Answer
🧪 Methane Gas Volume Change with Temperature and Pressure Reduction
Problem Summary
A methane gas sample undergoes the following changes:
- Temperature drops from −53.0 °C to −64.0 °C
- Pressure decreases by 15.0%
Determine if the gas volume increases, decreases, or stays the same, and calculate the percentage change in volume.
Step 1: Apply the Combined Gas Law
(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂
Rearranged to solve for the volume ratio:
V₂ / V₁ = (T₂ / T₁) × (P₁ / P₂)
Step 2: Convert Temperatures to Kelvin
T₁ = −53.0 + 273.15 = 220.15 K
T₂ = −64.0 + 273.15 = 209.15 K
T₂ = −64.0 + 273.15 = 209.15 K
Step 3: Apply Pressure Change
P₂ = 0.85 × P₁ (15% decrease)
Step 4: Substitute into the Equation
V₂ / V₁ = (209.15 / 220.15) × (1 / 0.85)
≈ 0.950 × 1.176 ≈ 1.117
≈ 0.950 × 1.176 ≈ 1.117
Step 5: Calculate Percentage Change in Volume
Percentage change = (1.117 − 1) × 100 = 11.7% ≈ 12%
✅ Final Answer:
The methane gas volume increases by approximately 12%.
Key Concepts Recap:
- The ideal gas law simplifies to the combined gas law when n and R are constant.
- Decreased temperature would normally reduce volume.
- However, a significant pressure drop allows the gas to expand.
- Here, the pressure effect dominates, resulting in a net increase in volume.