Question

A gas with a volume of 4.0 L at a pressure of 2.02 atm is allowed to expand to a volume of 12.0

L. What is the pressure in the container if the temperature remains constant

Answer 1

P₂ = 0.67 atm

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Equality Properties

• Multiplication Property of Equality
• Division Property of Equality
• Addition Property of Equality
• Subtraction Property of Equality

Chemistry

Gas Laws

Boyle's Law: P₁V₁ = P₂V₂

• P₁ is pressure 1
• V₁ is volume 1
• P₂ is pressure 2
• V₂ is volume 2

Step-by-step explanation:

Step 1: Define

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

Step 2: Solve

1. Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
2. [Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
3. [Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
4. [Pressure] Rewrite: P₂ = 0.673333 atm

Step 3: Check

Follow sig fig rules and round. We are given 2 sig figs as our smallest.

0.673333 atm ≈ 0.67 atm