This question concerns the ideal gas law.

Question

Blimps hold approximately 4960 m3 of helium. Assume the gas is at 17 degrees C and 1.00 atm.

a. What mass of helium is contained in the blimp?

b. If the blimp were full of hydrogen instead of helium, what would be the mass of hydrogen in the blimp?

Answer 1

PV = nRT

P = pressure = 1.00 atm

V = volume in liters = 4960 m³ x 1000 L / m³ = 4.96x10⁶ L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 17ºC + 273 = 290K

Solving for moles (n) ..

n = PV/RT = (1.00)(4.96x10⁶) / (0.0821)(290)

n = 208325 mols of helium

Converting this to grams, we have..

208325 mols He x 4.00 g / mol = 833300 g = 8.33x10⁵ g (3 sig.figs.) = 8.3x10⁵ g (2 sig.figs.)

If hydrogen replaced the helium, the moles of gas would be the same, i.e. 208325 mols

mass of H₂ = 208325 mols H₂ x 2.00 g / mol = 416650 g = 4.17x10⁵ g (3 sig.figs.) = 4.2x10⁵g (2 sig.figs.)