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

P₂ = 0.67 atm

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition 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 1V₁ is volume 1P₂ is pressure 2V₂ is volume 2Explanation:

Step 1: Define

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

Step 2: Solve

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