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

Option E. 9.39 * 10⁵g =

Explanation:

1) dа ta:

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:

mass in grams = 4.003 g / mol*2.34495 * 10⁵ moles = 9.39 * 10⁵g

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answer