How many grams of hydrogen are in the container?

Question

http://i35.tinypic.com/5mlkc7.jpg

Temp = 25 C

Answer 1

First figure out the pressure in the hydrogen:

P = Patm + (rho)*g*h    (make sure you use the rho of water)

Then use ideal gas law:

M = (rho)*R*T / P        (make sure you use the rho of hydrogen)

Then us the simple chemistry formula (who's name I now forget):

m = M * n

You should have all those numbers there yourself, that's how you do it, but I'm too lazy to do the math for you

Answer 2

Assume it's an ideal gas for the moment--we can correct that later if desired by giving the particles volume and considering VdW forces.

At any point, the ideal gas law applies:

PV = NkT

So in a volume dV = A dh, the number of mollecules is:

dN = (PA / kT) dh

You could assume pressure to be constant (which it is, pretty much, which is why the answers above are probably fine) or that it varies with height.  Just for fun, we'll solve for the pressure as a function of height.  For one meter, I don't think this will matter, but in general it could.

The pressure at the bottom of the cylinder is:

P(h=0) = Patm + rhowater g H,

where H is the height of the cylinder, so h runs from zero at the bottom to H at the top.  This assumes that the water is incompressible, so its density does not change.

Over a tiny interval, dh, the pressure in the gas changes by:

dP = -rho g dh

The mass density, rho, is just the mass of the H2 mollecule, which I'll call m, times the number density:

rho = mN/V

= mP/kT

So dP/P = -(mg/kT) dh

log(P(h) / P(0)) = -mgh/kT

P(h) = P(0) exp (-mgh / kT)

Plug that into our expression for the number of molecules of hydrogen in a little volume:

dN = (A P(0) exp (-mgh / kT) / kT) dh

And integrate that from h=0 to h=H

N = (A P(0) / mg) (1 - exp (-mgh / kT) )

And the mass is:

M = mN = (A P(0) / g ) (1 - exp (-mgh / kT) )

If mgh/kT is tiny, you can expand the exponential to first order and get:

M =  m P(0) V / kT, as was obvious in the first place if you just assume the pressure is constant.  I suspect the correction is much smaller than the correction you get from using the VdW equation instead of the ideal gas law:

(P + a (N/V)^2) (V - Nb) = NkT

You can look up the VdW coefficients and find the roots of the cubic equation in N numerically.

And assuming the book value of 1atm rather than measuring the actual pressure introduces more error than either of those things in most cases.

This is what happens when someone who usually asks interesting questions asks a trivial one--I try to make something out of nothing.

Answer 3

PV=nRT

n = PV/RT

T = 273+25 K

P=  1*10^5Pa since the depth of water is 1m and assuming this is on normal air  

n = 1*10^5*1/8.314*298

n = 40.36mol

w = n*2 ; since the atomic mas of H2 = 2

w = 80.72g

Hope this helps

Damitha