Question #18865

How much weight can be lifted by a Hydrogen (& also separately for Helium) filled balloon of 10 cubic Meter, at sea level & at a temp. of 20 Degree Centigrade. Pl. describe the inputs used.

Expert's answer

Question#18865

How much weight can be lifted by a Hydrogen (̄) also separately for Helium) filled balloon of 10 cubic Meter, at sea level ̄ at a temp. of 20 Degree Centigrade. Pl. describe the inputs used.

Solution:

Let:


V=10m3V = 10 \, \mathrm{m}^3T=20CT = 20{}^\circ \mathrm{C}


m–?

F=mgF = mg, m=Fgm = \frac{F}{g} were: F – buoyant force

According to the Archimedes’ principle:


F=(ρairρhydrogen)VgF = (\rho_{air} - \rho_{hydrogen}) V g


Were: ρair\rho_{air} – density of air, ρhydrogen\rho_{hydrogen} – density of hydrogen, g=9,8m/s2g = 9,8 \, \mathrm{m/s}^2

ρair=1.29kg/m3,ρhydrogen=0.09kg/m3\rho_{air} = 1.29 \, \mathrm{kg/m}^3, \quad \rho_{hydrogen} = 0.09 \, \mathrm{kg/m}^3m=(ρairρhydrogen)Vgg=(ρairρhydrogen)V=(1.290.09)×10=12kgm = \frac{(\rho_{air} - \rho_{hydrogen}) V g}{g} = (\rho_{air} - \rho_{hydrogen}) V = (1.29 - 0.09) \times 10 = 12 \, \mathrm{kg}


Answer: 12 kg.

For Helium: ρ=0.178kg/m3\rho = 0.178 \, \mathrm{kg/m}^3, m=(1.290.178)×10=11.12kgm = (1.29 - 0.178) \times 10 = 11.12 \, \mathrm{kg}

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