# What is the lifting force of a Helium balloon in air?

I am trying to solve the problem:

What is the lifting force of in kN for a 10 m diameter spherical balloon with helium inside at 101 kPa and 320 K surrounded by air at 101 kPa and 298.15 K?

The given answer is 5.28 kN.

I start from the Buoyancy Formula:

$$F_{b} = (p_{air} - p_{gas})*g*V$$

And Ideal Gas Law: $$PV = mRT$$ $$P/RT = m/V$$

Density of Air: $101/0.287*298.15=1.18$

Density of He: $101/0.287*320=1.10$

Volume of Helium Balloon: $(4\pi/3)*(10/2)^3=(500/3\pi)$

Substitute:

$$F_b = (1.18 - 1.10)*9.81*(500/3\pi)$$

I get 0.4101 N; am I missing something or is the given answer wrong?

• In the ideal gas law for helium, you have an incorrect value for the specific gas constant $R$. – John1024 Jul 4 '15 at 6:13

Firstly, Gas constant of $He$ equals $2077 \ \mathrm{J/kg\cdot K}$ (engineering toolbox), so density of $He$ becomes $101/2.077*320 = 0.152\ \mathrm{kg/m^3}$.
Secondly, $Volume = 523.6 \ \mathrm{m^3}$ not as you calculated.
Thirdly, you need to subtract balloon weight from the buoyancy force $$F_b = \rho_{air}*g*V$$ $$F_b = 1.18 *9.81*523.6 = 6061 \ \mathrm{N}$$ $$F_{weight} = \rho_{He} Vg = 0.152*523.6*9.81 = 780.75 \ \mathrm{N}$$ $$F_{net} = F_b - F_{weight} = 6061 - 780.75 = 5280\ \mathrm{N} = 5.28\ \mathrm{kN}$$