Hi Palmer!
To answer this question, we need to a) make sure we know the principle
behind the second law of thermodynamics and b) make sure that our
definitions of some basic chemistry/physics terms are accurate. So:
a)The second law of thermodynamics implies that the entropy of a
closed system shall never decrease, and therefore can only remain the
same or increase.
[ http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html ]
Entropy is a measure of the “randomness” of a system. This randomness
may be manifested as increased motion of an object (which, while
moving faster, will have a greater chance of hitting into something
else, thereby causing MORE random events, etc.).
On entropy: [http://hyperphysics.phy-astr.gsu.edu/hbase/therm/entrop.html
]
The Second Law can be demonstrated by observations of everyday events.
A common example would be the motion of a glass if it were to be throw
against a wall. We note that the glass always breaks apart into
smaller pieces. Thus the glass has moved from a state of relative
order (the whole glass) to a state of relative disorder (the
fragmented pieced). Now, if we were to pick up all of those pieces,
and throw THEM against the wall, it is theoretically possible that
they would reassemble themselves into the whole glass. Yet this is
never observed in nature because that would mean that the glass was
moving from a state of disorder to one of relative order. This state
of relative orderliness is what we mean by the “randomness” of the
system, and as was just demonstrated, processes in nature always move
from states of relative order to ones of greater randomness.
Just as an historical reference, much that was learned about entropy
that led to the second law was first devised while studying cycles of
internal combustion engines. It was observed that while most of the
energy released during combustion could be converted to work energy
(the end result of motion for our cars, for example, with internal
combustion engines), not ALL of the energy could be conserved. Some
of it was always lost to the surroundings, and this is interpreted as
increased entropy within the engine-system. You can see this visually
explained here: [http://www.mhhe.com/physsci/physical/jones/ol13-3.htm
]
b) Two scientific terms need to be better described. We already
better defined entropy above. Temperature is defined as “the average
kinetic energy of a system.”
To define and clarify: “Kinetic energy is an expression of the fact
that a moving object can do work on anything it hits; it quantifies
the amount of work the object could do as a result of its motion.”
[http://hyperphysics.phy-astr.gsu.edu/hbase/ke.html]
When a particle of fixed mass has increased kinetic energy, AKA it
moves faster, we can perceive this increased motion as it being “hot”.
This explains the first definition of temperature given above. The
important link to make here, however, is that when the temperature of
a system increases, the entropy must ALSO increase. This is because
the temperature is increased motion of the particles and the increased
motion implies increased entropy of the system of particles.
Now we can finally apply this background to our problem.
In this scenario, the water molecules are the “particles” described
above that each have a fixed amount of kinetic energy. I say it is
fixed because the system (the chamber, in this case) is a “closed”
system, meaning that energy nor mass cannot enter or leave it. The
water molecules must have some average amount of kinetic energy (read
“temperature”) because they are in a liquid state (if they had a much
lower amount of kinetic energy, they would be in the solid state
(ice)).
Now, let’s analyze what happens when the valve is opened.
____________________________
| |----------| |
H1 |........| | |
|xxxxxxxx| | |
|xxxxxxxx| |.......|H2
|xxxxxxxx| |xxxxxxx|
|xxxxxxxx| |xxxxxxx|
|--------|====><====|-------|
Valve
Using the nice diagram livioflores-ga drew (thank you!), and using
that notation, we would assume that gravity would exert its force on
the taller column more than the shorter column, such that water would
flow through the bottom tube and both H1 and H2 would level out to
some average central value, which would be equal to [ (H1+H2) / 2 ].
We’ll call this Heq (for “equilibrium height”). What would actually
be observed, and what you want to be proven, however, is that this
central value will in reality be something LESS than [ (H1+H2) / 2 ],
meaning a lower height of both, now equal, water columns. The main
question is: where did that water that makes up the volume difference
between Heq and the observed height go? Did it just vanish? What
happened?
When the valve was opened, most physical properties of the system
(being that it is “closed”) were not altered. The total amount of
energy and mass were conserved. The flow of water from one side to
the other, however, represents a change to the system. The fact that
the system has been changed indicates that the entropy of the system
has increased.
“If the volume and energy of a system are constant, then every change
to the system increases the entropy.”
[ http://www.cchem.berkeley.edu/~chem130a/sauer/outline/secondlaw.html
]
OK, so now we know that not much really changed in this chamber except
1) the water flowed through the bottom tube and leveled out and 2) the
entropy of the system increased.
(Just a note here, the reason for having the top tube is that, when
the bottom valve is opened, if that were the only connecting tube,
water would not flow freely between the two containers, because a
vacuum would be formed in one and pressure would build up in the
other. The top tube is only present so that the experiment can
proceed unhindered.)
The important fact is that the entropy of the system increased. Going
back to our definition of entropy, this means an increase in the
“randomness” of the system. This randomness of the water molecules is
precisely what temperature is. In other words, in this closed system,
the entropy increased, and the only way it is manifested is via an
increase in the temperature of the molecules.
So what effect does increased temperature have on the height of the
water columns that would produce the discrepancy between Heq and the
observed height? It has to do with the equilibrium between water in
the liquid phase and that in the gaseous phase. Although this article
is about atmospheric water vapor, the same principle applies, and the
description is very eloquent:
“...a liquid and vapor phase of the same substance in equilibrium are
actually in dynamic equilibrium, in which molecules in the gas phase
are combining and condensing to form liquid, and molecules in the
liquid phase are evaporating to form gas. The rates of these processes
are the same, so the total proportions in the two phases remain
constant.
When the temperature increases, that means that the molecules in both
phases have more kinetic energy. This makes it more difficult for
gas-phase molecules to condense to form the liquid (they bounce apart
instead of sticking together), and easier for liquid-phase molecules
to evaporate (they can fly out of the liquid phase more readily). So,
more liquid-phase molecules move into the gas phase than the reverse.
The concentration of the substance in the liquid phase increases until
the molecules being so close together that the condensation process
again is as fast as the evaporation process. This is the new
equilibrium concentration of the substance in the vapor phase, which
is usually expressed as a vapor pressure. Because of this interplay
between condensation and evaporation, the vapor pressure of any
substance is greater at higher temperatures.”
[ http://newton.dep.anl.gov/askasci/wea00/wea00016.htm ]
Thus the increase in temperature causes a shift in the equilibrium
ratio between liquid water and gaseous water, namely it shifts towards
more gaseous water and less liquid water.
As we can see, the discrepancy between the two water column heights
(“Where did the water go?” has been solved. A small portion of the
liquid water was converted to gas, and thus cannot be measured as part
of the “height”, via a chain of changes occurring in the system: flow
of water --> increased entropy --> increased temperature --> vapor
pressure equilibrium shift --> more gaseous water/less liquid water.
Thus to conclude your question, if there is less liquid water in the
two containers, then the average of their heights will also be lower,
as previously described by the principles of the Second Law of
Thermodynamics.
I hope this was nice and comprehensive. Please let me know if any
clarifications are necessary. Have a nice day!
Sincerely,
Andrewxmp
Textbook referenced:
Principles of Modern Chemistry, 3rd Edition, by David W. Oxtoby,
Norman H. Nachtrieb, published by Saunders College Publishing
(Saunders Golden Sunburst Series), Fort Worth, 1996, ISBN: 0030059046
Google Search Strategy:
second law of thermodynamics
second law of thermodynamics entropy
“entropy of a closed system”
“carnot cycle” |
Clarification of Answer by
andrewxmp-ga
on
20 Apr 2003 11:48 PDT
Hello again Palmer,
Sorry if the first explanation was only a theoretical one. Now let’s
try it with some numbers.
To quantify this problem, we simply need to follow the progression of
events described in the previous theoretical description, but give
arbitrary values. Also, we will need a number of standard physics and
thermodynamics formulas, as you requested.
First, let’s define our starting conditions. The two chambers will
have a volume of 1M^3 each. For the sake of simplicity, we will say
that the valve/tube along the bottom has zero volume, such that any
change in the volume of one chamber will be exactly inversely
reflected in the second chamber (the amount of water in the tub never
changes, it is always full, so we do not need to factor this into the
equations). Our starting temperature for the system will be 300
Kelvins ( equal to about 27 degrees C, a temperature at which water is
in it’s liquid form). Keep in mind that temperature is merely a
measurement on the average kinetic energy of particles, and the Kelvin
scale is an absolute scale of this measurement, meaning that 0K = zero
motion. Further defining our arbitrary initial conditions, we will
say that the initial height of column 1 is 0.7M tall and that of
column 2 is 0.3M tall.
In order to quantify the total initial amount of energy within the
system, we need the formula:
delta E = (delta Q) + (work done ON the system)
[ http://www.indstate.edu/thcme/mwking/thermodynamics.html#first ]
Where delta E is the change in energy of the system. Note that the
value for work done ON the system is added. Sometimes this value is
negative and defines the work done BY the system. For this problem,
however, this notation is more applicable.
Now, delta Q is the amount of heat added to the system; we added no
heat by opening the valve, thus this value is zero and drops out of
the equation. The amount of work done on the system, however, is
quantifiable. When the water level is shifted due to gravity, that
means that work has been done on the water. The force of gravity, in
this case, has caused a force and thereby has increased the energy of
the system. So in this one respect, it is not a closed system.
However, subsequent calculations do not depend on gravity, and thus is
must be treated as if it were closed.
The amount of work done on the water is reflective of a change in the
potential energy of the column due to its change in relative height.
The formula for potential energy of any object (our object here is the
water column) is given by:
E(pot.) = mgh
[http://studyphysics.iwarp.com/30/kepe.pdf ]
Where m is the mass, g is the acceleration due to gravity (which on
the earth is = to 9.8 meters/sec^2), and h is the height. The
variable here is the height, which changes when the valve opens. The
mass of water is 1gram/mL, therefore a full column of water (1 m^3 in
volume) would weigh one kilogram.
We will define the relative height of the water column as the average,
the height of the top of the column divided by 2. Plugging these
numbers into the above formula (potential energy = MGH), and solving
for the potential energies due to gravity of the two water columns, we
find that:
Column 1
E(pot.) = (0.7 kg) (9.8 m/sec^2)(.7m / 2) = 2.401 Joules
Column 2
E(pot.) = (0.3kg) (9.8 m/sec^2)(.3m / 2) = 0.441 Joules
Note that a) the measurement is in joules, the standard unit of work
and energy and b) that these values are only in reference to the base
height of the columns and also only account for the potential energy
due to gravity. The molecules themselves have other energies which
will come into play later.
Now let’s see what happens when the valve is opened. Just noting
volume change due to water flow, column 1 will lower to a height of
0.5 meters while column 2 will raise the same height, such that they
are level. If we do our calculations again for the values of
potential energy of the new columns we find:
Column 1
E(pot.) = (0.5 kg) (9.8 m/sec^2)(.5m / 2) = 1.225 Joules
Column 2
E(pot.) = (0.5kg) (9.8 m/sec^2)(.5m / 2) = 1.225 Joules
They have the same value for potential energy due to gravity. Now
lets look at the values for DELTA E(pot.). [Note: “delta” is the
Greek symbol that is used to denote “change in” a certain value]
Column 1 lost 1.176 joules (2.401 -> 1.225 = 1.176) of energy during
the change, while column 2 gained 0.784 joules (0.441 -> 1.225 =
0.784). Note that while all of the WATER that was lost from column 1
wound up in column 2, not all of the energy from column 1 did so.
This is quantified by the difference in delta E between the two
columns: 1.176 – 0.784 = 0.392 joules. This amount of energy was
apparently “lost” somehow in the process.
Let’s investigate where those 0.392 joules of energy "went".
This energy must have been either transferred or transformed, because
energy cannot be created nor destroyed (that is the basis of the FIRST
law of thermodynamics, by the way). Since this is a closed system we
know it has not been transferred, therefore it must have been
transformed. Because we presume that no more motion takes place after
the water levels have equalized, the only other manifestation of this
energy can be a change in heat (otherwise known as “enthalpy”), with
the symbol H. The formula for delta H is given by:
Delta H = delta E + Pressure(delta volume)
[http://www.indstate.edu/thcme/mwking/thermodynamics.html#enthalpy ]
Because there is no change in the pressure nor volume associated with
the movement of the water, the last term drops out, such that delta H
is equal to delta E. Thus the 0.392 joules of energy that came from
the movement of the columns has been converted to 0.392 joules of heat
energy.
This will be explained in a moment in detail, but the process by which
the volume of water changes is that, because the water columns contain
slightly increased heat energy, they have a greater tendency to
evaporate, increasing the ratio of gaseous water to liquid water. The
heat of vaporization of a substance is a standard physical constant,
and has been measured many times. For water, this value is 2256kJ per
kg.
[ http://hyperphysics.phy-astr.gsu.edu/hbase/tables/phase.html#c2]
In the two containers, we started with a total of 2 kg of water. We
then added 0.000392 kilojoules (0.392 joules x (1000 joules = 1 kJ))
of heat energy. Thus they are in proportion:
1 kg’s will be vaporized by 2256kJ of energy
x kg’s will be vaporized by 0.000392 joules of energy
Cross multiplying, we find that x = approximately 1.738x10^-7 kg,
which will be vaporized by this added heat. Yes, this is a very very
small number. We will qualitatively asses this value in a moment.
Because we are using two unit containers of 1M^3 volume, this
corresponds to an average height change of [1.738x10^-7) / 2] meters.
This is equal to 8.69x10^-8 meters less than the original average
height of 0.5 meters.
Used the scientific calculator available at:
[ http://www.ktf-split.hr/periodni/en/calc4chem.html ]
As we said before this is a very small value. If you mentally picture
this experiment happening, you wouldn’t expect to see an observable
change in the height, simply because only a very minimal amount of
heat will be produced and therefore only this minute amount of water
will be vaporized. Still, as demonstrated, it can be calculated via
the principles of the Second Law and theoretically could be observed.
I hope this better clarifies your question. Obviously chemistry,
physics, and other areas of science all apply to this problem of
thermodynamics, and it can be fairly confusing. Oftentimes there are
more than one way of stating the same thing. Also, my experience with
thermodynamics is that a lot of people throw around formulas of
entropy and heat and such, but they are not often actually calculated.
If there are any further questions, please let me know.
Regards,
Andrewxmp
Google search strategy (search terms used):
entropy law movement heat
entropy temperature heat Kelvin
thermodynamics Q work
vapor pressure heat
kinetic energy potential energy formula
|
