Answer to Question #207486 in Chemistry for sbuda

Question

A balloon holds 30.0 kg of helium. What is the volume of the balloon if its pressure is 1.20 atm and the temperature is 22˚C

Answer 1

P = 1.20 atm

V = unknown

m = (30.0 kg) × (1000 g / 1 kg) = 30000 g

R = 0.0821 atm L mol^-1 K^-1

T = 22˚C + 273 = 295 K

Solution:

The molar mass of helium (He) is 4.00 g/mol.

Therefore,

(30000 g He) × (1 mol He / 4.00 g He) = 7500 mol He

n = 7500 mol

The Ideal Gas equation can be used.

The Ideal Gas equation can be expressed as: PV = nRT

Therefore,

V = nRT / P

V = (7500 mol × 0.0821 atm L mol^-1 K^-1 × 295 K) / (1.20 atm) = 151371.875 L = 1.51×10⁵ L

V = 1.51×10⁵ L

Answer: 1.51×10⁵ L is the volume of the balloon.