For many purposes, we can treat methane (CH4) as an ideal gas at temperatures above its boiling point of -161. °C.
Answer
🧪 Methane Gas Volume Change Under Varying Conditions
Problem Statement
A sample of methane gas is subjected to a:
- Temperature decrease from 80.0 °C to 62.0 °C
- Pressure increase of 15.0%
We aim to determine if the volume increases, decreases, or remains constant and calculate the percentage change.
Step 1: Convert Temperatures to Kelvin
T₁ = 80.0 + 273.15 = 353.15 K
T₂ = 62.0 + 273.15 = 335.15 K
T₂ = 62.0 + 273.15 = 335.15 K
Step 2: Account for Pressure Increase
P₂ = 1.15 × P₁
Step 3: Use the Ideal Gas Law Proportion
(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂
Solve for V₂/V₁:
V₂ / V₁ = (T₂ / T₁) × (P₁ / P₂)
Substitute values:
V₂ / V₁ = (335.15 / 353.15) × (1 / 1.15) ≈ 0.949 × 0.870 ≈ 0.826
Step 4: Determine Volume Change
Since V₂ / V₁ = 0.826, the volume has decreased.
Percentage Change = (1 − 0.826) × 100% = 17.4%
✅ Final Answer:
The volume of methane gas decreases by approximately 17.4%.
Key Concepts:
- The ideal gas law: PV = nRT
- Decreasing temperature → less kinetic energy → volume decreases
- Increasing pressure → gas compression → volume decreases
- Both factors contribute to a net volume decrease in this case