Download Chem 452 – Homework # 1A

Chem 452B – HW 6 (A) Fall 2012
Homework 6A KEY
Q1) For the following processes state whether the listed quantities are greater, less than or
equal to zero. Indicate your reasoning in a brief sentence and state any assumptions you made.
Unless stated otherwise, assume all gases to be ideal.
a) A gas sample goes through a Carnot cycle: T, q, w, U, H, S, G and A.
By going through a cycle the system is back to its original state, so all state functions must
be zero, so this leaves only q and w to change, and q=-w is required. If the system did work
then q is positive. Heat flowed through and was used to produce work.
b) A sample of hot water is mixed with a sample of cold water in an insulated container of
fixed volume: q, w, U, H and S. This is adiabatic but irreversible, so q=0, DU=0=DH,
as the internal energy is conserved, and DS>0 because it goes to equilibrium spontaneously
by having all terms have the same final temperature. Presumably no work, w=0.
c) An ideal gas expands adiabatically and reversibly: VT, q, w, U, H and S
Adiabatic reversible change is DS=0 (by definition) and q=q_rev=0. Expansion implies the
system does work so w<0, and DV>0 (expands), DT<0, DU<0, DH<0, internal energy is
consumed.
Q2) We have examined cases where the entropy is always positive for irreversible adiabatic
transformations. Now we want to show there are several ways to compute the entropy because
it is path independent. Consider the adiabatic, irreversible transformation of one mole of an
ideal monatomic gas starting at P1 ,V1 , T1 ends up compressed in an irreversible one step process
against a constant external pressure, which is Pext  P2  2P1
The problem begins by equating the two parts of the first laws:
CV dT  dU  q  w  w   P2 V2  V1 
 T2 P2 
3  T2 
  1     
2  T1 
 T1 P1 
T2 7
V2 7

and

T1 5
V1 10
1A) Find the entropy change of the system for this process. There are at least two approaches to
solving this problem, which is a challenging one. The first approach is to generate a path that
helps one solve for the entropy change, which is defined for the reversible path. So we have the
starting and ending points of the process. From the Carnot cycle we learned that we can
connect any two states for an I.G. by an isothermal reversible process followed by an adiabatic
reversible process. We must find the point, state 3, that connects 1 and 2 by these two steps.
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

For the isothermal step: PV
.
3 3  PV
1 1 , and for the adiabatic process P3 V3   P2 V2   
3
V
These give 2 equations and 2 unknowns from which you can get 3 and all other terms. Find
V2
the ratio’s for state 3 relative to either 1 or 2.
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Chem 452B – HW 6 (A) Fall 2012
P3 V3   P2 V2   

P3  V3 
P2  V2 
    
P1  V1 
P1  V1 
PV
3 3
1
PV
1 1
 V3 
 
 V1 
 1
5
3


 53 
P2  V2 
    2   0.7 
P1  V1 


3
  53   2
 V3 
    2   0.7    1.16
 V1 

 
1B) The entropy change is the sum of the two terms. The entropy change over the reversible
adiabatic part is zero, so the total entropy is found from the reversible isothermal change from
T1 to T3. (I got S  nR  0.148 ; see if you get something similar). It is rather striking that
very little about the initial and final states are specified and yet we get a numerical value for the
entropy.
V3
V
w
PdV
S 

 nR ln 3  nR  0.148
T V1 T
V1
1C) There is an alternative path (among many) which is also possible. In this path the
temperature is raised from T1 to T2 at constant volume (held at V1). Then in the second step,
with the temperature held at T2 the volume is changed (i.e. decreased) from V1 to V2. This is
just one of several paths one might imagine and does not require finding an intermediate point
on the path. This works because the entropy is path independent. The answer lies in knowing
the derivatives and integrating to find the change. For example for the first part of this change:
V2
T2
 S 
 S 
.
And
for
the
second
part
of
the
path:
SV   
dT

S

V1

T

 dV T2 . The total


T

V



T
V
T1
V1
entropy is the sum of these two and should be the same as part 1B). Using the definition of
these partial derivatives (as applied to the I.G.) solve for this entropy change and compare with
1B. I find this method much easier because I don’t have to think hard about the path, and I
checked and got the same answer as above. Pretty slick.
T2 7

T1 5
and
V2 7

V1 10
S  SV  ST
 S 
S   
 dT
T V
T1 
 S 
 
 dV
V T
V1 
V2
T2
V1
S  nR 0.148
2
T2
3 T
V 
 nR  ln 2  ln 2 
V1 
 2 T1
Chem 452B – HW 6 (A) Fall 2012
2) Find the entropy change of the environment and the total entropy change for this process.
Because it is adiabatic (albeit and 1 step) q=0, so the environment DSenv=0, Therefore:
STot  SSys  nR  0.148  0
3) Verify the sign of the change of the total entropy and comment on the relation to the
fundamental inequality of the second law.
The total entropy change is positive indicating that this is a spontaneous process as written.
Q3) We discussed the difference between a rubber band (as an entropic spring) and a steel
spring (as an enthalpic spring).
a) Qualitatively describe why heat flows out of an enthalpic spring upon contraction but flows
into an entropic spring upon contraction. (Check your rubber band again to be sure of this.)
Enthalpic spring: As the spring is stretched the potential energy is stored into the spring as
internal (potential) energy. So when the spring is allowed to spring back that energy must come
out. If the spring is not doing work, then the energy comes out as heat, the contraction is
exothermic; negative q, heat out.
Entropic spring: When the rubber band is stretched the potential energy is stored in the rubber
band in entropic order. It is a more ordered system, the internal energy does not change.
Because the entropy goes down on stretching, the heat must be negative, so stretching is
exothermic. Now when the rubber band can relax it can still do work but the order goes to
disorder and so the entropy goes up, so the reversible heat must be positive, even done with no
work then maximal heat is absorbed because the rubber band is more disordered.
b) We rewrote dU in terms of the length of the rubber band and the temperature in the room:
dU  TdS  Fd . In this form, the force (the negative of the tension) is F  k . And we noted
that the Thermodynamic equation of state (T.EoS) should be found by replacing P by –F:
 U 
 F 

  F T 

  T
 T 
There are two possibilities for using this expression.
A) Use the T.EoS to verify that if the spring constant is independent of the temperature one
recovers the usual rule for a Hook’s Law spring. Explain and show how the internal energy of
the spring, U, depends on the length in this case. Explain this case in terms of the potential
energy.
 F 
 k 
T
 T 
 0
 T 
 T 
 U 
 k 

  F T 
 F k
  T
 T 
This shows that the internal energy increases as the spring is stretched.
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Chem 452B – HW 6 (A) Fall 2012
B) Assume that the spring constant depends on temperature: k  T where  is independent of
temperature. Find the dependence of the internal energy, U, for this case. Compare your answer
to that of an ideal gas for an isothermal process.
 F 
  T 
T
 T 
  T  F
 T 
 T 
 U 
 k 

  F T 
  FF 0
  T
 T 
This shows that the internal energy of a rubber band does not depend on the extension of the
rubber band.
C) We did not talk about what happens when we heat the rubber band when it has a fixed mass
attached. Will the rubber band contract or expand on heating? Use the cyclic rule to bolster
your thinking. In this case you must relate this change to the change in the force with changing
temperature (at fixed length) and how the length changes with changing force (at fixed
temperature).
 F 



T 
 

 0

   F 
T
T


 T  F


  T
This shows that the length must get shorter as the rubber band is warmed. Again,
counterintuitive, but just another way to represent the fact that the spring constant increases
with temperature.
Optional Problem
Q4) Consider a system undergoing a phase transition from phase  to phase . The phase
transition occurs at temperature T0 and pressure P0 . The heat absorbed per mole of the
substance undergoing the phase transition is q 0 and there is a molar volume change of V 0 .
The molar heat capacities (independent of temperature) for the phases  and  are C p , and
C p ,  respectively.
I will be assigning this problem on hw6B, so I have started it here and will finish it on the next
assignment.
a) Find expressions for w, U , H , S and G for converting 1 mole of phase  to phase
 at T0 and P0 .
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Chem 452B – HW 6 (A) Fall 2012
w   Po V 0
U  q0  w,
H  q0  q
S =
H
To
G=0
The rest to be continued. See if you can make progress on it before I post my answers.
b) Find expressions for H and S for converting 1 mole of phase  to phase  at
temperature, T * different from T0 , but at the same pressure, P0 .
c) The denaturation (unfolding) transition of a globular protein can be approximated as a
phase change of  to Calculate, H , S and G for
q0 =638 kJ mole-1 , T0 =70 C, P0=1atm and C p, -C p, =8.37 kJ mole-1 K 1 .
d) Calculate, H , S and G for the same transition at 37 °C and 1 atm. Assume that
C p , -C p ,  is independent of temperature.
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