For many purposes we can treat methane (CH) as an ideal gas at temperatures above its boiling point of – 161. °C. Suppose the temperature of
Answer
Methane Gas Volume Change Analysis
Context
We examine how the volume of methane gas changes under different combinations of pressure and temperature changes using the ideal gas law:
V₂ / V₁ = (P₁ / P₂) × (T₂ / T₁)
In the original scenario:
- Temperature: increased from −72.0°C (201.15 K) to −62.0°C (211.15 K)
- Pressure: increased by 15% → P₂ = 1.15 × P₁
- Result: Volume decreased by approximately 8.75%
✅ A similar change in magnitude, but with pressure decreasing instead of increasing, would lead to a volume increase.
Alternative Scenario
- Same temperature change: −72.0°C to −62.0°C
- Pressure decreases by 15%: P₂ = 0.85 × P₁
Calculation:
V₂ / V₁ = (1 / 0.85) × (211.15 / 201.15) ≈ 1.1765 × 1.0498 ≈ 1.235
Percentage change in volume:
(1.235 - 1) × 100 = 23.5%
Comparison Table
| Scenario | Pressure Change | Temperature Change | Volume Change |
|---|---|---|---|
| Original Case | +15% | −72°C → −62°C | −8.75% |
| Reversed Pressure | −15% | −72°C → −62°C | +23.5% |
Conclusion
A similar degree of pressure change in the opposite direction (from an increase to a decrease), combined with the same temperature increase, results in a volume increase rather than a decrease.