Comments re Chapter 7 Problems:

7-2

A problem in which the Gibbs phase rule must be used to determine the number of degrees of freedom.  The key is correctly counting phases (easier) and components (harder).  This problem has no complications;  it is straightforward.

7-4

Another problem based on the Gibbs phase rule.  This problem explores some of the complications of counting components.  The simple rule about being able to count components by working out how many bottles must be opened to make the system works, but writing down the material requested by Levine is a good review of CHE 226 material.
[NB/  It is not important to memorize equation 7.10 in Levine’s textbook.  It is not an equation that I remember.  The easiest way to count components is to think about how many bottles must be opened to make the system without any constraints on the concentrations..  So H2SO4(aq) has two components (H2SO4and H2O) even though there are more than two species present:  H2SO4(aq), H2O, H+(aq), HSO4-(aq).and SO42-(aq).  This simple method for counting components works most of the time.]

7-5

A third problem based on the Gibbs phase rule.  Part (e) is tricky  It is an example of a situation in which the shortcut to counting components does not work;  the correct number of components can only be obtained by writing down all the chemicals present and then finding all the relationships between them.  The point of this problem is to show that counting independent components is occasionally difficult.

7-10

Another T/F problem. This one concerns phase equilibria. A triple point is a pressure, temperature combination at which three phases (e.g., solid/liquid/gas or solid1/solid2/liquid) are in equilibrium.  If c=1 then f=0 at a triple point.
For part (d) see the top of pg. 14 in the textbook.  Part (i) is true, as would be predicted by Le Chatelier’s principle, but the temperature difference is too small to matter.  At the triple point of water (0.01° C and 4.6 torr) the melting point of ice has risen only 0.01 deg when the pressure is reduced 760 torr to much less than 100 torr.

7-11

A problem based on the common substance water.  A number of sets of temperatures and pressures are given.  Which phase of water is stable (i.e., has the lowest chemical potential = molar free energy) in each case?

7-13

A basic problem involving the vaporization equilibrium of water.

7-16

A simple thought problem that requires a little knowledge of chemistry. The quantities DvapHo and DvapUo measure the strength of intermolecular attractions in the liquid (i.e., in a condensed) phase.  The stronger the intermolecular attractions the larger DvapHo and DvapUo.  Attractions increase with the total number of electrons (which is why the boiling point is correlated with molar mass) and with the number of hydrogen bonds.

 

 

7-21

Short, but very important T/F problem. 
It is necessary to know the relationship between  the Clapeyron Equation (dP/dT = DtransS/ DtransV for the equilibrium between two phases) and the Clausius-Clapeyron equation [ln(Pvap at T2/Pvap at T1) = -(DvapH°/R)(1/T2 – 1/T1)].  The Clausius-Clapeyron equation is just the equation for the change in Kp = Pvap with T, but the equation can also be derived from the Clapeyron equation by remembering that DS = DH/T if DG = 0 and making the assumption that DtransV = VvapVliq @ Vvap.
Part (d) of this problem requires knowing that no gas behaves perfectly near its critical point, which is the point at which the distinction between the liquid and the gas is lost.  (The pressure at the critical point is normally very high;  for CO2 it is 73 atm and for water it is 218 atm).

7-22

Calculate the vapor pressure of an organic liquid given the normal boiling point and DvapHo at the normal boiling point using the Clausius-Clapeyron equation.  Remember that DtransS = DtransH/Ttrans.because DtransG = 0 if the two phases are in equilibrium.

7-23

Standard problem: Estimate the pressure that must be applied to lower the melting point of water n deg C.
The equation needed is DP = ̣(DS/DV)dT =  ̣(DH/TDV)dT @ (DH/DV)ln (T2/T1).  The more approximate equation DP = ̣(DS/DV)dT =  ̣(DH/TDV)dT @ (DH/TDV)DT works fine in part (a) where DT is only ‑1 K. [DT/T = -0.00366 while ln(T2/T1) = -0.00367].  The approximate equation also works well enough part (b) [DT = -10 K;  DT/T = -0.0366 while ln(T2/T1) = ‑0.0373] because the other assumptions made (no change in DH and DV) are worse.  The assumption that DV is constant is especially bad when the pressure change is very large. 
(Perhaps the most important point of this problem is to see that very large pressure changes are needed to make small changes in the melting point).

7-60

A small thought problem designed to help with understanding of chemical potential.  Vapor pressure is a direct measure of chemical potential.

7-65

A very short question about the properties of the triple point.

 

 

7-37

Calculation of the transition pressure for graphite/diamond at 1000 K rather than at 298 K.  The increased temperature is desirable because it means the transition will take place faster. 
Calculate also the pressure needed for the transition at 298 K.  At 298 K the densities of diamond and graphite are 3.51 and 2.16 g/cm3 and DfG° for diamond is 2.900 kJ/mol.  Is the pressure needed for the transition at 1000 K higher or lower than the pressure needed at 298 K?

7-38

At what temperature are two solid forms of tin at equilibrium? The difference in entropy between the two forms is more important that the difference in density.

7-39

How does including the difference in the compressibilities of graphite and diamond affect the estimated transition pressure?

 

 

7-35

A problem in which the temperature and pressure of the triple point are estimated by using the Clausius-Clapeyron equations for both sublimation and vaporization.  At the triple point the T,P combination must be the same for sublimation as for vaporization.

7-73

Another of Levine's very useful thought problems.  When answering part (f) look at Fig. 7.9(b), which is more complete than Fig. 7.1(b).  When thinking about part (h) look at the first paragraph on pg. 222 and the associated Fig. 7.10.  The question in part (h) is whether water can be found as a liquid at 150° if Pwater = Pexternal = 1 atm rather than whether it should be possible to find liquid water under those conditions.  (See first paragraph on pg. 222).  In part (i) the answer if c=1 is different from the answer if c>1.  Since the number of components is not specified it has to be assumed that c might be greater than 1.


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