Question

A container initially holds 5.67 x 10^-2 mol of propane and has a volume of V1. The volume of the container was increased by adding an additional 2.95 x 10^-2 mol if propane to the container, so that the container has a final volume of 1.93 L. If the temperature and pressure are constant, what was the initial volume of the container?

Answer 1

Answer: 1.27 L

Step-by-step explanation:

First, calculate the final number of moles of propane (n2) in the container.

n2 = n1 + nadded = 5.67 × 10^−2 mol + 2.95 × 10^−2 mol = 8.62 × 10^−2 mol

Rearrange Avogadro's law to solve for V1.

V1 = V2 × n1 / n2

Substitute the known values of n1, n2, and V2,

V1 = 1.93 L × 5.67 × 10^−2 mol / 8.62 × 10^−2 mol = 1.27 L

Answer 2

Initial volume of the container (V1) = 1.27 L (Approx)

Step-by-step explanation:

Given:

Number of mol (n1) = 5.67 x 10⁻²

Number of mol (n2) = (5.67 +2.95) x 10⁻² = 8.62 x 10⁻²

New volume (V2) = 1.93 L

Find:

Initial volume of the container (V1)

Computation:

Using Avogadro's law

V1 / n1 = V2 / n2

V1 / 5.67 x 10⁻² = 1.93 / 8.62 x 10⁻²

V1 = 10.9431 / 8.62

Initial volume of the container (V1) = 1.2695

Initial volume of the container (V1) = 1.27 L (Approx)