• Demons of Physics.
• The Ideal Gas Law from Boyle, Mariotte, Charles, Gay-Lussac, and Avogadro.
• Viscosity is the ratio of a shear stress to the shear strain rate it induces.
• Permeability and permeance: Vapor barriers and porous materials.
• Resonant frequencies of air in a box.
• The Earth's atmosphere. Pressure at sea-level and total mass above.
• Raising the Titanic, with (a lot of) hydrogen.

Related articles on this site:

• Physics.
• Steam Engines.
• Cosmology 101: Basic concepts for studying the physical Universe as a whole.
• Physical Units: A tribute to the late physicist Richard P. Feynman (Nobel 1965).
• Escutcheons of Science (Armorial):  Coats of arms of some famous scientists.

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(2003-05-21)
Who discovered the ideal gas law?     [ pV = nRT ]

Emile Clapeyron (1799-1864; X1816) gave the law its modern form in 1834.  Louis-Joseph Gay-Lussac (1778-1850) is the one who first put all aspects of this relation together around 1802, after key contributions from several other people.

This deceptively simple equation states that the product of the pressure (p) and volume (V) of an ideal gas is proportional to its absolute temperature (T) and to the number of molecules in it;  "n" is actually the number of moles, which is the number of molecules divided by a numerical constant called Avogadro's number, whereas "R" is a constant of proportionality equal to  8.314472(15) J/K/mol.

The law was instrumental in the clarification of all the concepts involved, with the sole exception of "volume", of course...  The law can be broken down into four statements which were historically studied separately (some, or all, of these may be invalid in elaborate models of actual fluids).  When the full law was finally formulated, it became possible to quantify the discrepancies between the "ideal" behavior it described and the actual properties of real gases...

Avogadro's Law :

At the same temperature and [low] pressure, equal volumes
of different gases contain the same number of molecules.

Like other "ideal gas" law, this law is only a fairly good approximation under ordinary conditions.  However, it becomes extremely accurate at very low pressures, and has even be stated as a rigorous law in the vocabulary of limits.

Boyle-Mariotte Law :  Isothermal

At a given temperature, the volume of a gas
is inversely proportional to its pressure.

This law was first formulated by Robert Boyle (1627-1691) in 1662, without the important isothermal specification which was added in 1676 by Edme Mariotte (1620-1684) when he rediscovered the whole thing independently...

Charles' Law :  Isochoric

When confined to a given volume, the pressure of a gas
is proportional to its absolute temperature.

This isochoric (or isovolumic) law has been named after Jacques [Alexandre César] Charles (1746-1823) at the suggestion of Gay-Lussac (after whom the isobaric law, stated below, is named).  This is meant to acknowledge the unpublished work on the subject which Charles performed in 1787.

Gay-Lussac's Law :  Isobaric

At a given pressure, the volume of a gas is
proportional to its absolute temperature.

---

(2002-06-07)
What's viscosity?

A viscous deformation is a lateral deformation which occurs at a rate that's proportional to shear stress.  Such a deformation is most readily observed with thick paste, but it occurs much faster in not-so-thick liquids (low viscosity).  Technically, tar pitch is actually a thick liquid of extremely high viscosity.  On the other hand, contrary to urban legends, glasses are amorphous solids at room temperature.  They do not behave like "supercooled liquids" (unless heated above a certain transition point) but they may "creep" like crystalline solids do.

For crystalline solids and/or rocks, what occurs [over geological periods of time] is a so-called "plastic" deformation.  The rate of plastic flow is not directly proportional to stress but rather to some increasing function of stress with a zero derivative at the origin (so that there's no plastic deformation at all when the stress is low).  For example, in what's called "Power Law Creep", the strain rate is proportional to some n-th power of stress (n>1) and to a temperature factor exp(-Q/RT), which indicates how easily crystal dislocations may move at an absolute temperature T.  For typical rocks, the exponent n is around 3 (it ranges from slightly below 2 up to 8 or so) and the "activation energy" Q is roughly 200 kJ/mol (it may range from below 100 kJ/mol up to about 500 kJ/mol).  Viscous flow is a special case (n = 1 and Q = 0) of Power Law Creep...

Quantitatively, viscosity is the ratio of shear stress to shear strain rate.  (These two are proportional in the case of what's called a Newtonian fluid.) Units of viscosity are thus homogeneous to the product of a stress (or pressure) by a time.  Therefore, the SI unit of viscosity is the pascal-second (Pa×s), also called poiseuille (Pl), which is 10 times as large as the corresponding [deprecated] CGS unit, the poise (P) of 1 dyne-second per square centimeter.  The centipoise (cP) is a surviving synonym of the millipascal-second (mPa×s), which happens to be the viscosity of water at a temperature of about 20.2°C.

• At 20°C, the viscosity of water is about 1.005 mPa×s and decreases with temperature:  At  t°C, the viscosity in millipascal-second is within 0.2% of
  1.005 exp(-0.024475 (t-20) (1 - 0.8787 (t-20) / (t+98.5) )).
• At 20°C, the viscosity of air is about 0.01808 mPa×s and increases with temperature; it's about 0.01709 mPa×s at 0°C.

The above is called absolute or dynamic viscosity, as opposed to the relative or kinematic viscosity, which is defined as the quotient of the dynamic viscosity by the mass density, expressed in units that are homogeneous to the ratio of an area and a time (m²/s).  The [deprecated] CGS unit of kinematic viscosity is the stoke (St), which is 10000 times smaller than the SI unit (m²/s).  A kinematic viscosity is routinely obtained with a [Saybolt or Engler] viscometer, by timing the passage of a calibrated volume of liquid under its own weight through a small aperture.  The reciprocal of [either flavor of] viscosity is called fluidity.

Laminar Flow

---

Michel Couvreux (TransMineral USA.  2003-04-29)       Permeability :
Mortar made from our natural hydraulic lime  (NHL 5, with twice its volume of sand)  has vapor-exchange properties listed as 0.55 gram of air per hour, per square meter, per mmHg [for one-coat renders].
What's its permeability?

Short Answer:   10 perm-inches (US)   or   14.5  ng / s.m.Pa.  Read on...

Note:   Besides the obvious problem with units, this question raises the issue of converting a rating about the diffusion of air into the kind of water-vapor rating which is clearly expected in the construction business.  We must also dispose of the fact that the stated rating is about a thickness of the material which is not explicitely specified:
      In the US, the thickness of "one-coat renders" is traditionally quoted as 3/8"  (there are normally 3 coats, including a thin outer coat, totalling 7/8" in thickness: 3/8+3/8+1/8).  However, the link quoted in the question makes explicit reference to a "French standard" (which we've been unable to locate).  This seems to imply that "one-coat renders" is shorthand for a metric reference thickness of 1 cm (1 cm is only 5% more than 3/8").  We'll be assuming this much in the following article.  If you know better, please let us know...

There are two kinds of permeability.  Geologists who consider the flow of water or oil through rocks deal with hydraulic permeability (which we'll examine last) based on "Darcy's Law".  On the other hand, the construction industry refers to diffusion permeability, which is based on "Fick's Law", the statement that the flux of a diffusing substance is proportional to its concentration gradient.  For gases and vapors, partial pressures (more precisely,  fugacities) are used instead of "concentrations"...  Air and vapor ratings are quite different.  A layer of given thickness is rated according to its permeance (= permeability times thickness):

Permeance, for "Water Vapor Transmission" (WVT):

The permeance of a vapor barrier [more precisely, a "vapor diffusion retarder" or VDR] is a measure of how fast it lets water vapor through, when its two sides have different partial pressures of water vapor.  Permeance may be stated either in perms [US perms] or in metric perms (which are about 52% larger than US perms).  These units are defined as follows:

• 1 US perm  =  1 grain (gn) of water vapor per hour, per ft², per inHg.
• 1 metric perm  =  1 gram (g) of water vapor per day, per m², per mmHg.

A perm is exactly  32399455 / 49161192, or about 0.659 of a metric perm.

Note that this "metric perm" unit is not metric at all, since it's based on nonmetric units like the day for time, and the millimeter of mercury (mmHg) for pressure.  The strict metric equivalents of both of the above units are:

• 57.213494670945101394-  ng / s.m².Pa   =   1 US perm.
• 86.812682389543557170+  ng / s.m².Pa   =   1 metric perm.

The above SI unit (ng / s.m².Pa) is homogeneous to a time divided by a length and actually boils down to a  "picosecond per meter"  (ps / m)  but it's never expressed this way.  Unfortunately, this very metric unit has [wrongly] been dubbed a "metric perm" by some sources...

The value of a perm in SI units does not depend on temperature !  Please ignore all those well-meaning Internet sources [ 1 | 2 | 3 | 4 ] which state otherwise by listing the above equivalences as valid at 0°C, while giving slightly higher values (+0.418 %) for 23°C.  This nonsense comes from a poor understanding of units of pressure:  The millimeter of mercury (mmHg) and the inch of mercury (inHg) are units based on the conventional density of mercury, which is exactly 13595.1 g/L...  The fact that this is very close to the actual density of mercury at 0°C is not directly relevant to unit conversions...  Even less relevant is the density at 23°C (about 13538.5 g/L) which happens to be 0.418 % less [here we go].  This only means that, if we were measuring pressure with an actual column of mercury at 23°C, we would have to divide the readings in millimeters by about 1.00418 to obtain actual pressures expressed in so-called mmHg.  Got it?  (If you didn't, you're in good company, since even standardization organizations have been known to introduce such silly "units" of pressures, whose values vary with temperature, just like the actual densities of mercury and water do.)

Permeance is occasionally given in g/m²/day (DIN 53122) where a "standard" vapor-pressure difference is understood, which corresponds to a difference in relative humidity of 85% at 23°C between both sides (for example, 0%||85% or 15%||100%).  As the saturation vapor-pressure over liquid water (100% r.h.) at 23°C is about 21.080 mmHg, the standard vapor-pressure difference is 17.918 mmHg, making the above unit equivalent to about (1/17.918) of a metric perm, or 0.084683 of a US perm  (the conversion factor is about 11.8).

"Water-Vapor" Permeability:

The above permeance is equal to the [diffusion] permeability of the material divided by its thickness.  Permeability is a characteristic of the material itself, which may be expressed in "perm-inches":  If a material has a permeability of one perm-inch, a one-inch layer of it has a permeance of one perm, a two-inch layer has a permeance of  ½ perm, etc.  (For the record, units of permeability are homogeneous to a time, and a perm-inch is about 1.45322 picoseconds.)

The mass diffusivity of water vapor in air (D) is known to be 2.42 10^-5 m²/s...  This is the coefficient to use in Fick's equation  Jx = -D (dC/dx), which relates the mass flux (J) to the gradient of mass concentration (C).  The partial pressure (p) of a gas is obtained from C by introducing the molar mass (M) into the ideal gas law:  p = RTC/M.  Thus we have:  J = -(DM/RT) (dp/dx).  In other words, the diffusion permeability of ideal water vapor is DM/RT, where R is the molar gas constant: 8.314 472(15) J/K/mol.  In the case of water vapor (M = 0.018 kg) diffusing in air at T = 300 K, this gives a permeability of about 175 ng/s.m.Pa, or 120 perm-inches, as shown in the first line of the above table.

Concerning NHL mortar, the unit given in Michel's question seems equivalent to 24 metric perms (since the time basis is an hour instead of a day), so that 0.55 of such a unit would be exactly 13.2 metric perms (about 20 US perms) if it was a true WVT rating.  Unfortunately it's explicitely specified as an air rating instead, so we must somehow convert that into a water-vapor rating by considering the different diffusion properties of air and water vapor...

If we assume that most pores in the mortar are much larger than the dimensions of the molecules of the gases involved, we have microscopic spaces in which the molecules bounce in thermal equilibrium, in much the same way they would in an open space.  Under this simplification, each molecule of an ideal gas of molar mass M would essentially diffuse at a rate proportional to its average speed, which is itself proportional to Ö(T/M) at temperature T.

• Water vapor (H₂O) has a molar mass of 18.0153 g.
• Dry air is essentially 79% N₂ (28.0135 g) and 21% O₂ (31.9988 g).

Ignoring interesting details, like the nitrogen enrichment of air diffusing through mortar, we'll consider that air is roughly a simple gas of molar mass 28.96456 g.  Such a gas would diffuse about 1.268 times slower than water vapor (this is the square root of the molar mass ratio).  Accordingly, the permeance is roughly 1.268 times larger for water vapor than for air, so the WVT permeance of a 1 cm layer of NHL mortar is about 25.40 US perms (16.74 metric perms);  the permeability of 1:2 NHL5 mortar is thus very nearly equal to 10 perm-inches.

Hydraulic Permeability

Darcy's Law (of which there's a more general tensor form) states that a porous material lets a fluid through at a rate inversely proportional to its viscosity and directly proportional to both the cross-sectional area and the gradient of the applied pressure (also called hydraulic gradient).  The material's [intrinsic] permeability is simply the coefficient of proportionality in this law:

(flow rate)   =   (permeability) (area) (pressure gradient)  /  (viscosity)

The pressure gradient between two parallel sides is their pressure difference divided by the thickness of the material between them.

The permeability coefficient happens to be homogeneous to a surface area.

ASTM E 96 Standard Test   |   ASTM E 96 Testing
J.C. Alvarez Master's Thesis: Evaluation of moisture diffusion theories in porous materials
Fick's Law   |   Laws of Gas Transport   |   Plastics and Elastomers

---

(Mark Barnes, UK. 2000-12-05)
What are the resonant frequencies of a given volume of enclosed air?

The amplitude U of a wave that propagates at celerity V verifies the equation:

1		¶ 2 U	=	¶ 2 U	+	¶ 2 U	+	¶ 2 U
V 2		¶ t 2		¶ x 2		¶ y 2		¶ z 2
			=	DU		[D is the Laplacian operator]

As used above, the term "celerity" is best reserved for the "phase speed" which appears in this equation.  Here V is simply the speed of sound in air.

The elementary solutions of this wave equation whose amplitudes are zero on the walls of a rectangular box are obtained as the products of single-dimensional solutions for a segment (namely, sine waves whose half-wavelengths are a submultiple of the length of the segment, so they are zero at both ends).  Such an elementary solution of frequency n has therefore the following form for a rectangular box of dimensions A, B and C, when the origin of coordinates is one corner of the box (in this, a, b and c are integers):

U(x,y,z,t) = sin(apx/A) sin(bpy/B) sin(cpz/C) exp(2ip n t)

The Laplacian DU is thus   -[(a/A)²+(b/B)²+(c/C)²]p²U   whereas the second time derivative of U is simply   -[4n²]p²U  . As the above wave equation tells us, the ratio of those two square brackets is equal to the square of the speed of sound V.  This gives us the following formula for a resonant frequency (n) of air in a rectangular box (where a, b and c are any positive integers):

n   =   ½V Ö	a2/A2 + b2/B2 + c2/C2

For a cylindrical box, say, you would solve a 2-dimensional equation for the disk (using Bessel functions) and combine things as above into another formula for n.

The lowest resonant frequency is, of course, obtained for a = b = c = 1.  The next frequencies are not multiples of this one...  The general solution of the wave equation for the cavity is a superposition of the above vibrations (à la Fourier).

We assumed perfectly rigid walls.  The relative correction for vibrating walls will typically be as small as the ratio of the vibration amplitude (probably less than a millimeter) to the wavelength in air (about a meter at 343 Hz, for air at 20°C)

The notion that resonant frequencies are directly related to volume is a myth, but we may investigate what happens to the lowest resonant frequency (a = b = c = 1) for a constant volume L³ = ABC  (A = xL, B = yL, C = L/xy):

n   =   no Ö	( 1/x2+1/y2+x2y2 ) / 3

This happens to have a minimum at n = no in the case of the cube (x = y = 1). For a cavity that's not too different from a cube (which may not help much with guitar design) n is thus pretty close to ½Ö3 V/L, where L is the cubic root of the volume, and V is the speed of sound...

To conclude, we may examine temperature dependency:  As the thermal dilatation of the instrument itself is very minute, the frequencies are essentially proportional to the speed of sound V, which is itself proportional to the square root of the absolute temperature T (for an ideal gas).  At room temperature (T = 295 K), this means that a variation of one kelvin (1°C) will induce a frequency change of about 0.17%.  This translates musically into (take your pick):  0.00244 octave, 0.0293 semitones, 0.735 savart, 1.465 centitone, 2.93 cents or 2.44 millioctaves.  That's not much:  It takes a difference of about 10°C for wind instruments to be off by 30 cents (30% of a semitone). The shift is thus about one semitone between a warm interior and freezing weather outdoors.  In an orchestra, all wind instruments are affected the same way, and string instruments could adjust and/or retune, so everybody could still be in tune.  Pianos are a problem, keep them indoors...

---

(J. L. of Canada. 2000-11-23)
The atmosphere is made of molecules, piled up on top of each other. It starts at earth's surface and becomes indistinguishable from the vacuum of space a few hundred kilometers up. Now, consider a 1 cm by 1 cm square sitting at sea level and all the air that's sitting on top of that square, from the surface all the way up to outer space. What is the approximate mass of the air in that one square-centimeter column?

The short answer is that pressure comes from the weight of the column of air above the surface. The standard atmospheric pressure of 101325 Pa exerts a force of 10.1325 N on a surface of a square centimeter.  The problem is to estimate the mass from the weight. As we shall see, Earth's gravity does not vary much for most of the atmosphere, so we may use the standard value of gravity (9.80665 m/s² ) and come up with a mass of approximately 10.1325 / 9.80665, or roughly 1.033 kg over each square centimeter at sea level.

Indeed, Earth's gravity is approximately constant throughout the atmosphere:  At an altitude equal to 0.5% of the Earth radius (about 32km), it is only 1% less than at sea level, and most of the atmosphere is below that point (more than 99% of it is said to be below an altitude of about 70 km, where gravity is only down 2%).

Consider a column of air at rest (it's essentially part of a slim cone whose apex is at the center of the Earth).  The weight of all the molecules of air ultimately results in a single force pS exerted by the pressure p at sea-level on the surface area S at sea-level. Now the weight (that's force, not mass) of each molecule of mass m is slightly less than the weight at sea level mg (g being gravity at sea-level) so that the mass is slightly more than pS/g, but not much more, as previously observed.

Taking the standard values g=9.80665 m/s² and p=101325 Pa, a surface S=1 cm²= 0.0001 m² would have approximately 1.03323 kg of air above it. As we've seen, this is actually a slight underestimate of the true value, so we may state that a cm² at sea-level has slightly more than a kilogram of air above it.

The interesting thing is to estimate the total mass of the atmosphere. The surface of the Earth's is very close to 5.1 10¹⁴ m² (that's the surface of the reference ellipsoid and that's also very close to the surface area of a sphere whose radius is the "conventional" radius of the Earth, namely 6371 km).  Therefore, the above gives a total mass of about 5.27 10¹⁸ kg.

Now there's a problem: the mass of the atmosphere is listed as 5.136 10¹⁸ kg in my copy of the 1995 CRC Handbook of Chemistry and Physics (page 14-7).  We should not have been troubled at all about the small (2.6%) discrepancy from our crude estimate, except that it turns out to be in the wrong direction: If the variation of gravity with altitude was the only factor to correct, the true number should be higher, no lower.  You may remark that more air is packed at the poles (the temperature's lower, the density's higher) with a larger surface gravity (9.832186 m/s² at the poles).  This makes a correction in the right direction, but it cannot exceed 0.26% (that's how much the gravity at the pole exceeds the standard value of g) whereas our discrepancy is 10 times as large.  What else could be wrong?  Well, our value of the "standard" atmospheric pressure was precisely designed to make this particular computation work out right and it is not at fault...

The reason for the discrepancy is that we overlooked land masses in the computation! Whatever volume is occupied by land is not available for air and does decrease the total mass of the atmosphere.  To get a 2.6% difference in the mass of the atmosphere, we would need the average land (30% of the surface of the globe) to decrease by about 9% the mass of air above it, as compared to sea level. That looks like a pretty reasonable figure, and so does the CRC value of 5.136 10¹⁸ kg for the mass of the entire Earth's atmosphere.

The molar mass of air is about 28.96456 g/mol, so there are about 1.733 10²⁰ moles in the atmosphere, that's a grand total of about 1.068 10⁴⁴ individual molecules. The square root of that is 1.034 10²², which is 0.01716 moles, or the number of molecules in 0.497 g of air. This occupies a volume of about 400cc at 11°C under 1 atm. In the folklore of physics, this volume was once known as "Caesar's last breath".  This volume is about equal to a human breath, so each time we inhale, we take in about one of the molecules that Julius Caesar last exhaled more than 2000 years ago (the 2000-year delay implies that air has been thoroughly mixed since then).  This is so because the whole atmosphere is to a human breath (400cc) what a human breath is to a single molecule.  We should not spoil everybody's fun by pointing out that quantum mechanics tells us that identical molecules cannot be distinguished, even in principle, so that the whole concern is really fallacious (you just can't paint a molecule red)...

---

(D. M. of Titusville, FL. 2001-02-11)
How much hydrogen would you need to raise the Titanic?

How do you want to do it? I suppose you do not want the wreck airborne (that would be both silly and unpractical). I am guessing that you are not considering inflating some device at the surface of the ocean with long (3800 meters!) cables to pull the wreck. (Besides, hydrogen is not much better that air for this application.) So, I assume that your idea is probably along the following lines:

Tie some kind of deflated balloons to the wreck (or put them inside the hull) and let some kind of liquefied gas evaporate into these until the whole thing lifts up. Lowering the gas in liquid form (in containers) avoids difficult pumping problems when it comes to overcoming the tremendous pressure difference (about 380 times the atmospheric pressure!). Once the balloon starts going up, its volume will expand greatly because the surrounding pressure decreases. In practice (?), you'd have to let some gas out during the ascent so the balloons won't explode. The maximum amount of gas would be needed at the very bottom, at a depth of about 3800 meter (12500 ft).

Let's assume the necessary work can be carried out at such depth. The problem is that the gas is under a pressure equal to about 3800*1025*9.8 (depth x density x gravitation) or about 38.2 MPa (377 atm). What's the density of hydrogen at such a pressure? Well, hydrogen cannot be considered an ideal gas (pV=RT) at that pressure, but we may still use the Van der Waals approximation, which reduces to (p+3/V²)(3V-1)=8T if p,V,T are expressed using as units the p,V,T values at the critical point of hydrogen (p=1.293 MPa and T=32.98 K). The bottom of the ocean is probably at a temperature near the densest point for water (3.98°C) under normal pressure, so let's use T=277.13 K. The volume V of a mole of hydrogen at P=38.2 MPa and T=277.13 K is thus about x times the critical molar volume of hydrogen, if x is solution of:

(38.2/1.293+3/x²)(3x-1) = 8(277.13/32.98)

Which gives x = 1.024948... The density of hydrogen near the Titanic wreck would thus be 1/x times the density at the critical point of hydrogen, which I found listed as 0.031 g/cc. It's therefore roughly equal to 0.03 g/cc (which is more than 300 times the density of hydrogen at 1 atm and 0°C).

The density of seawater at the surface is about 1.025 g/cc. Under 377 atm of pressure, this would increase only by about 2% to about 1.043 g/cc. All told, 0.03 grams of hydrogen generate 1.013 grams of buoyancy at the bottom of the ocean and the mass of hydrogen required is 30/1013 or about 3% of the mass of the whole wreck, if we neglect the self-buoyancy of the wreck. Taking into account the self-buoyancy of the wreck would probably bring the above number down to about 2.5%. I am guessing the mass of the Titanic to be about 10,000 metric tons (if anybody knows better, please let me know); 2.5% of that is 250 metric tons. That's a lot of hydrogen! It would be a lot less if the Titanic was in shallower waters, but the huge pressure is the killer here!

On 2001-03-08, Bob Cat (USA) made the following suggestion: Rather than using a balloon to lift the wreck, run an electrical cable into the ship and use [electrolysis] to fill it with your choice of hydrogen or oxygen.

You do need the same amount of hydrogen whether you put it in a balloon or in the hull of the Titanic (assuming a broken hull at such a depth could somehow be fixed into something airtight). You would also need to let gas out during the ascent, or the hull would almost immediately burst open (just like a balloon would). The solubility of hydrogen in water is rather low under normal pressure (saturating at 2.14 volumes of hydrogen in 100 volumes of water), but I would venture a guess that it's much higher at 380 atm of pressure. If hydrogen is significantly soluble, it can't be used to expel water from the hull, since whatever hydrogen is produced gets dissolved immediately (and oozes out of the wreck through whatever opening is left for water itself to be expelled). Let's not dwell on that particular objection, which may or may not be valid, and focus only on the main aspect of Bob's comment:

Bob's interesting proposition is to use electricity to make 250 metric tons of hydrogen from seawater, instead of bringing it to the site in liquid form. Let's see how much electricity would be needed: To make a mole of  H ₂ gas, it takes 2 faradays of electric charge, or 192970 C [a coulomb (C) is the charge delivered by a current of one ampere in a time of one second ] which is roughly 53.6 A-h (ampere-hour). What this gives you is just 2 grams of hydrogen. What you need is about 250 000 000 grams, corresponding to 6.7 GA-h (6700 000 000 A-h).

To complete the job in a century (876600 hours) you'd have to run a current of about 7643 A, 24 hour a day 7 days a week.

The delivery voltage at the sea floor should be at least a few volts (I'll be more precise if/when I find the time to derive the proper dependence on pressure for the electrolysis of seawater; I've not done this in 20 years and I am rusty on this particular issue).  Using transformers and AC to DC converters at the sea floor is something you could consider (it would not be technologically easy) but let's say we stick with low-tech delivery of straight DC current from the surface...

A length of 3800 m of hefty copper cable ½" in diameter (12.7 mm) would have a resistance of about 0.5 W.  Even if we use the ocean as a low-resistance return path (so that we only need a single wire) this means a voltage drop of about 3800 volts!  So, you're gonna use about 30 megawatts (30 MW) of power, most of it spent warming up seawater...  This is 150% of the power delivered by each of the 4 engines of a 747 jetliner, 24 hours a day for a hundred years!

A Jumbo Jet for 10 Years

If you have the power of the 4 engines of a jumbo jet (80 MW) you'll run a current 63.3% larger (1.633 is the square root of 80/30) and do the job in "only" 61 years.  You may want to save time and energy by running N identical cables fed with 1/N of your power:  Each cable would be running a current ÖN smaller than before under a voltage also ÖN times smaller. The total current in the N cables is thus ÖN bigger for the same power than it would be with a single cable.  Therefore, the whole job may be completed ÖN times as fast. With N=100 cables, the job would thus apparently be completed in "only" 6 years (and 45 days).  Actually, the power lost to the return path becomes very much relevant at this point (see next comment by Edgar Bonet) so we would need at least 10 solid years, even with a bigger boat to lessen the return resistance...

After viewing this page (on 2004-06-14) French physicist Edgar Bonet remarked that the resistance of the ocean on the return path could be estimated by considering the hull of the surface ship through which it goes to be roughly a half-sphere of radius R = 5 m (we are gold-plating the thing to avoid corrosion   ).  This gives a resistance of about 6 mW (namely, r/2pR, where r » 0.2 W×m is the resistivity of seawater).
 
This is negligible when we have but a single cable, but roughly doubles the total resistance when we have 100 of them in parallel.  This also means that there's little to gain by adding many more cables beyond this point, unless we're willing to use about half of them on the return path...  Lest they become a critical bottleneck, the bottom electrodes should be quite large too...

Note that each cable has a mass of about 4.3 metric tons, so that the total mass of copper you need to bring on site to produce the hydrogen (in ten years) is about twice the mass of the liquid hydrogen you could have brought.

A Jumbo Jet for One Hour

Notice also how terribly inefficient the whole thing is. Lifting a wreck of 10000 metric tons from a depth of 3800 meters requires "only" about 300 gigajoules (allowing for helpful self-buoyancy). Using the power of a jumbo jet (80 MW) for about an hour would do the trick. (Actually you don't want to do it that fast because you'd have to fight quite a drag if the wreck is being lifted at 3.8 km/h!) The hydrogen idea is very wasteful to begin with (because you end up releasing on the way up almost all the hydrogen you needed to get some buoyancy at the bottom). If, on top of that, you insist on running long copper cables (presumably from a nuclear-powered ship) to produce the hydrogen where it's needed, you end up with a combined efficiency of roughly 0.001%.

Well, that was a nice excuse for some "fantasy engineering"... Thanks, Bob.