Answer
🧪 Methane Gas Volume Change with Decreased Temperature and Pressure
Problem Summary
A methane gas sample undergoes the following changes:
- Temperature decreases from 13.0 °C to −1.0 °C
- Pressure decreases by 15.0%
We are to determine whether the volume increases, decreases, or stays the same, and calculate the percentage change in volume.
Step 1: Convert Temperatures to Kelvin
T₁ = 13.0 + 273.15 = 286.15 K
T₂ = −1.0 + 273.15 = 272.15 K
T₂ = −1.0 + 273.15 = 272.15 K
Step 2: Apply Pressure Change
P₂ = 0.85 × P₁ (15% pressure decrease)
Step 3: Use the Combined Gas Law
(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂
Rearranged: V₂ / V₁ = (T₂ / T₁) × (P₁ / P₂)
= (272.15 / 286.15) × (1 / 0.85) ≈ 0.9511 × 1.1765 ≈ 1.119
Rearranged: V₂ / V₁ = (T₂ / T₁) × (P₁ / P₂)
= (272.15 / 286.15) × (1 / 0.85) ≈ 0.9511 × 1.1765 ≈ 1.119
Step 4: Calculate Percentage Change
Percentage Change = (1.119 − 1) × 100 = 11.9% ≈ 12%
✅ Final Answer:
The volume of methane gas increases by approximately 12%.
Key Concepts:
- Decreased pressure causes the gas to expand.
- Decreased temperature tends to shrink the gas.
- In this case, pressure reduction has a stronger effect, resulting in an overall volume increase.