Question

Buoyancy for the Goodyear blimp Spirit of Innovation comes from 203,000 ft3 of helium. Calculate the mass of this much helium at 22.00 °C and 0.987 atm pressure.?

Calculate the buoyancy of the balloon as "grams lift"–that is, the difference between the mass of an equal volume of dry air at the same temperature and pressure and the mass of the He in the balloon. Assume the the molar mass of air is 28.8 g/mol.

Answer 1

Well first we need Kosher units.....

Explanation:

`1*ft^3-=(30.5*cm)^3xx10^-3*L*cm^-3-=28.4*L`

And thus......

`"volume"=203000*ft^3xx28.4*L*ft^-3-=5.76xx10^6*L`

An now we use the ideal gas equation....

`n=(PV)/(RT)=(0.987*atmxx5.76xx10^6*L)/(0.0821*(L*atm)/(K*mol)xx295.15*K)=23.5xx10^4*mol`

And so `m_"He"=23.5xx10^4*molxx4.0*g*mol^-1=938.5*kg`.

And so `m_"air"=23.5xx10^4*molxx28.8*g*mol^-1=6768*kg`.

And so your `"grams lift"=underbrace(6768*kg)_"mass of air"-underbrace(938.5*kg)_"mass of helium"=??*kg`.

Anyway please review my arithmetic. All care taken but no responsibility admitted! And I bet this problem would be much easier if we used `"cubic metres"`, mass of air at normal pressures, `1*atm=1.20*kg*m^-3`, i.e. approx. `1*kg`...