QUIZ 9
Physics 101, Fall 2007
December 7, 2007
SOLUTIONS
A thermodynamic model of a bit of computer memory is a pair of boxes each of volume V = 10−12 m3
that are in contact with a thermal bath at temperature T = 25C. There is exactly one monoatomic
gas molecule present. If it is in the left box, we say the bit is a 0, while if the molecule is in the
right box we say the bit is a 1.
If the computer needs to change 0 to 1 or vice versa during a computation, it could do so, in principle, at
no cost of energy or entropy by rotating the pair of boxes by 180◦ about the midpoint of the partition
between them.
The single-molecule “gas” obeys the ideal gas law if the pressure is taken to be the average effect
of many collisions of the molecule with the wall of the box.
Thermodynamics of Memory Loss. If the partition is removed (and not reinserted), the usefulness
of the memory is irreversibly lost. This operation results in the free expansion of the gas from volume V to
2V . Although no heat flows during a free expansion,
we cannot say that there is zero entropy change, because free expansion is not a quasi-equilibrium process
during which the laws of thermodynamics apply.
A reversible process during which these laws apply and
which produces the same final state, and same entropy
change, of the gas as the free expansion is to suppose
the partition were a piston that is retracted (sideways)
slowly and isothermally until the volume doubles.
1. [2 pts] How much work is done by the single
molecule during this isothermal expansion?
W = nRT ln(Vf /Vi ) = (RT ln 2)/NA = kT ln 2,
where NA = 6.023 × 1023 = Avagadro’s number
= number of molecules in a mole. Noting that
T = 273 + 25 = 297K, we have
W = 1.38 × 10−23 · 295 · 0.69 = 2.82 × 10−21 J.
2. [1 pt] How much heat flows into the gas during
the isothermal expansion?
The internal energy remains constant during an
isothermal process, so ΔU = 0 = Q − W . Hence,
Q = W = kT ln 2 = 2.82 × 10−21 J.
The only formulas needed on this quiz are
ΔS = Q/T ,
P V = nRT ,
ΔU = nCΔT = Q − W ,
where Cmono = 3R/2,
R = 8.31 J/◦ K,
W = nRT ln(Vf /Vi ) during an isothermal
process.
Factoid: k = R/NA = 1.38 × 10−23 J/◦ K
= Boltzmann’s constant.
3. [2 pts] What is the entropy change ΔSgas of the gas/memory, and ΔSbath of the outside thermal
bath during the isothermal expansion?
ΔSgas = Q/T = k ln 2 = 9.6 × 10−24 J/◦ K. Heat Q flows out of the thermal bath into the memory
box during this process, so ΔSbath = −Q/T = −ΔSgas = −9.6 × 10−24 J/◦ K.
The total entropy change during the isothermal expansion is 0, as required for a process to be reversible.
QUIZ 9
Physics 101, Fall 2007
December 7, 2007
4. [2 pts] Now consider the case of memory loss due to removal of the partition. What is the entropy
change ΔSgas of the gas/memory, and ΔSbath of the outside thermal bath during the subsequent
(irreversible) free expansion?
No heat flows from the bath to the gas during the free expansion, so ΔSbath = 0. . However, the
entropy change of the gas is the same for the free expansion and for the isothermal compressions since
their initial and final states are the same. Hence, ΔSgas = Q/T = k ln 2 = 9.6 × 10−24 J/◦ K.
The total entropy change, ΔSgas + ΔSbath = k ln 2, during the irreversible loss of memory is positive,
which is consistent with the interpretation of entropy as a measure of disorder.
Thermodynamics of Memory Erasure. We can
partially recover from the “loss of memory” considered
in part 4 if, after the free expansion, the righthand
wall/piston compresses the volume 2V back down
to V , the partition is reinserted, and the righthand
wall/piston is retracted creating an empty volume V
on the right. The memory is now in the well-defined
state 0, so we have erased the memory (which is more
useful than merely “losing” the memory).
5. [2 pts] What is the entropy change ΔSgas
of the gas/memory, and ΔSbath
of the outside thermal
bath during the isothermal compression that resets the memory to 0?
The isothermal compression of this part is just the reverse of the isothermal expansion considered in
parts 1-3, so the entropy changes are just the reverse: ΔSgas
= −Q/T = −k ln 2 = −9.6 × 10−24 J/◦ K.
ΔSbath
= Q/T = k ln 2 = 9.6 × 10−24 J/◦ K.
6. [2 pts] What is the total entropy change of the system of gas + bath during the erasure of the
memory consisting of the free expansion (part 4) followed by the isothermal compression (part 5)?
The total entropy change is the sum of that in parts 4 and 5. Hence,
ΔStotal = ΔSgas + ΔSbath + ΔSgas
+ ΔSbath
= k ln 2 + 0 − k ln 2 + k ln 2 = k ln 2 = 9.6 × 10−24 J/◦ K.
This method of erasing the memory presumes no knowledge/information as to the state of the memory
prior to its erasure, and the entropy of the system increased during the erasure. But if we knew the
state of the memory, we could reset it to 0 with no increase of entropy. Consistency with the 2nd las
of thermodynamics implies that knowledge/information has a value that can be measured in terms of
entropy. In particular, a system that contains knowledge/information is in a higher state of entropy
than one which does not.
In mechanistic terms, the outside world has knowledge of the computer memory only if it has a a
memory of its own that contains a copy of the state of the computer memory. This knowledge could
be used to erase the computer memory at no entropy cost, but the outside system still retains in its
memory the knowledge/copy of the prior state of the computer memory. If the outside system is to free
up its memory by erasing this now-useless knowledge, it must then pay the entropy cost of doing so.
This problem was first posed in 1867 by Maxwell, who remarked that an intelligent demon could use
its knowledge to reduce the entropy of a system by timely control of the partition. However, the
knowledge of the “demon” implies that the total entropy of the system + demon is higher than if no
knowledge/information were available. The action of the demon does not result in a lower total entropy
of the universe than for the demon-free case.