The goodyear blimp contains 5.74×10^6 l of helium at 25 °c and 1.00 atm. what is the mass in grams of the helium inside the blimp?

a. 2.30*10^6 g
b. 1.12*10^7 g
c. 2.34*10^5 g
d. 2.80*10^6 g
e. 9.39*10^5 g

Explanation:

1) Data:

a) V = 5.74×10⁶ liter

b) T = 25°C = 25 + 273.15 K = 298.15 K

c) p = 1.00 atm

d) Helium ⇒ molar mass = 4.003 g/mol

2) Formulae:

a) pV = nRT

b) molar mass = mass in grams / n

3) Solution:

a) n = (pV) / (RT) =

       =  (1.00 tam × 5.74×10⁶ liter ) / (0.0821 atm-liter/mol-K) × 298.15K)

        = 2.34495 ×10⁵ moles

b) Convert moles to mass in grams:

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