A 110.0 L weather balloon filled with 6.30 mol of helium has a small leak. If the helium leaks at a rate of 10.0 mmol/hr, what is the volume of the balloon after 52.0 hours?

Answer 1

Hi, Elizabeth,

This is an unusual application of the Ideal Gas Law where: PV = nRT (P = pressure of the gas, V= volume of the container, n = "amount" (typically moles or millimoles) of gas, R = ideal gas constant, and T = temperature of the system in Kelvin). Since P, R, and T are unchanged, this becomes:

V₁/n₁ = V₂/n₂

You are provided with V₁ (110.0 L), and n₁ (6.30 mol), so you first have to solve for the new "amount" (n₂). This is simply n₂ = n₁ - (10.0 mmol/hr x 52 hours). In this case, substituting in the value for n₁: n₂ = 6.30 mol - 520 mmol, or 6.30 mol - 0.52 mol, or finally n₂ = 5.78 mol. Note that 1 mol = 1000 mmol.

To determine V₂, we can now plug these values into the original equation as so:

110.0L/6.30 mol = V₂/5.78 mol

V₂ = (110)(5.78)/6.30)

V₂ = 100.9 L The volume of the balloon after 52.0 hours is 100.9 liters.

'Hope that helps,

Chris