CHM 3410 – Problem Set 3a
Due date: NONE. These are a few additional problems covering some of the things we can obtain from the Maxwell
equations and associated relationships. They cover material at the end of Chapter 3 that will also be covered on our
first exam on Friday (9/24).
Do all of the following problems. Show your work.
1) We derived one of the “T dS” equations in class.
a) Using the same methods, derive the second T dS equation
T dS = T (p/T)V dV + CV dT
(1.1)
b) Starting with equn 1.1 find the change in entropy when the volume of a fixed amount of a van der Waals
gas is changed isothermlly from an initial value Vi to a final volume Vf.
c) Under what conditions, if any, will q for the above process be equal to T S?
2) Thermodynamic equations of state can be derived from the Maxwell equations and associated relationships, and
can be applied to substances more complicated than ideal gases.
a) Derive the following thermodynamic equation of state
(H/p)T = V - T (V/T)p
(2.1)
b) Starting with equn 2.1 find the change in enthalpy when the pressure of a fixed amount of a solid is
changed isothermally from an initial value pi to a final value pf. Assume the following for the equation of state of the
solid
V = V0 [ 1 +  (T – T0) -  (p - p0) ]
where V0 is the volume of the solid at some reference temperature and pressure T 0 and p0, and  and  are constants.
Solutions.
1)
a) Since the right hand side of the equation to be proved has terms depending on dV and dT, that suggests
that we begin by writing S as a function of these variables, that is, as S  S(V,T). From the general properties of
prtial derivatives we can then say
dS = (S/V)T dV + (S/T)V dT
If we multiply both sides of the above by T, we get
T dS = T (S/V)T dV + T (S/T)V dT
But (S/V)T = (p/T)V (Handout #3, equn 1.21)
So
T dS = T (p/T)V dV + T (S/T)V dT
But T = (U/S)V
(Handout #3, equn 1.11)
Substituting for the second T on the right hand side gives
T dS = T (p/T)V dV + (U/S)V (S/T)V dT
Since both partial derivatives in the second term have V being held constant, they can be combined to give
(U/S)V (S/T)V = (U/T)V = CV
Substitution leads to our final result
T dS = T (p/VT)V dV + CV dT
b) The process is isothermal, and so T = constant, and dT = 0. Therefore
T dS = T (p/T)V dV
or
dS = (p/T)V dV
For a van der Waals gas
p = nRT/(V - nb) - an2V2
and so
(p/T)V = nR/(V - nb)
So
dS = nR/(V – nb) dV
Integrating from Vi to Vf gives
if dS = if nR/(V - nb) dV
Sf – Si = S = nR if dV/(V - nb) = nR ln{(Vf - nb)/(Vi - nb)}
c) If the process in part b is reversible, then, since dS = (dq) rev/T
S = q/T, and so q = T S
2)
a) A good strategy for a problem like this is to see if there is a partial derivative that looks similar to the one
in the problem. There is (Handout #3, equn 1.12), and so using one of the partial derivative relationships from
Handout #1, we may say
(H/p)T = (H/p)S + (H/S)p (S/p)T
But
(H/p)S = V
(Handout #3, equn 1.12)
(H/S)p = T
(Handout #3, equn 1.12)
(S/p)T = - (V/T)p
(Handout #3, equn 1.22)
and so
(H/p)T = V + T [ - (V/T)p ] = V - T (V/T)p
b) The process is isothermal, so T = constant.
(H/p)T = V - T (V/T)p
and so
dH = { V - T (V/T)p } dp
Integrating both sides from pi to pf gives
if dH = Hf – Hi = H = if { V - T (V/T)p } dp
Since
V = V0 [ 1 +  (T - T0) -  (p - p0) ]
then
(V/T)p = V0
and so
H = if { V0 [1 +  (T - T0) -  (p - p0) ] - T (V0) } dp
= if [ V0 + V0T - V0T0 - V0p + V0p0 - V0T ] dp
where we have multiplied out the term in the integral. The second and last terms in the integral cancel, and so, after
rearranging to put the the term involving p last, we get
H = if [ V0 - V0T0 + V0p0 - V0p ] dp
= (V0 - V0T0 + V0p0) (pf - pi) – (V0/2) (pf2 - pi2)