
The phase diagram is a variation of that shown in Fig. The diagram there- fore separates into four sections, as shown schematically in Fig. Note that no information is given on the temperature or composition of the eutectics, so these have simply been selected arbitrarily. The diagram has four eutectics labelled e 1 , e 2 , e 3 , and e 4 ; eight two-phase liquid-solid regions, t 1 through t 8 ; and four two-phase solid regions, S1 , S2 , S3 , and S4.
The composition and physical state of the regions are as follows:. The cooling curve is shown to the right of the phase diagram in Fig. For Mg2 Cu the percentage by mass of Mg is This solution freezes without further change. The cooling curve is shown next to the phase diagram. The solid points are the data given in the Exercise, and the lines are simply plausible connections be- tween these points; the open circle is the incongruent melting of K2 FeCl4. An expanded section of the phase diagram is shown as this includes the part of interest.
The solid consists of K2 FeCl4 , and as the temperature falls further the liquid grows progressively richer in FeCl2. Answers to discussion questions D5E. For binary systems the tie lines to which the rule appplies are always horizontal and so can be added to the phase diagram at will. In contrast, for a ternary system the tie lines have no such simple orientation and have to be determined experimentally.
Thus the lever rule applies, but in order to use it additional information is needed about the tie lines at the composition of interest. Solutions to exercises E5E. This line intersects the NaCl axis at a mole fraction corresponding to a mixture by mass of the two salts.
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