Comments re Chapter 4 Problems:

4-1

Another T/F problem. You will need to remember that the definition of A is
A = U TS so that DA = DG - D(PV).
Remember that for a change carried out reversibly DStot(which is DSuniv) is zero. For a reversible change carried out at constant T and P, DG (= DGsys) is zero. If T and P are constant (and only if T and P are constant) G (which is Gsys) is minimized at equilibrium. Similarly, for a reversible change carried out at constant T and V, DA (= DAsys) is zero. If T and V are constant (and only if T and V are constant) A (which is Asys) is minimized at equilibrium.

4-2

Calculation of DStot, DG, and DA for two phase changes (one melting, one vaporization) and for adiabatic expansion of a perfect gas into a vacuum. Note the similarities and differences between the three quantities. Note that the difference between DG and DA is often only a few kJ/mol (just as the difference between DH and DU is often only a few kJ/mol). It is very important to remember that if two phases are in equilibrium at constant T and P then they have the same molar free energy per mole (ie, the same Gm).

4-8

Numerical evaluation for a typical liquid of some of the derivatives for which formulas can be derived and for which it is useful to look at approximate values. A good exercise in cancelling units. Please look at the magnitudes of the values.
Formulas (it is not necessary to remember or be able to work out the 2nd, 3rd, or 6th):
(
H/T)P = CP
(
H/P)T = V (1 - Ta) [which reduces to 0 for a perfect gas]
(
U/V)T = [(aT)/k] P [which reduces to 0 for a perfect gas]
(
S/T)P = CP/T
(
S/P)T = -aV
CV = CP (
a2TV)/k [which reduces to nR for a perfect gas)
(A/V)T = P
Please note that the result should include an energy unit (J or cal) in all cases. Please do not leave (
A/V)T = P in atm or bar. The conversions can be done using values for R, the gas constant.
Note/ It is not necessary to know any of these equations except the first and the fourth.
After the next class you should be able to derive the last equation very quickly.
You should be able to derive the last equation quite quickly using the square of letters discussed in class.

 

 

4-8

Numerical evaluation for a typical liquid of some of the derivatives for which formulas can be derived and for which it is useful to look at approximate values. A good exercise in cancelling units. Please look at the magnitudes of the values.
Formulas (it is not necessary to remember or be able to work out the 2nd, 3rd, or 6th):
(H/T)P = CP
(H/P)T = V (1 - Ta) [which reduces to 0 for a perfect gas]
(U/V)T = [(aT)/k] P [which reduces to 0 for a perfect gas]
(S/T)P = CP/T
(S/P)T = -aV
CV = CP (a2TV)/k [which reduces to nR for a perfect gas)
(A/V)T = P
Please note that the result should include an energy unit (J or cal) in all cases. Please do not leave (A/V)T = P in atm or bar. The conversions can be done using values for R, the gas constant.
Note/ It is not necessary to know any of these equations except the first and the fourth.
After the next class you should be able to derive the last equation very quickly.
You should be able to derive the last equation quite quickly using the square of letters discussed in class.

4-25

Calculation of DG, and DA for the isothermal expansion of a perfect gas.

4-27

Can be done very quickly. Emphasizes a very important point.

4-28

Part (a) can be done very quickly. Emphasizes a very important point.
Part (b) concerns (again) the freezing of supercooled water. The process of freezing of supercooled water illustrates many important ideas.

4-29

A problem based on the mixing of gases, which is another classic problem in chemical thermodynamics.

4-43

Straightforward application of an equation derived in class. The problem does not state that P and T are constant, but if the system is at equilibrium and the free energy variable G is being considered then P and T must be constant and dG must be zero. Note that it cannot be guaranteed that dξ = 0 because the reaction could be shifted slightly forward or backwards while still being essentially at equilibrium.

4-44

Calculate ξ , which was called x in General Chemistry.

 

 

4-40

Another question about chemical potentials. If the chemical potentials of a component are the same in two phases then that component is in equilibrium If the chemical potentials of a component are different in different phases then that component in the phase in which its chemical potential is higher is unstable relative to the phase in which its chemical potential is lower (ie, ice at 298 K and 1 bar has a higher chemical potential than liquid water under the same conditions). Remember that for pure solids, pure liquids, and perfect gases the chemical potential is just the molar free energy.


Return to Course Listing

Description: Description: Description: Description: Description: UK Chemistry Home PageReturn to the UK Chemistry Home Page