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Teacher End of the School Year Gifts. The perfect gift for teacher appreciation week. Report DMCA. Facebook Twitter Whatsapp. You may also like. Organic Chemistry 8th Edition. Handbook of Oral Disease. Leave a Comment Cancel Reply You must be logged in to post a comment. For a change in V and T, dU is given by [2D.
For a reversible expansion pex is the pressure of the gas, hence. Substituting in the expression for the pressure of a van der Waals gas, [1C. The expansion coefficient is defined in [2D. This expression gives the relationship between the change in pressure and the change in density.
Solutions to problems P2D. Answers to discussion questions D2E. This equation is the point from which the relationships between pressure, volume and temperature for a re- versible adiabatic expansion are found: the heat capacity comes into the final expressions via this route.
In words, the key thing here is that in an adiabatic process there is a change in temperature, so it is not surprising that the properties of such a process are related to the heat capacity because this quantity relates the energy and the temperature rise.
Solutions to exercises E2E. For a perfect gas, [2B. C p,m C p,m This indicates that vibrations may be con- tributing somewhat - for example the lower frequency bending modes. Using the value of C p,m from the Resource section gives. The initial volume of the sample is. For this gas. The initial and final states in a reversible adia- batic expansion are related by [2E. Solutions to problems P2E. It is evident that the agreement is quite good.
It follows that. The average molecular weight of air is approximately Answers to discussion questions D3A. For example, for a gas expanding freely and spontaneously into a vacuum, the process is accompanied by a dispersal of energy and matter. For a perfect gas this entropy change of such an expansion is derived in Section 3A. The entropy change is clearly positive if Vf is greater than Vi. The molecular interpretation of this thermodynamic result is based on the iden- tification of entropy with molecular disorder.
An increase in disorder results from the chaotic dispersal of matter and energy and the only changes that can take place within an isolated system the universe are those in which this kind of dispersal occurs.
Solutions to exercises E3A. These two are combined and rearranged into an expression for the temperature of the hot source. This is rearranged to give an expression for the temperature of the cold sink.
Solutions to problems P3A. The work will be in joules if the pressure is expressed in pascals and the volume in m3. The First Law [2A.
As explained in Section 3A. The entropy of the surroundings is given by [3A. This implies that Stages 1 and 3 of the Carnot cycle are horizontal straight lines, whereas Stages 2 and 4 are vertical straight lines on the temperature—entropy diagram, as shown in Fig. Stage 1 A B Temperature, T. Stage 4 Stage 2. 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.
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