Question

A balloon having 8gm Ne gas and radius 20 cm is pricked and 7 gm of Ne is allowed to effuse from it. What will be the radius of the balloon?

Answer 1

Answer:

Case 1:

The initial mass of the balloon, m = 8 gm

Radius, r₁ = 20 cm

Molecular mass of Neon, M = 20.17 g/mol

Using the Ideal Gas Law,

PV = nRT

⇒ P * [4/3 * π * r₁³] = m/M * RT ….. [since Volume = 4/3πr³]

⇒ P * [4/3 * π * r₁³] = m/M * RT

⇒ P * [4/3 * 3.14 * (20)³] = 8/20.17 * RT

⇒ RT / P = [3.14 * (20³) * 10.085] / 3 = 84445.06 ….. (i)

Case 2:

After 7 gm of Ne is effused, remaining mass of Ne in the balloon = 8 gm – 7 gm = 1 gm

∴ PV = nRT  

Or, V = (1/20) * RT/P

Or, V = 84445.06 / 20 ….. [from (i)]

Or, V = 4222.25 cm³

Let the radius of the balloon after effusion be “r₂”

Now, we know,  

V = 4/3 * π * r₂³

Or, 4222.25 = 4/3 * 3.14 * r₂³

Or, r₂³ = 12666.75 / 12.56 = 1008.499

Or, r₂ = ∛(1008.49) = 10.02 cm ≈ 10 cm

Hence, the radius of the balloon will be 10 cm.