What is the maximum height the ballon will rise & its potential energy at that height?

Question

If the atmosphere were isothermal the pressure at a given height would vary as follows:

P = P' e^Mgh/RT

where P' = 1 ATM, M is the average molar mass (0.80*28 + 0.20*32), g = 9.81m/s^2, R = 0.08206 l atm/mole degree K,
T = degree K

Given 100,000 L He @ 25*C & 1 ATM in a rigid ballon carrying a 100 kg load how high would it rise till it reaches equilibrium?
I've tried this one for 3 weeks and I'm getting nowhere.

OK, OK, it's BALLOON.
Height of a Rigid Balloon Containing 100m3 He and Suspending 100kgs @ 25*C

h = 102.49m
h = -RT/gμ(air) ln[(1+M100kg / MHe) e)/μ(air)]

h = 102.49 m h = -RT/gμ(air) ln(P2/P0)

h = 102.49 m h = -(8774.3) ln(P2/P0)

P0 101,325 kg/m-s2
VHe 100 m3
R 8.31451 kg-m2/s2-mol-K (.001)m3/l 101,325 kg/ms2/atm * 0.0821 l-atm/mole-K
T298.15*K 298.15 degrees K Isothermal Process
nHe 4087.4 moles n = P0 V / R T

g 9.81 m/s2
μ(air) 0.029 kgs/mole
μ(He) 0.004 kgs/mole
M100kg 100 kgs
MHe 16.35 kgs MHe = μ(He) n = μ(He) P0 V / R T
Mair derived Mair = μ(air) n = μ(air) Ph V / R T
Ph = P0exp(-μ(air)gh/RT)
Mair = μ(air) P0V/RTexp(-μ(air)gh/RT)
Mair = MHe μ(air)/μ(He)exp(-μ(air)gh/RT)

Bouyancy Equilibrium Condition Mair = MHe + M100kg Mair/MHe = 1 + M100kg/MHe
(1+M100kg/MHe)μ(He)/μ(air) = exp(-μ(air)gh/RT)

Answer 1

First of all your units are a little questionable.
R = 8.3143 J/mol K
Also P = P' * exp(-Mgh / RT)
The pressure decreases with height
For air M = 0.0288 kg/mol
So P = P' exp(-h / 8769.6)

We assume the helium pressure inside the balloon does not change.
ρHe = PM/RT = 1.013e5*4e-3 / 8.3143*298
= 0.1635 kg/m^3

ρAir = P*0.0288/ 8.3143*298 = P/86030

Volume of air displaced = 100 m^3
Weight of air displaced at altitude, h = Pg/860.3

Weight of He = 16.35*g
Weight of object = 100*g

So P/860.3 = 100 + 16.35
P = 100096 Pa
= 1.013e5 * exp(-h/8769.6)

h = 104.9 m
This is the altitude where the air pressure (hence density) is low enough such that the upthrust = weight, assuming isothermal conditions.

Answer 2

We have V = 100 m^3, T = 298.15 K, P = 101325 Pa, so N = PV / RT = 4087.4 moles.
The helium mass = 4 * 4087.4/1000 = 16.35 kg.
Total mass is 100 + 16.35 = 116.35 kg.
Displaced air mass at sea level = 29 * 4087.4/1000 = 118.54 kg.
With isothermal assumption, P / density = constant, thus equilibrium P = 101325 * 116.35 / 118.54 = 99453 Pa.
The inverse of the pressure equation is h = ln(P / P0) * RT / (gM) where M is molecular weight/1000.
Solving yields an equilibrium altitude of 162.7 m.

Answer 3

When the density of air at certain altitude becomes equal to the density of Helium at that altitude, then your balloon will stop rising any more. You have to consider the 100 kg weight plus your own weight to reach to equilibrium. Therefore, just before the density of Helium reaches the density of air, your equilibrium will be met due to your weight and the 100 kg of weight you are carrying with your balloon.

The best way to calculate the height, at which your balloon will reach equilibrium, is find the weight of the 100,000 L of helium. Then from your equation calculate the density of air at which both the Helium and air becomes equal. Then calculate your weight plus the 100 kg you are carrying with you. At this point you will reach equilibrium before the density of air and helium become equal due to your weight and the 100 kg you taking along.

Find out the density of air using the equation provided in your question. You can go from there on!

Answer 4

P = Po exp(-μ(air)g/RT h)
note 'minus' sign in parenthesis

μ(air) = 29 g/mole
g = 9.81 m/s²
R = 0.082 Po liter / mole K
(please, do not say 'degree Kelvin', it sounds unprofessional)

V = 100,000 L
μ(He) = 4 g/mole
T = 25 C = 298 K
Mo = 100 kg

Mass m of helium:
m = μ(He) n = μ(He) PoV / RT

Mass of displaced air:
Ma = μ(air) n = μ(air) P(h)V / RT
Ma = μ(air) PoV / RT exp(-μ(air)g/RT h)
Ma = μ(air)/μ(He) m exp(-μ(air)g/RT h)


Buoyancy equilibrium:
M + m = Ma
M/m + 1 = μ(air)/μ(He) exp(-μ(air)g/RT h)
μ(He)/μ(air) (1 + M/m) = exp(-μ(air)g/RT h)

h = - RT/μ(air)g ln [μ(He)/μ(air) (1 + M/m)]

Now:
m = 16.37kg
ln [μ(He)/μ(air) (1 + M/m)] = -0.171
h = [86 atm x liter / Newton] x 0.171 = 1484 m


Answer:
h = - RT/μ(air)g ln [μ(He)/μ(air) (1 + M/m)] = 1.50 km