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Juliette [100K]
3 years ago
7

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

Chemistry
1 answer:
Maurinko [17]3 years ago
4 0
<h3>Answer:</h3>

P₂ = 0.67 atm

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

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<u> </u>

<u>Chemistry</u>

<u>Gas Laws</u>

Boyle's Law: P₁V₁ = P₂V₂

  • P₁ is pressure 1
  • V₁ is volume 1
  • P₂ is pressure 2
  • V₂ is volume 2
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

<u>Step 2: Solve</u>

  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

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

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