Worksheet More Boyles Law and Charless Law | StudyX - Free Printable

I'll solve each problem step by step, showing all work and checking my calculations.

## Problem 1: Celsius to Kelvin Conversion

Part A: Converting Celsius to Kelvin
To convert from degrees Celsius to Kelvin, you must ADD 273 (or more precisely, 273.15) to the Celsius temperature.

Formula: K = °C + 273

Part B: Why use Kelvin in gas law problems?
We must use the Kelvin scale in gas law problems because:
- Gas laws involve proportional relationships between variables
- The Kelvin scale starts at absolute zero (0 K), where molecular motion theoretically stops
- Using Kelvin ensures we never have negative temperatures in our calculations
- If we used Celsius, we could get negative values which would make the mathematical relationships incorrect (you can't have negative volume or pressure ratios)

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## Problem 2: Charles's Law Problem

Given:
- Initial volume (V₁) = 2.00 L
- Initial temperature (T₁) = 11.0°C
- Final volume (V₂) = 3.00 L
- Pressure is constant
- Find: Final temperature (T₂)

Step 1: Convert initial temperature to Kelvin
T₁ = 11.0°C + 273 = 284 K

Step 2: Apply Charles's Law
Charles's Law states: V₁/T₁ = V₂/T₂ (at constant pressure)

Rearranging to solve for T₂:
T₂ = V₂ × T₁ / V₁

Step 3: Substitute values
T₂ = 3.00 L × 284 K / 2.00 L
T₂ = 852 / 2.00
T₂ = 426 K

Step 4: Convert back to Celsius (if needed)
T₂ in °C = 426 K - 273 = 153°C

Let me verify: V₁/T₁ = 2.00/284 = 0.00704
V₂/T₂ = 3.00/426 = 0.00704 ✓

This is an example of Charles's Law.

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## Problem 3: Boyle's Law Problem

Given:
- Initial volume (V₁) = 480 mL
- Initial pressure (P₁) = 115 kPa
- Final volume (V₂) = 650 mL
- Temperature is constant (implied)
- Find: Final pressure (P₂)

Step 1: Apply Boyle's Law
Boyle's Law states: P₁V₁ = P₂V₂ (at constant temperature)

Rearranging to solve for P₂:
P₂ = P₁V₁ / V₂

Step 2: Substitute values
P₂ = 115 kPa × 480 mL / 650 mL
P₂ = 55,200 / 650
P₂ = 84.92 kPa

Let me verify this calculation:
115 × 480 = 55,200
55,200 ÷ 650 = 84.923... ≈ 84.9 kPa

Check: P₁V₁ = 115 × 480 = 55,200
P₂V₂ = 84.92 × 650 = 55,198 ≈ 55,200 ✓

This is an example of Boyle's Law.

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## Problem 4: Boyle's Law Problem

Given:
- Initial volume (V₁) = 240.0 mL
- Initial pressure (P₁) = 1.20 atm
- Final pressure (P₂) = 0.860 atm
- Temperature is constant
- Find: Final volume (V₂)

Step 1: Apply Boyle's Law
Boyle's Law states: P₁V₁ = P₂V₂ (at constant temperature)

Rearranging to solve for V₂:
V₂ = P₁V₁ / P₂

Step 2: Substitute values
V₂ = 1.20 atm × 240.0 mL / 0.860 atm
V₂ = 288 / 0.860
V₂ = 334.88 mL

Let me verify this calculation:
1.20 × 240.0 = 288
288 ÷ 0.860 = 334.883... ≈ 335 mL

Check: P₁V₁ = 1.20 × 240.0 = 288
P₂V₂ = 0.860 × 334.88 = 287.997 ≈ 288 ✓

Since pressure decreased, volume should increase (inverse relationship). 335 mL > 240 mL ✓

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Final Answer:

Problem 1:
- To convert Celsius to Kelvin: Add 273 to the Celsius temperature
- Why use Kelvin: Kelvin starts at absolute zero, preventing negative temperatures that would make gas law calculations incorrect

Problem 2:
- Temperature changed to: 426 K (or 153°C)
- This is an example of Charles's Law

Problem 3:
- Pressure required: 84.9 kPa
- This is an example of Boyle's Law

Problem 4:
- Volume will be: 335 mL