
#1
Dec1112, 01:44 PM

P: 3

The Helium balloon of total mass=0.200kg and volume 1.00M^3 is attached a a rope with mass density of 1.00g/m. How high will the balloon rise?
I know that air density is 1.204kg/m^3 and that the balloon will stop rising after the density on the inside equals the density on the outside. I tried calculated that the density of Helium inside the balloon would be .1797g/L, but I am unsure how to calculate when it would reach a height of 1.204kg/m^3? 



#2
Dec1112, 02:17 PM

P: 1,030





#3
Dec1112, 02:26 PM

P: 3





#4
Dec1112, 02:53 PM

Sci Advisor
PF Gold
P: 2,247

How high will a balloon rise?Dave 



#5
Dec1112, 04:05 PM

P: 1,030

If the volume of the balloon and the density of air did not change with height, one could hang up 1004 g of weight to the balloon > then 1004 g + 200 g will balance the buoyant force on 1 m^3 large balloon. This means the rope could be 1 km long (1004 m).
However, the density of air and the volume of the balloon will change with height, so the actual height of the balloon may be slightly different than the above crude estimate. The ballon will expand, which could increase the buoyancy, but the density of air will decrease, which has the opposite effect. If you want to go into that, here is some help: For density of air, one can use the (as always, approximate) barometric formula [tex] \rho(h) = \rho_0 e^{\frac{M_a gh}{RT}}. [/tex] where [itex]M_a = 28[/itex] g/mol is approx. the molar mass of the air (mostly nitrogen) and R = 8.3 J/mol is the gas constant. The volume of the balloon V can be estimated from the equation of state for helium gas [tex] PV = m_{\mathrm He}/M_{\mathrm He} R T [/tex] where [itex]M_{\mathrm He} = 4[/itex] g/mol. The pressure P inside the balloon can be approximated by the atmospheric pressure at height h: [tex] P\approx p_a(h) = p_0 e^{\frac{M_agh}{RT}}. [/tex] With this, you can find out the buoyancy force [tex] F_B(h) = \rho(h) V(h) g [/tex] You have to find also how the force due to rope depend on the height (linear in h); the force of weight is constant = mg. The balloon will be in final equilibrium when the sum of forces is zero. 



#6
Dec1112, 05:24 PM

Sci Advisor
P: 2,470




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