Question

The pressure of a sample of helium in a 3.40 L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.15 L container?

Answer 1

[AI-generated answer — Step-by-Step Bot]

To find the new pressure when the sample of helium is placed in a smaller container, we can use Boyle's Law, which states that pressure and volume are inversely proportional at constant temperature.

Boyle's Law equation can be written as:

P1 * V1 = P2 * V2

where
P1 = initial pressure
V1 = initial volume
P2 = final pressure (new pressure)
V2 = final volume

Given:
P1 = 0.988 atm
V1 = 3.40 L
V2 = 2.15 L

Now, let's plug in the values into the equation and solve for P2:

P1 * V1 = P2 * V2

0.988 atm * 3.40 L = P2 * 2.15 L

3.3552 atm*L = P2 * 2.15 L

Now, to solve for P2 (new pressure), divide both sides by 2.15 L:

P2 = 3.3552 atm*L / 2.15 L

P2 = 1.5613 atm

Therefore, the new pressure, when the sample of helium is placed in a 2.15 L container, is approximately 1.5613 atm.

Answer 2

[AI-generated answer — Explain Bot]

To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. Boyle's Law can be expressed as:

P₁V₁ = P₂V₂

where P₁ is the initial pressure, V₁ is the initial volume, P₂ is the new pressure, and V₂ is the new volume.

Given:
P₁ = 0.988 atm
V₁ = 3.40 L
V₂ = 2.15 L

We can rearrange the equation and solve for P₂:

P₂ = (P₁ * V₁) / V₂

Plugging in the given values, we get:

P₂ = (0.988 atm * 3.40 L) / 2.15 L

Calculating this expression, we find:

P₂ ≈ 1.56 atm

Therefore, the new pressure of the helium sample in the 2.15 L container is approximately 1.56 atm.