Question

The Goodyear blimps, which frequently fly over sporting events, hold approximately 1.50×10⁵ ft3 of helium.

If the gas is at 23 ∘C and 1.0 atm, what mass of helium is in the blimp?

Answer 1

Ans- Given: Temperature, t = 23 ^{$^{\circ}$} C

Temperature in kelvin, T = 273 + 23 = 296 K

Pressure, P = 1.0 atm = 101325 Pa

Volume, V = 1.50 * 10⁵ ft³

As 1 ft³ = 0.0283 m³, so

V = 1.50 * 10⁵ * 0.0283 m³

V = 42450 m³

As we know ideal gas equation : PV = nRT

here, R = 8.314 JK^-1mol^-1

101325 * 42450 = n * 8.314 * 296

4.3 * 10⁹ = n * 2460.944

n = (4301246250) / 2460.944

n = 1747803.38 mol

As we know, n = mass / molar mass

So, mass = n * molar mass of helium

molar mass of helium = 4.0 g mol^-1

So, mass of helium = 1747803.38 *4

mass of helium in Goodyear blimps = 6991213.52 grams